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2025-12-31 19:37:45 +08:00
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commit 3c9d5689c4
279 changed files with 40895 additions and 16087 deletions
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docs/chapter_array_and_linkedlist/array.md
-105
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@@ -132,14 +132,6 @@ comments: true
nums = [1, 3, 2, 5, 4]
```
=== "Zig"
```zig title="array.zig"
// 初始化数组
const arr = [_]i32{0} ** 5; // { 0, 0, 0, 0, 0 }
const nums = [_]i32{ 1, 3, 2, 5, 4 };
```
??? pythontutor "可视化运行"
<div style="height: 243px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%E6%95%B0%E7%BB%84%0Aarr%20%3D%20%5B0%5D%20*%205%20%20%23%20%5B%200,%200,%200,%200,%200%20%5D%0Anums%20%3D%20%5B1,%203,%202,%205,%204%5D&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=0&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
@@ -326,19 +318,6 @@ comments: true
end
```
=== "Zig"
```zig title="array.zig"
// 随机访问元素
fn randomAccess(nums: []const i32) i32 {
// 在区间 [0, nums.len) 中随机抽取一个整数
const random_index = std.crypto.random.intRangeLessThan(usize, 0, nums.len);
// 获取并返回随机元素
const randomNum = nums[random_index];
return randomNum;
}
```
??? pythontutor "可视化运行"
<div style="height: 531px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=import%20random%0A%0Adef%20random_access%28nums%3A%20list%5Bint%5D%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E9%9A%8F%E6%9C%BA%E8%AE%BF%E9%97%AE%E5%85%83%E7%B4%A0%22%22%22%0A%20%20%20%20%23%20%E5%9C%A8%E5%8C%BA%E9%97%B4%20%5B0,%20len%28nums%29-1%5D%20%E4%B8%AD%E9%9A%8F%E6%9C%BA%E6%8A%BD%E5%8F%96%E4%B8%80%E4%B8%AA%E6%95%B0%E5%AD%97%0A%20%20%20%20random_index%20%3D%20random.randint%280,%20len%28nums%29%20-%201%29%0A%20%20%20%20%23%20%E8%8E%B7%E5%8F%96%E5%B9%B6%E8%BF%94%E5%9B%9E%E9%9A%8F%E6%9C%BA%E5%85%83%E7%B4%A0%0A%20%20%20%20random_num%20%3D%20nums%5Brandom_index%5D%0A%20%20%20%20return%20random_num%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%E6%95%B0%E7%BB%84%0A%20%20%20%20nums%20%3D%20%5B1,%203,%202,%205,%204%5D%0A%20%20%20%20print%28%22%E6%95%B0%E7%BB%84%20nums%20%3D%22,%20nums%29%0A%0A%20%20%20%20%23%20%E9%9A%8F%E6%9C%BA%E8%AE%BF%E9%97%AE%0A%20%20%20%20random_num%3A%20int%20%3D%20random_access%28nums%29%0A%20%20%20%20print%28%22%E5%9C%A8%20nums%20%E4%B8%AD%E8%8E%B7%E5%8F%96%E9%9A%8F%E6%9C%BA%E5%85%83%E7%B4%A0%22,%20random_num%29%0A&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=7&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
@@ -535,21 +514,6 @@ comments: true
end
```
=== "Zig"
```zig title="array.zig"
// 在数组的索引 index 处插入元素 num
fn insert(nums: []i32, num: i32, index: usize) void {
// 把索引 index 以及之后的所有元素向后移动一位
var i = nums.len - 1;
while (i > index) : (i -= 1) {
nums[i] = nums[i - 1];
}
// 将 num 赋给 index 处的元素
nums[index] = num;
}
```
??? pythontutor "可视化运行"
<div style="height: 495px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20insert%28nums%3A%20list%5Bint%5D,%20num%3A%20int,%20index%3A%20int%29%3A%0A%20%20%20%20%22%22%22%E5%9C%A8%E6%95%B0%E7%BB%84%E7%9A%84%E7%B4%A2%E5%BC%95%20index%20%E5%A4%84%E6%8F%92%E5%85%A5%E5%85%83%E7%B4%A0%20num%22%22%22%0A%20%20%20%20%23%20%E6%8A%8A%E7%B4%A2%E5%BC%95%20index%20%E4%BB%A5%E5%8F%8A%E4%B9%8B%E5%90%8E%E7%9A%84%E6%89%80%E6%9C%89%E5%85%83%E7%B4%A0%E5%90%91%E5%90%8E%E7%A7%BB%E5%8A%A8%E4%B8%80%E4%BD%8D%0A%20%20%20%20for%20i%20in%20range%28len%28nums%29%20-%201,%20index,%20-1%29%3A%0A%20%20%20%20%20%20%20%20nums%5Bi%5D%20%3D%20nums%5Bi%20-%201%5D%0A%20%20%20%20%23%20%E5%B0%86%20num%20%E8%B5%8B%E7%BB%99%20index%20%E5%A4%84%E7%9A%84%E5%85%83%E7%B4%A0%0A%20%20%20%20nums%5Bindex%5D%20%3D%20num%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%E6%95%B0%E7%BB%84%0A%20%20%20%20nums%20%3D%20%5B1,%203,%202,%205,%204%5D%0A%20%20%20%20print%28%22%E6%95%B0%E7%BB%84%20nums%20%3D%22,%20nums%29%0A%0A%20%20%20%20%23%20%E6%8F%92%E5%85%A5%E5%85%83%E7%B4%A0%0A%20%20%20%20insert%28nums,%206,%203%29%0A%20%20%20%20print%28%22%E5%9C%A8%E7%B4%A2%E5%BC%95%203%20%E5%A4%84%E6%8F%92%E5%85%A5%E6%95%B0%E5%AD%97%206%20%EF%BC%8C%E5%BE%97%E5%88%B0%20nums%20%3D%22,%20nums%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=6&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
@@ -720,19 +684,6 @@ comments: true
end
```
=== "Zig"
```zig title="array.zig"
// 删除索引 index 处的元素
fn remove(nums: []i32, index: usize) void {
// 把索引 index 之后的所有元素向前移动一位
var i = index;
while (i < nums.len - 1) : (i += 1) {
nums[i] = nums[i + 1];
}
}
```
??? pythontutor "可视化运行"
<div style="height: 459px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20remove%28nums%3A%20list%5Bint%5D,%20index%3A%20int%29%3A%0A%20%20%20%20%22%22%22%E5%88%A0%E9%99%A4%E7%B4%A2%E5%BC%95%20index%20%E5%A4%84%E7%9A%84%E5%85%83%E7%B4%A0%22%22%22%0A%20%20%20%20%23%20%E6%8A%8A%E7%B4%A2%E5%BC%95%20index%20%E4%B9%8B%E5%90%8E%E7%9A%84%E6%89%80%E6%9C%89%E5%85%83%E7%B4%A0%E5%90%91%E5%89%8D%E7%A7%BB%E5%8A%A8%E4%B8%80%E4%BD%8D%0A%20%20%20%20for%20i%20in%20range%28index,%20len%28nums%29%20-%201%29%3A%0A%20%20%20%20%20%20%20%20nums%5Bi%5D%20%3D%20nums%5Bi%20%2B%201%5D%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%E6%95%B0%E7%BB%84%0A%20%20%20%20nums%20%3D%20%5B1,%203,%202,%205,%204%5D%0A%20%20%20%20print%28%22%E6%95%B0%E7%BB%84%20nums%20%3D%22,%20nums%29%0A%0A%20%20%20%20%23%20%E5%88%A0%E9%99%A4%E5%85%83%E7%B4%A0%0A%20%20%20%20remove%28nums,%202%29%0A%20%20%20%20print%28%22%E5%88%A0%E9%99%A4%E7%B4%A2%E5%BC%95%202%20%E5%A4%84%E7%9A%84%E5%85%83%E7%B4%A0%EF%BC%8C%E5%BE%97%E5%88%B0%20nums%20%3D%22,%20nums%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=6&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
@@ -980,33 +931,6 @@ comments: true
end
```
=== "Zig"
```zig title="array.zig"
// 遍历数组
fn traverse(nums: []const i32) void {
var count: i32 = 0;
// 通过索引遍历数组
var i: usize = 0;
while (i < nums.len) : (i += 1) {
count += nums[i];
}
// 直接遍历数组元素
count = 0;
for (nums) |num| {
count += num;
}
// 同时遍历数据索引和元素
for (nums, 0..) |num, index| {
count += nums[index];
count += num;
}
}
```
??? pythontutor "可视化运行"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20traverse%28nums%3A%20list%5Bint%5D%29%3A%0A%20%20%20%20%22%22%22%E9%81%8D%E5%8E%86%E6%95%B0%E7%BB%84%22%22%22%0A%20%20%20%20count%20%3D%200%0A%20%20%20%20%23%20%E9%80%9A%E8%BF%87%E7%B4%A2%E5%BC%95%E9%81%8D%E5%8E%86%E6%95%B0%E7%BB%84%0A%20%20%20%20for%20i%20in%20range%28len%28nums%29%29%3A%0A%20%20%20%20%20%20%20%20count%20%2B%3D%20nums%5Bi%5D%0A%20%20%20%20%23%20%E7%9B%B4%E6%8E%A5%E9%81%8D%E5%8E%86%E6%95%B0%E7%BB%84%E5%85%83%E7%B4%A0%0A%20%20%20%20for%20num%20in%20nums%3A%0A%20%20%20%20%20%20%20%20count%20%2B%3D%20num%0A%20%20%20%20%23%20%E5%90%8C%E6%97%B6%E9%81%8D%E5%8E%86%E6%95%B0%E6%8D%AE%E7%B4%A2%E5%BC%95%E5%92%8C%E5%85%83%E7%B4%A0%0A%20%20%20%20for%20i,%20num%20in%20enumerate%28nums%29%3A%0A%20%20%20%20%20%20%20%20count%20%2B%3D%20nums%5Bi%5D%0A%20%20%20%20%20%20%20%20count%20%2B%3D%20num%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%E6%95%B0%E7%BB%84%0A%20%20%20%20nums%20%3D%20%5B1,%203,%202,%205,%204%5D%0A%20%20%20%20print%28%22%E6%95%B0%E7%BB%84%20nums%20%3D%22,%20nums%29%0A%0A%20%20%20%20%23%20%E9%81%8D%E5%8E%86%E6%95%B0%E7%BB%84%0A%20%20%20%20traverse%28nums%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=6&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
@@ -1189,18 +1113,6 @@ comments: true
end
```
=== "Zig"
```zig title="array.zig"
// 在数组中查找指定元素
fn find(nums: []i32, target: i32) i32 {
for (nums, 0..) |num, i| {
if (num == target) return @intCast(i);
}
return -1;
}
```
??? pythontutor "可视化运行"
<div style="height: 477px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20find%28nums%3A%20list%5Bint%5D,%20target%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E5%9C%A8%E6%95%B0%E7%BB%84%E4%B8%AD%E6%9F%A5%E6%89%BE%E6%8C%87%E5%AE%9A%E5%85%83%E7%B4%A0%22%22%22%0A%20%20%20%20for%20i%20in%20range%28len%28nums%29%29%3A%0A%20%20%20%20%20%20%20%20if%20nums%5Bi%5D%20%3D%3D%20target%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20return%20i%0A%20%20%20%20return%20-1%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%E6%95%B0%E7%BB%84%0A%20%20%20%20nums%20%3D%20%5B1,%203,%202,%205,%204%5D%0A%20%20%20%20print%28%22%E6%95%B0%E7%BB%84%20nums%20%3D%22,%20nums%29%0A%0A%20%20%20%20%23%20%E6%9F%A5%E6%89%BE%E5%85%83%E7%B4%A0%0A%20%20%20%20index%3A%20int%20%3D%20find%28nums,%203%29%0A%20%20%20%20print%28%22%E5%9C%A8%20nums%20%E4%B8%AD%E6%9F%A5%E6%89%BE%E5%85%83%E7%B4%A0%203%20%EF%BC%8C%E5%BE%97%E5%88%B0%E7%B4%A2%E5%BC%95%20%3D%22,%20index%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=6&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
@@ -1431,23 +1343,6 @@ comments: true
end
```
=== "Zig"
```zig title="array.zig"
// 扩展数组长度
fn extend(allocator: std.mem.Allocator, nums: []const i32, enlarge: usize) ![]i32 {
// 初始化一个扩展长度后的数组
const res = try allocator.alloc(i32, nums.len + enlarge);
@memset(res, 0);
// 将原数组中的所有元素复制到新数组
std.mem.copyForwards(i32, res, nums);
// 返回扩展后的新数组
return res;
}
```
??? pythontutor "可视化运行"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=%23%20%E8%AF%B7%E6%B3%A8%E6%84%8F%EF%BC%8CPython%20%E7%9A%84%20list%20%E6%98%AF%E5%8A%A8%E6%80%81%E6%95%B0%E7%BB%84%EF%BC%8C%E5%8F%AF%E4%BB%A5%E7%9B%B4%E6%8E%A5%E6%89%A9%E5%B1%95%0A%23%20%E4%B8%BA%E4%BA%86%E6%96%B9%E4%BE%BF%E5%AD%A6%E4%B9%A0%EF%BC%8C%E6%9C%AC%E5%87%BD%E6%95%B0%E5%B0%86%20list%20%E7%9C%8B%E4%BD%9C%E9%95%BF%E5%BA%A6%E4%B8%8D%E5%8F%AF%E5%8F%98%E7%9A%84%E6%95%B0%E7%BB%84%0Adef%20extend%28nums%3A%20list%5Bint%5D,%20enlarge%3A%20int%29%20-%3E%20list%5Bint%5D%3A%0A%20%20%20%20%22%22%22%E6%89%A9%E5%B1%95%E6%95%B0%E7%BB%84%E9%95%BF%E5%BA%A6%22%22%22%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%E4%B8%80%E4%B8%AA%E6%89%A9%E5%B1%95%E9%95%BF%E5%BA%A6%E5%90%8E%E7%9A%84%E6%95%B0%E7%BB%84%0A%20%20%20%20res%20%3D%20%5B0%5D%20*%20%28len%28nums%29%20%2B%20enlarge%29%0A%20%20%20%20%23%20%E5%B0%86%E5%8E%9F%E6%95%B0%E7%BB%84%E4%B8%AD%E7%9A%84%E6%89%80%E6%9C%89%E5%85%83%E7%B4%A0%E5%A4%8D%E5%88%B6%E5%88%B0%E6%96%B0%E6%95%B0%E7%BB%84%0A%20%20%20%20for%20i%20in%20range%28len%28nums%29%29%3A%0A%20%20%20%20%20%20%20%20res%5Bi%5D%20%3D%20nums%5Bi%5D%0A%20%20%20%20%23%20%E8%BF%94%E5%9B%9E%E6%89%A9%E5%B1%95%E5%90%8E%E7%9A%84%E6%96%B0%E6%95%B0%E7%BB%84%0A%20%20%20%20return%20res%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%E6%95%B0%E7%BB%84%0A%20%20%20%20nums%20%3D%20%5B1,%203,%202,%205,%204%5D%0A%20%20%20%20print%28%22%E6%95%B0%E7%BB%84%20nums%20%3D%22,%20nums%29%0A%0A%20%20%20%20%23%20%E9%95%BF%E5%BA%A6%E6%89%A9%E5%B1%95%0A%20%20%20%20nums%20%3D%20extend%28nums,%203%29%0A%20%20%20%20print%28%22%E5%B0%86%E6%95%B0%E7%BB%84%E9%95%BF%E5%BA%A6%E6%89%A9%E5%B1%95%E8%87%B3%208%20%EF%BC%8C%E5%BE%97%E5%88%B0%20nums%20%3D%22,%20nums%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=6&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
docs/chapter_array_and_linkedlist/linked_list.md
-116
View File
@@ -191,26 +191,6 @@ comments: true
end
```
=== "Zig"
```zig title=""
// 链表节点类
pub fn ListNode(comptime T: type) type {
return struct {
const Self = @This();
val: T = 0, // 节点值
next: ?*Self = null, // 指向下一节点的指针
// 构造函数
pub fn init(self: *Self, x: i32) void {
self.val = x;
self.next = null;
}
};
}
```
## 4.2.1 &nbsp; 链表常用操作
### 1. &nbsp; 初始化链表
@@ -439,23 +419,6 @@ comments: true
n3.next = n4
```
=== "Zig"
```zig title="linked_list.zig"
// 初始化链表
// 初始化各个节点
var n0 = inc.ListNode(i32){.val = 1};
var n1 = inc.ListNode(i32){.val = 3};
var n2 = inc.ListNode(i32){.val = 2};
var n3 = inc.ListNode(i32){.val = 5};
var n4 = inc.ListNode(i32){.val = 4};
// 构建节点之间的引用
n0.next = &n1;
n1.next = &n2;
n2.next = &n3;
n3.next = &n4;
```
??? pythontutor "可视化运行"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=class%20ListNode%3A%0A%20%20%20%20%22%22%22%E9%93%BE%E8%A1%A8%E8%8A%82%E7%82%B9%E7%B1%BB%22%22%22%0A%20%20%20%20def%20__init__%28self,%20val%3A%20int%29%3A%0A%20%20%20%20%20%20%20%20self.val%3A%20int%20%3D%20val%20%20%23%20%E8%8A%82%E7%82%B9%E5%80%BC%0A%20%20%20%20%20%20%20%20self.next%3A%20ListNode%20%7C%20None%20%3D%20None%20%20%23%20%E5%90%8E%E7%BB%A7%E8%8A%82%E7%82%B9%E5%BC%95%E7%94%A8%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%E9%93%BE%E8%A1%A8%201%20-%3E%203%20-%3E%202%20-%3E%205%20-%3E%204%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%E5%90%84%E4%B8%AA%E8%8A%82%E7%82%B9%0A%20%20%20%20n0%20%3D%20ListNode%281%29%0A%20%20%20%20n1%20%3D%20ListNode%283%29%0A%20%20%20%20n2%20%3D%20ListNode%282%29%0A%20%20%20%20n3%20%3D%20ListNode%285%29%0A%20%20%20%20n4%20%3D%20ListNode%284%29%0A%20%20%20%20%23%20%E6%9E%84%E5%BB%BA%E8%8A%82%E7%82%B9%E4%B9%8B%E9%97%B4%E7%9A%84%E5%BC%95%E7%94%A8%0A%20%20%20%20n0.next%20%3D%20n1%0A%20%20%20%20n1.next%20%3D%20n2%0A%20%20%20%20n2.next%20%3D%20n3%0A%20%20%20%20n3.next%20%3D%20n4&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
@@ -617,17 +580,6 @@ comments: true
end
```
=== "Zig"
```zig title="linked_list.zig"
// 在链表的节点 n0 之后插入节点 P
fn insert(comptime T: type, n0: *ListNode(T), P: *ListNode(T)) void {
const n1 = n0.next;
P.next = n1;
n0.next = P;
}
```
??? pythontutor "可视化运行"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=class%20ListNode%3A%0A%20%20%20%20%22%22%22%E9%93%BE%E8%A1%A8%E8%8A%82%E7%82%B9%E7%B1%BB%22%22%22%0A%20%20%20%20def%20__init__%28self,%20val%3A%20int%29%3A%0A%20%20%20%20%20%20%20%20self.val%3A%20int%20%3D%20val%20%20%23%20%E8%8A%82%E7%82%B9%E5%80%BC%0A%20%20%20%20%20%20%20%20self.next%3A%20ListNode%20%7C%20None%20%3D%20None%20%20%23%20%E5%90%8E%E7%BB%A7%E8%8A%82%E7%82%B9%E5%BC%95%E7%94%A8%0A%0Adef%20insert%28n0%3A%20ListNode,%20P%3A%20ListNode%29%3A%0A%20%20%20%20%22%22%22%E5%9C%A8%E9%93%BE%E8%A1%A8%E7%9A%84%E8%8A%82%E7%82%B9%20n0%20%E4%B9%8B%E5%90%8E%E6%8F%92%E5%85%A5%E8%8A%82%E7%82%B9%20P%22%22%22%0A%20%20%20%20n1%20%3D%20n0.next%0A%20%20%20%20P.next%20%3D%20n1%0A%20%20%20%20n0.next%20%3D%20P%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%E9%93%BE%E8%A1%A8%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%E5%90%84%E4%B8%AA%E8%8A%82%E7%82%B9%0A%20%20%20%20n0%20%3D%20ListNode%281%29%0A%20%20%20%20n1%20%3D%20ListNode%283%29%0A%20%20%20%20n2%20%3D%20ListNode%282%29%0A%20%20%20%20n3%20%3D%20ListNode%285%29%0A%20%20%20%20n4%20%3D%20ListNode%284%29%0A%20%20%20%20%23%20%E6%9E%84%E5%BB%BA%E8%8A%82%E7%82%B9%E4%B9%8B%E9%97%B4%E7%9A%84%E5%BC%95%E7%94%A8%0A%20%20%20%20n0.next%20%3D%20n1%0A%20%20%20%20n1.next%20%3D%20n2%0A%20%20%20%20n2.next%20%3D%20n3%0A%20%20%20%20n3.next%20%3D%20n4%0A%0A%20%20%20%20%23%20%E6%8F%92%E5%85%A5%E8%8A%82%E7%82%B9%0A%20%20%20%20p%20%3D%20ListNode%280%29%0A%20%20%20%20insert%28n0,%20p%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=39&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
@@ -831,18 +783,6 @@ comments: true
end
```
=== "Zig"
```zig title="linked_list.zig"
// 删除链表的节点 n0 之后的首个节点
fn remove(comptime T: type, n0: *ListNode(T)) void {
// n0 -> P -> n1 => n0 -> n1
const P = n0.next;
const n1 = P.?.next;
n0.next = n1;
}
```
??? pythontutor "可视化运行"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=class%20ListNode%3A%0A%20%20%20%20%22%22%22%E9%93%BE%E8%A1%A8%E8%8A%82%E7%82%B9%E7%B1%BB%22%22%22%0A%20%20%20%20def%20__init__%28self,%20val%3A%20int%29%3A%0A%20%20%20%20%20%20%20%20self.val%3A%20int%20%3D%20val%20%20%23%20%E8%8A%82%E7%82%B9%E5%80%BC%0A%20%20%20%20%20%20%20%20self.next%3A%20ListNode%20%7C%20None%20%3D%20None%20%20%23%20%E5%90%8E%E7%BB%A7%E8%8A%82%E7%82%B9%E5%BC%95%E7%94%A8%0A%0Adef%20remove%28n0%3A%20ListNode%29%3A%0A%20%20%20%20%22%22%22%E5%88%A0%E9%99%A4%E9%93%BE%E8%A1%A8%E7%9A%84%E8%8A%82%E7%82%B9%20n0%20%E4%B9%8B%E5%90%8E%E7%9A%84%E9%A6%96%E4%B8%AA%E8%8A%82%E7%82%B9%22%22%22%0A%20%20%20%20if%20not%20n0.next%3A%0A%20%20%20%20%20%20%20%20return%0A%20%20%20%20%23%20n0%20-%3E%20P%20-%3E%20n1%0A%20%20%20%20P%20%3D%20n0.next%0A%20%20%20%20n1%20%3D%20P.next%0A%20%20%20%20n0.next%20%3D%20n1%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%E9%93%BE%E8%A1%A8%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%E5%90%84%E4%B8%AA%E8%8A%82%E7%82%B9%0A%20%20%20%20n0%20%3D%20ListNode%281%29%0A%20%20%20%20n1%20%3D%20ListNode%283%29%0A%20%20%20%20n2%20%3D%20ListNode%282%29%0A%20%20%20%20n3%20%3D%20ListNode%285%29%0A%20%20%20%20n4%20%3D%20ListNode%284%29%0A%20%20%20%20%23%20%E6%9E%84%E5%BB%BA%E8%8A%82%E7%82%B9%E4%B9%8B%E9%97%B4%E7%9A%84%E5%BC%95%E7%94%A8%0A%20%20%20%20n0.next%20%3D%20n1%0A%20%20%20%20n1.next%20%3D%20n2%0A%20%20%20%20n2.next%20%3D%20n3%0A%20%20%20%20n3.next%20%3D%20n4%0A%0A%20%20%20%20%23%20%E5%88%A0%E9%99%A4%E8%8A%82%E7%82%B9%0A%20%20%20%20remove%28n0%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=34&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
@@ -1047,24 +987,6 @@ comments: true
end
```
=== "Zig"
```zig title="linked_list.zig"
// 访问链表中索引为 index 的节点
fn access(comptime T: type, node: *ListNode(T), index: i32) ?*ListNode(T) {
var head: ?*ListNode(T) = node;
var i: i32 = 0;
while (i < index) : (i += 1) {
if (head) |cur| {
head = cur.next;
} else {
return null;
}
}
return head;
}
```
??? pythontutor "可视化运行"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=class%20ListNode%3A%0A%20%20%20%20%22%22%22%E9%93%BE%E8%A1%A8%E8%8A%82%E7%82%B9%E7%B1%BB%22%22%22%0A%20%20%20%20def%20__init__%28self,%20val%3A%20int%29%3A%0A%20%20%20%20%20%20%20%20self.val%3A%20int%20%3D%20val%20%20%23%20%E8%8A%82%E7%82%B9%E5%80%BC%0A%20%20%20%20%20%20%20%20self.next%3A%20ListNode%20%7C%20None%20%3D%20None%20%20%23%20%E5%90%8E%E7%BB%A7%E8%8A%82%E7%82%B9%E5%BC%95%E7%94%A8%0A%0Adef%20access%28head%3A%20ListNode,%20index%3A%20int%29%20-%3E%20ListNode%20%7C%20None%3A%0A%20%20%20%20%22%22%22%E8%AE%BF%E9%97%AE%E9%93%BE%E8%A1%A8%E4%B8%AD%E7%B4%A2%E5%BC%95%E4%B8%BA%20index%20%E7%9A%84%E8%8A%82%E7%82%B9%22%22%22%0A%20%20%20%20for%20_%20in%20range%28index%29%3A%0A%20%20%20%20%20%20%20%20if%20not%20head%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20return%20None%0A%20%20%20%20%20%20%20%20head%20%3D%20head.next%0A%20%20%20%20return%20head%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%E9%93%BE%E8%A1%A8%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%E5%90%84%E4%B8%AA%E8%8A%82%E7%82%B9%0A%20%20%20%20n0%20%3D%20ListNode%281%29%0A%20%20%20%20n1%20%3D%20ListNode%283%29%0A%20%20%20%20n2%20%3D%20ListNode%282%29%0A%20%20%20%20n3%20%3D%20ListNode%285%29%0A%20%20%20%20n4%20%3D%20ListNode%284%29%0A%20%20%20%20%23%20%E6%9E%84%E5%BB%BA%E8%8A%82%E7%82%B9%E4%B9%8B%E9%97%B4%E7%9A%84%E5%BC%95%E7%94%A8%0A%20%20%20%20n0.next%20%3D%20n1%0A%20%20%20%20n1.next%20%3D%20n2%0A%20%20%20%20n2.next%20%3D%20n3%0A%20%20%20%20n3.next%20%3D%20n4%0A%0A%20%20%20%20%23%20%E8%AE%BF%E9%97%AE%E8%8A%82%E7%82%B9%0A%20%20%20%20node%20%3D%20access%28n0,%203%29%0A%20%20%20%20print%28%22%E9%93%BE%E8%A1%A8%E4%B8%AD%E7%B4%A2%E5%BC%95%203%20%E5%A4%84%E7%9A%84%E8%8A%82%E7%82%B9%E7%9A%84%E5%80%BC%20%3D%20%7B%7D%22.format%28node.val%29%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=34&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
@@ -1291,22 +1213,6 @@ comments: true
end
```
=== "Zig"
```zig title="linked_list.zig"
// 在链表中查找值为 target 的首个节点
fn find(comptime T: type, node: *ListNode(T), target: T) i32 {
var head: ?*ListNode(T) = node;
var index: i32 = 0;
while (head) |cur| {
if (cur.val == target) return index;
head = cur.next;
index += 1;
}
return -1;
}
```
??? pythontutor "可视化运行"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=class%20ListNode%3A%0A%20%20%20%20%22%22%22%E9%93%BE%E8%A1%A8%E8%8A%82%E7%82%B9%E7%B1%BB%22%22%22%0A%20%20%20%20def%20__init__%28self,%20val%3A%20int%29%3A%0A%20%20%20%20%20%20%20%20self.val%3A%20int%20%3D%20val%20%20%23%20%E8%8A%82%E7%82%B9%E5%80%BC%0A%20%20%20%20%20%20%20%20self.next%3A%20ListNode%20%7C%20None%20%3D%20None%20%20%23%20%E5%90%8E%E7%BB%A7%E8%8A%82%E7%82%B9%E5%BC%95%E7%94%A8%0A%0Adef%20find%28head%3A%20ListNode,%20target%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E5%9C%A8%E9%93%BE%E8%A1%A8%E4%B8%AD%E6%9F%A5%E6%89%BE%E5%80%BC%E4%B8%BA%20target%20%E7%9A%84%E9%A6%96%E4%B8%AA%E8%8A%82%E7%82%B9%22%22%22%0A%20%20%20%20index%20%3D%200%0A%20%20%20%20while%20head%3A%0A%20%20%20%20%20%20%20%20if%20head.val%20%3D%3D%20target%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20return%20index%0A%20%20%20%20%20%20%20%20head%20%3D%20head.next%0A%20%20%20%20%20%20%20%20index%20%2B%3D%201%0A%20%20%20%20return%20-1%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%E9%93%BE%E8%A1%A8%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%E5%90%84%E4%B8%AA%E8%8A%82%E7%82%B9%0A%20%20%20%20n0%20%3D%20ListNode%281%29%0A%20%20%20%20n1%20%3D%20ListNode%283%29%0A%20%20%20%20n2%20%3D%20ListNode%282%29%0A%20%20%20%20n3%20%3D%20ListNode%285%29%0A%20%20%20%20n4%20%3D%20ListNode%284%29%0A%20%20%20%20%23%20%E6%9E%84%E5%BB%BA%E8%8A%82%E7%82%B9%E4%B9%8B%E9%97%B4%E7%9A%84%E5%BC%95%E7%94%A8%0A%20%20%20%20n0.next%20%3D%20n1%0A%20%20%20%20n1.next%20%3D%20n2%0A%20%20%20%20n2.next%20%3D%20n3%0A%20%20%20%20n3.next%20%3D%20n4%0A%0A%20%20%20%20%23%20%E6%9F%A5%E6%89%BE%E8%8A%82%E7%82%B9%0A%20%20%20%20index%20%3D%20find%28n0,%202%29%0A%20%20%20%20print%28%22%E9%93%BE%E8%A1%A8%E4%B8%AD%E5%80%BC%E4%B8%BA%202%20%E7%9A%84%E8%8A%82%E7%82%B9%E7%9A%84%E7%B4%A2%E5%BC%95%20%3D%20%7B%7D%22.format%28index%29%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=34&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
@@ -1537,28 +1443,6 @@ comments: true
end
```
=== "Zig"
```zig title=""
// 双向链表节点类
pub fn ListNode(comptime T: type) type {
return struct {
const Self = @This();
val: T = 0, // 节点值
next: ?*Self = null, // 指向后继节点的指针
prev: ?*Self = null, // 指向前驱节点的指针
// 构造函数
pub fn init(self: *Self, x: i32) void {
self.val = x;
self.next = null;
self.prev = null;
}
};
}
```
![常见链表种类](linked_list.assets/linkedlist_common_types.png){ class="animation-figure" }
<p align="center"> 图 4-8 &nbsp; 常见链表种类 </p>
docs/chapter_array_and_linkedlist/list.md
-230
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docs/chapter_backtracking/backtracking_algorithm.md
-40
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docs/chapter_backtracking/n_queens_problem.md
-8
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@@ -744,14 +744,6 @@ comments: true
end
```
=== "Zig"
```zig title="n_queens.zig"
[class]{}-[func]{backtrack}
[class]{}-[func]{nQueens}
```
??? pythontutor "可视化运行"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20backtrack%28%0A%20%20%20%20row%3A%20int,%0A%20%20%20%20n%3A%20int,%0A%20%20%20%20state%3A%20list%5Blist%5Bstr%5D%5D,%0A%20%20%20%20res%3A%20list%5Blist%5Blist%5Bstr%5D%5D%5D,%0A%20%20%20%20cols%3A%20list%5Bbool%5D,%0A%20%20%20%20diags1%3A%20list%5Bbool%5D,%0A%20%20%20%20diags2%3A%20list%5Bbool%5D,%0A%29%3A%0A%20%20%20%20%22%22%22%E5%9B%9E%E6%BA%AF%E7%AE%97%E6%B3%95%EF%BC%9AN%20%E7%9A%87%E5%90%8E%22%22%22%0A%20%20%20%20%23%20%E5%BD%93%E6%94%BE%E7%BD%AE%E5%AE%8C%E6%89%80%E6%9C%89%E8%A1%8C%E6%97%B6%EF%BC%8C%E8%AE%B0%E5%BD%95%E8%A7%A3%0A%20%20%20%20if%20row%20%3D%3D%20n%3A%0A%20%20%20%20%20%20%20%20res.append%28%5Blist%28row%29%20for%20row%20in%20state%5D%29%0A%20%20%20%20%20%20%20%20return%0A%20%20%20%20%23%20%E9%81%8D%E5%8E%86%E6%89%80%E6%9C%89%E5%88%97%0A%20%20%20%20for%20col%20in%20range%28n%29%3A%0A%20%20%20%20%20%20%20%20%23%20%E8%AE%A1%E7%AE%97%E8%AF%A5%E6%A0%BC%E5%AD%90%E5%AF%B9%E5%BA%94%E7%9A%84%E4%B8%BB%E5%AF%B9%E8%A7%92%E7%BA%BF%E5%92%8C%E6%AC%A1%E5%AF%B9%E8%A7%92%E7%BA%BF%0A%20%20%20%20%20%20%20%20diag1%20%3D%20row%20-%20col%20%2B%20n%20-%201%0A%20%20%20%20%20%20%20%20diag2%20%3D%20row%20%2B%20col%0A%20%20%20%20%20%20%20%20%23%20%E5%89%AA%E6%9E%9D%EF%BC%9A%E4%B8%8D%E5%85%81%E8%AE%B8%E8%AF%A5%E6%A0%BC%E5%AD%90%E6%89%80%E5%9C%A8%E5%88%97%E3%80%81%E4%B8%BB%E5%AF%B9%E8%A7%92%E7%BA%BF%E3%80%81%E6%AC%A1%E5%AF%B9%E8%A7%92%E7%BA%BF%E4%B8%8A%E5%AD%98%E5%9C%A8%E7%9A%87%E5%90%8E%0A%20%20%20%20%20%20%20%20if%20not%20cols%5Bcol%5D%20and%20not%20diags1%5Bdiag1%5D%20and%20not%20diags2%5Bdiag2%5D%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20%23%20%E5%B0%9D%E8%AF%95%EF%BC%9A%E5%B0%86%E7%9A%87%E5%90%8E%E6%94%BE%E7%BD%AE%E5%9C%A8%E8%AF%A5%E6%A0%BC%E5%AD%90%0A%20%20%20%20%20%20%20%20%20%20%20%20state%5Brow%5D%5Bcol%5D%20%3D%20%22Q%22%0A%20%20%20%20%20%20%20%20%20%20%20%20cols%5Bcol%5D%20%3D%20diags1%5Bdiag1%5D%20%3D%20diags2%5Bdiag2%5D%20%3D%20True%0A%20%20%20%20%20%20%20%20%20%20%20%20%23%20%E6%94%BE%E7%BD%AE%E4%B8%8B%E4%B8%80%E8%A1%8C%0A%20%20%20%20%20%20%20%20%20%20%20%20backtrack%28row%20%2B%201,%20n,%20state,%20res,%20cols,%20diags1,%20diags2%29%0A%20%20%20%20%20%20%20%20%20%20%20%20%23%20%E5%9B%9E%E9%80%80%EF%BC%9A%E5%B0%86%E8%AF%A5%E6%A0%BC%E5%AD%90%E6%81%A2%E5%A4%8D%E4%B8%BA%E7%A9%BA%E4%BD%8D%0A%20%20%20%20%20%20%20%20%20%20%20%20state%5Brow%5D%5Bcol%5D%20%3D%20%22%23%22%0A%20%20%20%20%20%20%20%20%20%20%20%20cols%5Bcol%5D%20%3D%20diags1%5Bdiag1%5D%20%3D%20diags2%5Bdiag2%5D%20%3D%20False%0A%0A%0Adef%20n_queens%28n%3A%20int%29%20-%3E%20list%5Blist%5Blist%5Bstr%5D%5D%5D%3A%0A%20%20%20%20%22%22%22%E6%B1%82%E8%A7%A3%20N%20%E7%9A%87%E5%90%8E%22%22%22%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%20n*n%20%E5%A4%A7%E5%B0%8F%E7%9A%84%E6%A3%8B%E7%9B%98%EF%BC%8C%E5%85%B6%E4%B8%AD%20'Q'%20%E4%BB%A3%E8%A1%A8%E7%9A%87%E5%90%8E%EF%BC%8C'%23'%20%E4%BB%A3%E8%A1%A8%E7%A9%BA%E4%BD%8D%0A%20%20%20%20state%20%3D%20%5B%5B%22%23%22%20for%20_%20in%20range%28n%29%5D%20for%20_%20in%20range%28n%29%5D%0A%20%20%20%20cols%20%3D%20%5BFalse%5D%20*%20n%20%20%23%20%E8%AE%B0%E5%BD%95%E5%88%97%E6%98%AF%E5%90%A6%E6%9C%89%E7%9A%87%E5%90%8E%0A%20%20%20%20diags1%20%3D%20%5BFalse%5D%20*%20%282%20*%20n%20-%201%29%20%20%23%20%E8%AE%B0%E5%BD%95%E4%B8%BB%E5%AF%B9%E8%A7%92%E7%BA%BF%E4%B8%8A%E6%98%AF%E5%90%A6%E6%9C%89%E7%9A%87%E5%90%8E%0A%20%20%20%20diags2%20%3D%20%5BFalse%5D%20*%20%282%20*%20n%20-%201%29%20%20%23%20%E8%AE%B0%E5%BD%95%E6%AC%A1%E5%AF%B9%E8%A7%92%E7%BA%BF%E4%B8%8A%E6%98%AF%E5%90%A6%E6%9C%89%E7%9A%87%E5%90%8E%0A%20%20%20%20res%20%3D%20%5B%5D%0A%20%20%20%20backtrack%280,%20n,%20state,%20res,%20cols,%20diags1,%20diags2%29%0A%0A%20%20%20%20return%20res%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%204%0A%20%20%20%20res%20%3D%20n_queens%28n%29%0A%0A%20%20%20%20print%28f%22%E8%BE%93%E5%85%A5%E6%A3%8B%E7%9B%98%E9%95%BF%E5%AE%BD%E4%B8%BA%20%7Bn%7D%22%29%0A%20%20%20%20print%28f%22%E7%9A%87%E5%90%8E%E6%94%BE%E7%BD%AE%E6%96%B9%E6%A1%88%E5%85%B1%E6%9C%89%20%7Blen%28res%29%7D%20%E7%A7%8D%22%29%0A%20%20%20%20for%20state%20in%20res%3A%0A%20%20%20%20%20%20%20%20print%28%22--------------------%22%29%0A%20%20%20%20%20%20%20%20for%20row%20in%20state%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20print%28row%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=61&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
docs/chapter_backtracking/permutations_problem.md
-16
View File
@@ -538,14 +538,6 @@ comments: true
end
```
=== "Zig"
```zig title="permutations_i.zig"
[class]{}-[func]{backtrack}
[class]{}-[func]{permutationsI}
```
??? pythontutor "可视化运行"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20backtrack%28%0A%20%20%20%20state%3A%20list%5Bint%5D,%20choices%3A%20list%5Bint%5D,%20selected%3A%20list%5Bbool%5D,%20res%3A%20list%5Blist%5Bint%5D%5D%0A%29%3A%0A%20%20%20%20%22%22%22%E5%9B%9E%E6%BA%AF%E7%AE%97%E6%B3%95%EF%BC%9A%E5%85%A8%E6%8E%92%E5%88%97%20I%22%22%22%0A%20%20%20%20%23%20%E5%BD%93%E7%8A%B6%E6%80%81%E9%95%BF%E5%BA%A6%E7%AD%89%E4%BA%8E%E5%85%83%E7%B4%A0%E6%95%B0%E9%87%8F%E6%97%B6%EF%BC%8C%E8%AE%B0%E5%BD%95%E8%A7%A3%0A%20%20%20%20if%20len%28state%29%20%3D%3D%20len%28choices%29%3A%0A%20%20%20%20%20%20%20%20res.append%28list%28state%29%29%0A%20%20%20%20%20%20%20%20return%0A%20%20%20%20%23%20%E9%81%8D%E5%8E%86%E6%89%80%E6%9C%89%E9%80%89%E6%8B%A9%0A%20%20%20%20for%20i,%20choice%20in%20enumerate%28choices%29%3A%0A%20%20%20%20%20%20%20%20%23%20%E5%89%AA%E6%9E%9D%EF%BC%9A%E4%B8%8D%E5%85%81%E8%AE%B8%E9%87%8D%E5%A4%8D%E9%80%89%E6%8B%A9%E5%85%83%E7%B4%A0%0A%20%20%20%20%20%20%20%20if%20not%20selected%5Bi%5D%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20%23%20%E5%B0%9D%E8%AF%95%EF%BC%9A%E5%81%9A%E5%87%BA%E9%80%89%E6%8B%A9%EF%BC%8C%E6%9B%B4%E6%96%B0%E7%8A%B6%E6%80%81%0A%20%20%20%20%20%20%20%20%20%20%20%20selected%5Bi%5D%20%3D%20True%0A%20%20%20%20%20%20%20%20%20%20%20%20state.append%28choice%29%0A%20%20%20%20%20%20%20%20%20%20%20%20%23%20%E8%BF%9B%E8%A1%8C%E4%B8%8B%E4%B8%80%E8%BD%AE%E9%80%89%E6%8B%A9%0A%20%20%20%20%20%20%20%20%20%20%20%20backtrack%28state,%20choices,%20selected,%20res%29%0A%20%20%20%20%20%20%20%20%20%20%20%20%23%20%E5%9B%9E%E9%80%80%EF%BC%9A%E6%92%A4%E9%94%80%E9%80%89%E6%8B%A9%EF%BC%8C%E6%81%A2%E5%A4%8D%E5%88%B0%E4%B9%8B%E5%89%8D%E7%9A%84%E7%8A%B6%E6%80%81%0A%20%20%20%20%20%20%20%20%20%20%20%20selected%5Bi%5D%20%3D%20False%0A%20%20%20%20%20%20%20%20%20%20%20%20state.pop%28%29%0A%0A%0Adef%20permutations_i%28nums%3A%20list%5Bint%5D%29%20-%3E%20list%5Blist%5Bint%5D%5D%3A%0A%20%20%20%20%22%22%22%E5%85%A8%E6%8E%92%E5%88%97%20I%22%22%22%0A%20%20%20%20res%20%3D%20%5B%5D%0A%20%20%20%20backtrack%28state%3D%5B%5D,%20choices%3Dnums,%20selected%3D%5BFalse%5D%20*%20len%28nums%29,%20res%3Dres%29%0A%20%20%20%20return%20res%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20nums%20%3D%20%5B1,%202,%203%5D%0A%0A%20%20%20%20res%20%3D%20permutations_i%28nums%29%0A%0A%20%20%20%20print%28f%22%E8%BE%93%E5%85%A5%E6%95%B0%E7%BB%84%20nums%20%3D%20%7Bnums%7D%22%29%0A%20%20%20%20print%28f%22%E6%89%80%E6%9C%89%E6%8E%92%E5%88%97%20res%20%3D%20%7Bres%7D%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=13&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
@@ -1093,14 +1085,6 @@ comments: true
end
```
=== "Zig"
```zig title="permutations_ii.zig"
[class]{}-[func]{backtrack}
[class]{}-[func]{permutationsII}
```
??? pythontutor "可视化运行"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20backtrack%28%0A%20%20%20%20state%3A%20list%5Bint%5D,%20choices%3A%20list%5Bint%5D,%20selected%3A%20list%5Bbool%5D,%20res%3A%20list%5Blist%5Bint%5D%5D%0A%29%3A%0A%20%20%20%20%22%22%22%E5%9B%9E%E6%BA%AF%E7%AE%97%E6%B3%95%EF%BC%9A%E5%85%A8%E6%8E%92%E5%88%97%20II%22%22%22%0A%20%20%20%20%23%20%E5%BD%93%E7%8A%B6%E6%80%81%E9%95%BF%E5%BA%A6%E7%AD%89%E4%BA%8E%E5%85%83%E7%B4%A0%E6%95%B0%E9%87%8F%E6%97%B6%EF%BC%8C%E8%AE%B0%E5%BD%95%E8%A7%A3%0A%20%20%20%20if%20len%28state%29%20%3D%3D%20len%28choices%29%3A%0A%20%20%20%20%20%20%20%20res.append%28list%28state%29%29%0A%20%20%20%20%20%20%20%20return%0A%20%20%20%20%23%20%E9%81%8D%E5%8E%86%E6%89%80%E6%9C%89%E9%80%89%E6%8B%A9%0A%20%20%20%20duplicated%20%3D%20set%5Bint%5D%28%29%0A%20%20%20%20for%20i,%20choice%20in%20enumerate%28choices%29%3A%0A%20%20%20%20%20%20%20%20%23%20%E5%89%AA%E6%9E%9D%EF%BC%9A%E4%B8%8D%E5%85%81%E8%AE%B8%E9%87%8D%E5%A4%8D%E9%80%89%E6%8B%A9%E5%85%83%E7%B4%A0%20%E4%B8%94%20%E4%B8%8D%E5%85%81%E8%AE%B8%E9%87%8D%E5%A4%8D%E9%80%89%E6%8B%A9%E7%9B%B8%E7%AD%89%E5%85%83%E7%B4%A0%0A%20%20%20%20%20%20%20%20if%20not%20selected%5Bi%5D%20and%20choice%20not%20in%20duplicated%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20%23%20%E5%B0%9D%E8%AF%95%EF%BC%9A%E5%81%9A%E5%87%BA%E9%80%89%E6%8B%A9%EF%BC%8C%E6%9B%B4%E6%96%B0%E7%8A%B6%E6%80%81%0A%20%20%20%20%20%20%20%20%20%20%20%20duplicated.add%28choice%29%20%20%23%20%E8%AE%B0%E5%BD%95%E9%80%89%E6%8B%A9%E8%BF%87%E7%9A%84%E5%85%83%E7%B4%A0%E5%80%BC%0A%20%20%20%20%20%20%20%20%20%20%20%20selected%5Bi%5D%20%3D%20True%0A%20%20%20%20%20%20%20%20%20%20%20%20state.append%28choice%29%0A%20%20%20%20%20%20%20%20%20%20%20%20%23%20%E8%BF%9B%E8%A1%8C%E4%B8%8B%E4%B8%80%E8%BD%AE%E9%80%89%E6%8B%A9%0A%20%20%20%20%20%20%20%20%20%20%20%20backtrack%28state,%20choices,%20selected,%20res%29%0A%20%20%20%20%20%20%20%20%20%20%20%20%23%20%E5%9B%9E%E9%80%80%EF%BC%9A%E6%92%A4%E9%94%80%E9%80%89%E6%8B%A9%EF%BC%8C%E6%81%A2%E5%A4%8D%E5%88%B0%E4%B9%8B%E5%89%8D%E7%9A%84%E7%8A%B6%E6%80%81%0A%20%20%20%20%20%20%20%20%20%20%20%20selected%5Bi%5D%20%3D%20False%0A%20%20%20%20%20%20%20%20%20%20%20%20state.pop%28%29%0A%0A%0Adef%20permutations_ii%28nums%3A%20list%5Bint%5D%29%20-%3E%20list%5Blist%5Bint%5D%5D%3A%0A%20%20%20%20%22%22%22%E5%85%A8%E6%8E%92%E5%88%97%20II%22%22%22%0A%20%20%20%20res%20%3D%20%5B%5D%0A%20%20%20%20backtrack%28state%3D%5B%5D,%20choices%3Dnums,%20selected%3D%5BFalse%5D%20*%20len%28nums%29,%20res%3Dres%29%0A%20%20%20%20return%20res%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20nums%20%3D%20%5B1,%202,%202%5D%0A%0A%20%20%20%20res%20%3D%20permutations_ii%28nums%29%0A%0A%20%20%20%20print%28f%22%E8%BE%93%E5%85%A5%E6%95%B0%E7%BB%84%20nums%20%3D%20%7Bnums%7D%22%29%0A%20%20%20%20print%28f%22%E6%89%80%E6%9C%89%E6%8E%92%E5%88%97%20res%20%3D%20%7Bres%7D%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=13&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
docs/chapter_backtracking/subset_sum_problem.md
-24
View File
@@ -501,14 +501,6 @@ comments: true
end
```
=== "Zig"
```zig title="subset_sum_i_naive.zig"
[class]{}-[func]{backtrack}
[class]{}-[func]{subsetSumINaive}
```
??? pythontutor "可视化运行"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20backtrack%28%0A%20%20%20%20state%3A%20list%5Bint%5D,%0A%20%20%20%20target%3A%20int,%0A%20%20%20%20total%3A%20int,%0A%20%20%20%20choices%3A%20list%5Bint%5D,%0A%20%20%20%20res%3A%20list%5Blist%5Bint%5D%5D,%0A%29%3A%0A%20%20%20%20%22%22%22%E5%9B%9E%E6%BA%AF%E7%AE%97%E6%B3%95%EF%BC%9A%E5%AD%90%E9%9B%86%E5%92%8C%20I%22%22%22%0A%20%20%20%20%23%20%E5%AD%90%E9%9B%86%E5%92%8C%E7%AD%89%E4%BA%8E%20target%20%E6%97%B6%EF%BC%8C%E8%AE%B0%E5%BD%95%E8%A7%A3%0A%20%20%20%20if%20total%20%3D%3D%20target%3A%0A%20%20%20%20%20%20%20%20res.append%28list%28state%29%29%0A%20%20%20%20%20%20%20%20return%0A%20%20%20%20%23%20%E9%81%8D%E5%8E%86%E6%89%80%E6%9C%89%E9%80%89%E6%8B%A9%0A%20%20%20%20for%20i%20in%20range%28len%28choices%29%29%3A%0A%20%20%20%20%20%20%20%20%23%20%E5%89%AA%E6%9E%9D%EF%BC%9A%E8%8B%A5%E5%AD%90%E9%9B%86%E5%92%8C%E8%B6%85%E8%BF%87%20target%20%EF%BC%8C%E5%88%99%E8%B7%B3%E8%BF%87%E8%AF%A5%E9%80%89%E6%8B%A9%0A%20%20%20%20%20%20%20%20if%20total%20%2B%20choices%5Bi%5D%20%3E%20target%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20continue%0A%20%20%20%20%20%20%20%20%23%20%E5%B0%9D%E8%AF%95%EF%BC%9A%E5%81%9A%E5%87%BA%E9%80%89%E6%8B%A9%EF%BC%8C%E6%9B%B4%E6%96%B0%E5%85%83%E7%B4%A0%E5%92%8C%20total%0A%20%20%20%20%20%20%20%20state.append%28choices%5Bi%5D%29%0A%20%20%20%20%20%20%20%20%23%20%E8%BF%9B%E8%A1%8C%E4%B8%8B%E4%B8%80%E8%BD%AE%E9%80%89%E6%8B%A9%0A%20%20%20%20%20%20%20%20backtrack%28state,%20target,%20total%20%2B%20choices%5Bi%5D,%20choices,%20res%29%0A%20%20%20%20%20%20%20%20%23%20%E5%9B%9E%E9%80%80%EF%BC%9A%E6%92%A4%E9%94%80%E9%80%89%E6%8B%A9%EF%BC%8C%E6%81%A2%E5%A4%8D%E5%88%B0%E4%B9%8B%E5%89%8D%E7%9A%84%E7%8A%B6%E6%80%81%0A%20%20%20%20%20%20%20%20state.pop%28%29%0A%0A%0Adef%20subset_sum_i_naive%28nums%3A%20list%5Bint%5D,%20target%3A%20int%29%20-%3E%20list%5Blist%5Bint%5D%5D%3A%0A%20%20%20%20%22%22%22%E6%B1%82%E8%A7%A3%E5%AD%90%E9%9B%86%E5%92%8C%20I%EF%BC%88%E5%8C%85%E5%90%AB%E9%87%8D%E5%A4%8D%E5%AD%90%E9%9B%86%EF%BC%89%22%22%22%0A%20%20%20%20state%20%3D%20%5B%5D%20%20%23%20%E7%8A%B6%E6%80%81%EF%BC%88%E5%AD%90%E9%9B%86%EF%BC%89%0A%20%20%20%20total%20%3D%200%20%20%23%20%E5%AD%90%E9%9B%86%E5%92%8C%0A%20%20%20%20res%20%3D%20%5B%5D%20%20%23%20%E7%BB%93%E6%9E%9C%E5%88%97%E8%A1%A8%EF%BC%88%E5%AD%90%E9%9B%86%E5%88%97%E8%A1%A8%EF%BC%89%0A%20%20%20%20backtrack%28state,%20target,%20total,%20nums,%20res%29%0A%20%20%20%20return%20res%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20nums%20%3D%20%5B3,%204,%205%5D%0A%20%20%20%20target%20%3D%209%0A%20%20%20%20res%20%3D%20subset_sum_i_naive%28nums,%20target%29%0A%0A%20%20%20%20print%28f%22%E8%BE%93%E5%85%A5%E6%95%B0%E7%BB%84%20nums%20%3D%20%7Bnums%7D,%20target%20%3D%20%7Btarget%7D%22%29%0A%20%20%20%20print%28f%22%E6%89%80%E6%9C%89%E5%92%8C%E7%AD%89%E4%BA%8E%20%7Btarget%7D%20%E7%9A%84%E5%AD%90%E9%9B%86%20res%20%3D%20%7Bres%7D%22%29%0A%20%20%20%20print%28f%22%E8%AF%B7%E6%B3%A8%E6%84%8F%EF%BC%8C%E8%AF%A5%E6%96%B9%E6%B3%95%E8%BE%93%E5%87%BA%E7%9A%84%E7%BB%93%E6%9E%9C%E5%8C%85%E5%90%AB%E9%87%8D%E5%A4%8D%E9%9B%86%E5%90%88%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=16&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
@@ -1070,14 +1062,6 @@ comments: true
end
```
=== "Zig"
```zig title="subset_sum_i.zig"
[class]{}-[func]{backtrack}
[class]{}-[func]{subsetSumI}
```
??? pythontutor "可视化运行"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20backtrack%28%0A%20%20%20%20state%3A%20list%5Bint%5D,%20target%3A%20int,%20choices%3A%20list%5Bint%5D,%20start%3A%20int,%20res%3A%20list%5Blist%5Bint%5D%5D%0A%29%3A%0A%20%20%20%20%22%22%22%E5%9B%9E%E6%BA%AF%E7%AE%97%E6%B3%95%EF%BC%9A%E5%AD%90%E9%9B%86%E5%92%8C%20I%22%22%22%0A%20%20%20%20%23%20%E5%AD%90%E9%9B%86%E5%92%8C%E7%AD%89%E4%BA%8E%20target%20%E6%97%B6%EF%BC%8C%E8%AE%B0%E5%BD%95%E8%A7%A3%0A%20%20%20%20if%20target%20%3D%3D%200%3A%0A%20%20%20%20%20%20%20%20res.append%28list%28state%29%29%0A%20%20%20%20%20%20%20%20return%0A%20%20%20%20%23%20%E9%81%8D%E5%8E%86%E6%89%80%E6%9C%89%E9%80%89%E6%8B%A9%0A%20%20%20%20%23%20%E5%89%AA%E6%9E%9D%E4%BA%8C%EF%BC%9A%E4%BB%8E%20start%20%E5%BC%80%E5%A7%8B%E9%81%8D%E5%8E%86%EF%BC%8C%E9%81%BF%E5%85%8D%E7%94%9F%E6%88%90%E9%87%8D%E5%A4%8D%E5%AD%90%E9%9B%86%0A%20%20%20%20for%20i%20in%20range%28start,%20len%28choices%29%29%3A%0A%20%20%20%20%20%20%20%20%23%20%E5%89%AA%E6%9E%9D%E4%B8%80%EF%BC%9A%E8%8B%A5%E5%AD%90%E9%9B%86%E5%92%8C%E8%B6%85%E8%BF%87%20target%20%EF%BC%8C%E5%88%99%E7%9B%B4%E6%8E%A5%E7%BB%93%E6%9D%9F%E5%BE%AA%E7%8E%AF%0A%20%20%20%20%20%20%20%20%23%20%E8%BF%99%E6%98%AF%E5%9B%A0%E4%B8%BA%E6%95%B0%E7%BB%84%E5%B7%B2%E6%8E%92%E5%BA%8F%EF%BC%8C%E5%90%8E%E8%BE%B9%E5%85%83%E7%B4%A0%E6%9B%B4%E5%A4%A7%EF%BC%8C%E5%AD%90%E9%9B%86%E5%92%8C%E4%B8%80%E5%AE%9A%E8%B6%85%E8%BF%87%20target%0A%20%20%20%20%20%20%20%20if%20target%20-%20choices%5Bi%5D%20%3C%200%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20break%0A%20%20%20%20%20%20%20%20%23%20%E5%B0%9D%E8%AF%95%EF%BC%9A%E5%81%9A%E5%87%BA%E9%80%89%E6%8B%A9%EF%BC%8C%E6%9B%B4%E6%96%B0%20target,%20start%0A%20%20%20%20%20%20%20%20state.append%28choices%5Bi%5D%29%0A%20%20%20%20%20%20%20%20%23%20%E8%BF%9B%E8%A1%8C%E4%B8%8B%E4%B8%80%E8%BD%AE%E9%80%89%E6%8B%A9%0A%20%20%20%20%20%20%20%20backtrack%28state,%20target%20-%20choices%5Bi%5D,%20choices,%20i,%20res%29%0A%20%20%20%20%20%20%20%20%23%20%E5%9B%9E%E9%80%80%EF%BC%9A%E6%92%A4%E9%94%80%E9%80%89%E6%8B%A9%EF%BC%8C%E6%81%A2%E5%A4%8D%E5%88%B0%E4%B9%8B%E5%89%8D%E7%9A%84%E7%8A%B6%E6%80%81%0A%20%20%20%20%20%20%20%20state.pop%28%29%0A%0A%0Adef%20subset_sum_i%28nums%3A%20list%5Bint%5D,%20target%3A%20int%29%20-%3E%20list%5Blist%5Bint%5D%5D%3A%0A%20%20%20%20%22%22%22%E6%B1%82%E8%A7%A3%E5%AD%90%E9%9B%86%E5%92%8C%20I%22%22%22%0A%20%20%20%20state%20%3D%20%5B%5D%20%20%23%20%E7%8A%B6%E6%80%81%EF%BC%88%E5%AD%90%E9%9B%86%EF%BC%89%0A%20%20%20%20nums.sort%28%29%20%20%23%20%E5%AF%B9%20nums%20%E8%BF%9B%E8%A1%8C%E6%8E%92%E5%BA%8F%0A%20%20%20%20start%20%3D%200%20%20%23%20%E9%81%8D%E5%8E%86%E8%B5%B7%E5%A7%8B%E7%82%B9%0A%20%20%20%20res%20%3D%20%5B%5D%20%20%23%20%E7%BB%93%E6%9E%9C%E5%88%97%E8%A1%A8%EF%BC%88%E5%AD%90%E9%9B%86%E5%88%97%E8%A1%A8%EF%BC%89%0A%20%20%20%20backtrack%28state,%20target,%20nums,%20start,%20res%29%0A%20%20%20%20return%20res%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20nums%20%3D%20%5B3,%204,%205%5D%0A%20%20%20%20target%20%3D%209%0A%20%20%20%20res%20%3D%20subset_sum_i%28nums,%20target%29%0A%0A%20%20%20%20print%28f%22%E8%BE%93%E5%85%A5%E6%95%B0%E7%BB%84%20nums%20%3D%20%7Bnums%7D,%20target%20%3D%20%7Btarget%7D%22%29%0A%20%20%20%20print%28f%22%E6%89%80%E6%9C%89%E5%92%8C%E7%AD%89%E4%BA%8E%20%7Btarget%7D%20%E7%9A%84%E5%AD%90%E9%9B%86%20res%20%3D%20%7Bres%7D%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=16&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
@@ -1689,14 +1673,6 @@ comments: true
end
```
=== "Zig"
```zig title="subset_sum_ii.zig"
[class]{}-[func]{backtrack}
[class]{}-[func]{subsetSumII}
```
??? pythontutor "可视化运行"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20backtrack%28%0A%20%20%20%20state%3A%20list%5Bint%5D,%20target%3A%20int,%20choices%3A%20list%5Bint%5D,%20start%3A%20int,%20res%3A%20list%5Blist%5Bint%5D%5D%0A%29%3A%0A%20%20%20%20%22%22%22%E5%9B%9E%E6%BA%AF%E7%AE%97%E6%B3%95%EF%BC%9A%E5%AD%90%E9%9B%86%E5%92%8C%20II%22%22%22%0A%20%20%20%20%23%20%E5%AD%90%E9%9B%86%E5%92%8C%E7%AD%89%E4%BA%8E%20target%20%E6%97%B6%EF%BC%8C%E8%AE%B0%E5%BD%95%E8%A7%A3%0A%20%20%20%20if%20target%20%3D%3D%200%3A%0A%20%20%20%20%20%20%20%20res.append%28list%28state%29%29%0A%20%20%20%20%20%20%20%20return%0A%20%20%20%20%23%20%E9%81%8D%E5%8E%86%E6%89%80%E6%9C%89%E9%80%89%E6%8B%A9%0A%20%20%20%20%23%20%E5%89%AA%E6%9E%9D%E4%BA%8C%EF%BC%9A%E4%BB%8E%20start%20%E5%BC%80%E5%A7%8B%E9%81%8D%E5%8E%86%EF%BC%8C%E9%81%BF%E5%85%8D%E7%94%9F%E6%88%90%E9%87%8D%E5%A4%8D%E5%AD%90%E9%9B%86%0A%20%20%20%20%23%20%E5%89%AA%E6%9E%9D%E4%B8%89%EF%BC%9A%E4%BB%8E%20start%20%E5%BC%80%E5%A7%8B%E9%81%8D%E5%8E%86%EF%BC%8C%E9%81%BF%E5%85%8D%E9%87%8D%E5%A4%8D%E9%80%89%E6%8B%A9%E5%90%8C%E4%B8%80%E5%85%83%E7%B4%A0%0A%20%20%20%20for%20i%20in%20range%28start,%20len%28choices%29%29%3A%0A%20%20%20%20%20%20%20%20%23%20%E5%89%AA%E6%9E%9D%E4%B8%80%EF%BC%9A%E8%8B%A5%E5%AD%90%E9%9B%86%E5%92%8C%E8%B6%85%E8%BF%87%20target%20%EF%BC%8C%E5%88%99%E7%9B%B4%E6%8E%A5%E7%BB%93%E6%9D%9F%E5%BE%AA%E7%8E%AF%0A%20%20%20%20%20%20%20%20%23%20%E8%BF%99%E6%98%AF%E5%9B%A0%E4%B8%BA%E6%95%B0%E7%BB%84%E5%B7%B2%E6%8E%92%E5%BA%8F%EF%BC%8C%E5%90%8E%E8%BE%B9%E5%85%83%E7%B4%A0%E6%9B%B4%E5%A4%A7%EF%BC%8C%E5%AD%90%E9%9B%86%E5%92%8C%E4%B8%80%E5%AE%9A%E8%B6%85%E8%BF%87%20target%0A%20%20%20%20%20%20%20%20if%20target%20-%20choices%5Bi%5D%20%3C%200%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20break%0A%20%20%20%20%20%20%20%20%23%20%E5%89%AA%E6%9E%9D%E5%9B%9B%EF%BC%9A%E5%A6%82%E6%9E%9C%E8%AF%A5%E5%85%83%E7%B4%A0%E4%B8%8E%E5%B7%A6%E8%BE%B9%E5%85%83%E7%B4%A0%E7%9B%B8%E7%AD%89%EF%BC%8C%E8%AF%B4%E6%98%8E%E8%AF%A5%E6%90%9C%E7%B4%A2%E5%88%86%E6%94%AF%E9%87%8D%E5%A4%8D%EF%BC%8C%E7%9B%B4%E6%8E%A5%E8%B7%B3%E8%BF%87%0A%20%20%20%20%20%20%20%20if%20i%20%3E%20start%20and%20choices%5Bi%5D%20%3D%3D%20choices%5Bi%20-%201%5D%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20continue%0A%20%20%20%20%20%20%20%20%23%20%E5%B0%9D%E8%AF%95%EF%BC%9A%E5%81%9A%E5%87%BA%E9%80%89%E6%8B%A9%EF%BC%8C%E6%9B%B4%E6%96%B0%20target,%20start%0A%20%20%20%20%20%20%20%20state.append%28choices%5Bi%5D%29%0A%20%20%20%20%20%20%20%20%23%20%E8%BF%9B%E8%A1%8C%E4%B8%8B%E4%B8%80%E8%BD%AE%E9%80%89%E6%8B%A9%0A%20%20%20%20%20%20%20%20backtrack%28state,%20target%20-%20choices%5Bi%5D,%20choices,%20i%20%2B%201,%20res%29%0A%20%20%20%20%20%20%20%20%23%20%E5%9B%9E%E9%80%80%EF%BC%9A%E6%92%A4%E9%94%80%E9%80%89%E6%8B%A9%EF%BC%8C%E6%81%A2%E5%A4%8D%E5%88%B0%E4%B9%8B%E5%89%8D%E7%9A%84%E7%8A%B6%E6%80%81%0A%20%20%20%20%20%20%20%20state.pop%28%29%0A%0A%0Adef%20subset_sum_ii%28nums%3A%20list%5Bint%5D,%20target%3A%20int%29%20-%3E%20list%5Blist%5Bint%5D%5D%3A%0A%20%20%20%20%22%22%22%E6%B1%82%E8%A7%A3%E5%AD%90%E9%9B%86%E5%92%8C%20II%22%22%22%0A%20%20%20%20state%20%3D%20%5B%5D%20%20%23%20%E7%8A%B6%E6%80%81%EF%BC%88%E5%AD%90%E9%9B%86%EF%BC%89%0A%20%20%20%20nums.sort%28%29%20%20%23%20%E5%AF%B9%20nums%20%E8%BF%9B%E8%A1%8C%E6%8E%92%E5%BA%8F%0A%20%20%20%20start%20%3D%200%20%20%23%20%E9%81%8D%E5%8E%86%E8%B5%B7%E5%A7%8B%E7%82%B9%0A%20%20%20%20res%20%3D%20%5B%5D%20%20%23%20%E7%BB%93%E6%9E%9C%E5%88%97%E8%A1%A8%EF%BC%88%E5%AD%90%E9%9B%86%E5%88%97%E8%A1%A8%EF%BC%89%0A%20%20%20%20backtrack%28state,%20target,%20nums,%20start,%20res%29%0A%20%20%20%20return%20res%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20nums%20%3D%20%5B4,%204,%205%5D%0A%20%20%20%20target%20%3D%209%0A%20%20%20%20res%20%3D%20subset_sum_ii%28nums,%20target%29%0A%0A%20%20%20%20print%28f%22%E8%BE%93%E5%85%A5%E6%95%B0%E7%BB%84%20nums%20%3D%20%7Bnums%7D,%20target%20%3D%20%7Btarget%7D%22%29%0A%20%20%20%20print%28f%22%E6%89%80%E6%9C%89%E5%92%8C%E7%AD%89%E4%BA%8E%20%7Btarget%7D%20%E7%9A%84%E5%AD%90%E9%9B%86%20res%20%3D%20%7Bres%7D%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=16&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
docs/chapter_computational_complexity/iteration_and_recursion.md
-139
View File
@@ -198,20 +198,6 @@ comments: true
end
```
=== "Zig"
```zig title="iteration.zig"
// for 循环
fn forLoop(n: usize) i32 {
var res: i32 = 0;
// 循环求和 1, 2, ..., n-1, n
for (1..n + 1) |i| {
res += @intCast(i);
}
return res;
}
```
??? pythontutor "可视化运行"
<div style="height: 423px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20for_loop%28n%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22for%20%E5%BE%AA%E7%8E%AF%22%22%22%0A%20%20%20%20res%20%3D%200%0A%20%20%20%20%23%20%E5%BE%AA%E7%8E%AF%E6%B1%82%E5%92%8C%201,%202,%20...,%20n-1,%20n%0A%20%20%20%20for%20i%20in%20range%281,%20n%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20res%20%2B%3D%20i%0A%20%20%20%20return%20res%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%205%0A%20%20%20%20res%20%3D%20for_loop%28n%29%0A%20%20%20%20print%28f%22%5Cnfor%20%E5%BE%AA%E7%8E%AF%E7%9A%84%E6%B1%82%E5%92%8C%E7%BB%93%E6%9E%9C%20res%20%3D%20%7Bres%7D%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&"> </iframe></div>
@@ -442,21 +428,6 @@ comments: true
end
```
=== "Zig"
```zig title="iteration.zig"
// while 循环
fn whileLoop(n: i32) i32 {
var res: i32 = 0;
var i: i32 = 1; // 初始化条件变量
// 循环求和 1, 2, ..., n-1, n
while (i <= n) : (i += 1) {
res += @intCast(i);
}
return res;
}
```
??? pythontutor "可视化运行"
<div style="height: 459px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20while_loop%28n%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22while%20%E5%BE%AA%E7%8E%AF%22%22%22%0A%20%20%20%20res%20%3D%200%0A%20%20%20%20i%20%3D%201%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%E6%9D%A1%E4%BB%B6%E5%8F%98%E9%87%8F%0A%20%20%20%20%23%20%E5%BE%AA%E7%8E%AF%E6%B1%82%E5%92%8C%201,%202,%20...,%20n-1,%20n%0A%20%20%20%20while%20i%20%3C%3D%20n%3A%0A%20%20%20%20%20%20%20%20res%20%2B%3D%20i%0A%20%20%20%20%20%20%20%20i%20%2B%3D%201%20%20%23%20%E6%9B%B4%E6%96%B0%E6%9D%A1%E4%BB%B6%E5%8F%98%E9%87%8F%0A%20%20%20%20return%20res%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%205%0A%20%20%20%20res%20%3D%20while_loop%28n%29%0A%20%20%20%20print%28f%22%5Cnwhile%20%E5%BE%AA%E7%8E%AF%E7%9A%84%E6%B1%82%E5%92%8C%E7%BB%93%E6%9E%9C%20res%20%3D%20%7Bres%7D%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
@@ -702,25 +673,6 @@ comments: true
end
```
=== "Zig"
```zig title="iteration.zig"
// while 循环（两次更新）
fn whileLoopII(n: i32) i32 {
var res: i32 = 0;
var i: i32 = 1; // 初始化条件变量
// 循环求和 1, 4, 10, ...
while (i <= n) : ({
// 更新条件变量
i += 1;
i *= 2;
}) {
res += @intCast(i);
}
return res;
}
```
??? pythontutor "可视化运行"
<div style="height: 495px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20while_loop_ii%28n%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22while%20%E5%BE%AA%E7%8E%AF%EF%BC%88%E4%B8%A4%E6%AC%A1%E6%9B%B4%E6%96%B0%EF%BC%89%22%22%22%0A%20%20%20%20res%20%3D%200%0A%20%20%20%20i%20%3D%201%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%E6%9D%A1%E4%BB%B6%E5%8F%98%E9%87%8F%0A%20%20%20%20%23%20%E5%BE%AA%E7%8E%AF%E6%B1%82%E5%92%8C%201,%204,%2010,%20...%0A%20%20%20%20while%20i%20%3C%3D%20n%3A%0A%20%20%20%20%20%20%20%20res%20%2B%3D%20i%0A%20%20%20%20%20%20%20%20%23%20%E6%9B%B4%E6%96%B0%E6%9D%A1%E4%BB%B6%E5%8F%98%E9%87%8F%0A%20%20%20%20%20%20%20%20i%20%2B%3D%201%0A%20%20%20%20%20%20%20%20i%20*%3D%202%0A%20%20%20%20return%20res%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%205%0A%20%20%20%20res%20%3D%20while_loop_ii%28n%29%0A%20%20%20%20print%28f%22%5Cnwhile%20%E5%BE%AA%E7%8E%AF%EF%BC%88%E4%B8%A4%E6%AC%A1%E6%9B%B4%E6%96%B0%EF%BC%89%E6%B1%82%E5%92%8C%E7%BB%93%E6%9E%9C%20res%20%3D%20%7Bres%7D%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
@@ -956,26 +908,6 @@ comments: true
end
```
=== "Zig"
```zig title="iteration.zig"
// 双层 for 循环
fn nestedForLoop(allocator: Allocator, n: usize) ![]const u8 {
var res = std.ArrayList(u8).init(allocator);
defer res.deinit();
var buffer: [20]u8 = undefined;
// 循环 i = 1, 2, ..., n-1, n
for (1..n + 1) |i| {
// 循环 j = 1, 2, ..., n-1, n
for (1..n + 1) |j| {
const str = try std.fmt.bufPrint(&buffer, "({d}, {d}), ", .{ i, j });
try res.appendSlice(str);
}
}
return res.toOwnedSlice();
}
```
??? pythontutor "可视化运行"
<div style="height: 459px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20nested_for_loop%28n%3A%20int%29%20-%3E%20str%3A%0A%20%20%20%20%22%22%22%E5%8F%8C%E5%B1%82%20for%20%E5%BE%AA%E7%8E%AF%22%22%22%0A%20%20%20%20res%20%3D%20%22%22%0A%20%20%20%20%23%20%E5%BE%AA%E7%8E%AF%20i%20%3D%201,%202,%20...,%20n-1,%20n%0A%20%20%20%20for%20i%20in%20range%281,%20n%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20%23%20%E5%BE%AA%E7%8E%AF%20j%20%3D%201,%202,%20...,%20n-1,%20n%0A%20%20%20%20%20%20%20%20for%20j%20in%20range%281,%20n%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20res%20%2B%3D%20f%22%28%7Bi%7D,%20%7Bj%7D%29,%20%22%0A%20%20%20%20return%20res%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%205%0A%20%20%20%20res%20%3D%20nested_for_loop%28n%29%0A%20%20%20%20print%28f%22%5Cn%E5%8F%8C%E5%B1%82%20for%20%E5%BE%AA%E7%8E%AF%E7%9A%84%E9%81%8D%E5%8E%86%E7%BB%93%E6%9E%9C%20%7Bres%7D%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
@@ -1199,22 +1131,6 @@ comments: true
end
```
=== "Zig"
```zig title="recursion.zig"
// 递归函数
fn recur(n: i32) i32 {
// 终止条件
if (n == 1) {
return 1;
}
// 递：递归调用
const res = recur(n - 1);
// 归：返回结果
return n + res;
}
```
??? pythontutor "可视化运行"
<div style="height: 459px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20recur%28n%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E9%80%92%E5%BD%92%22%22%22%0A%20%20%20%20%23%20%E7%BB%88%E6%AD%A2%E6%9D%A1%E4%BB%B6%0A%20%20%20%20if%20n%20%3D%3D%201%3A%0A%20%20%20%20%20%20%20%20return%201%0A%20%20%20%20%23%20%E9%80%92%EF%BC%9A%E9%80%92%E5%BD%92%E8%B0%83%E7%94%A8%0A%20%20%20%20res%20%3D%20recur%28n%20-%201%29%0A%20%20%20%20%23%20%E5%BD%92%EF%BC%9A%E8%BF%94%E5%9B%9E%E7%BB%93%E6%9E%9C%0A%20%20%20%20return%20n%20%2B%20res%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%205%0A%20%20%20%20res%20%3D%20recur%28n%29%0A%20%20%20%20print%28f%22%5Cn%E9%80%92%E5%BD%92%E5%87%BD%E6%95%B0%E7%9A%84%E6%B1%82%E5%92%8C%E7%BB%93%E6%9E%9C%20res%20%3D%20%7Bres%7D%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
@@ -1428,20 +1344,6 @@ comments: true
end
```
=== "Zig"
```zig title="recursion.zig"
// 尾递归函数
fn tailRecur(n: i32, res: i32) i32 {
// 终止条件
if (n == 0) {
return res;
}
// 尾递归调用
return tailRecur(n - 1, res + n);
}
```
??? pythontutor "可视化运行"
<div style="height: 423px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20tail_recur%28n,%20res%29%3A%0A%20%20%20%20%22%22%22%E5%B0%BE%E9%80%92%E5%BD%92%22%22%22%0A%20%20%20%20%23%20%E7%BB%88%E6%AD%A2%E6%9D%A1%E4%BB%B6%0A%20%20%20%20if%20n%20%3D%3D%200%3A%0A%20%20%20%20%20%20%20%20return%20res%0A%20%20%20%20%23%20%E5%B0%BE%E9%80%92%E5%BD%92%E8%B0%83%E7%94%A8%0A%20%20%20%20return%20tail_recur%28n%20-%201,%20res%20%2B%20n%29%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%205%0A%20%20%20%20res%20%3D%20tail_recur%28n,%200%29%0A%20%20%20%20print%28f%22%5Cn%E5%B0%BE%E9%80%92%E5%BD%92%E5%87%BD%E6%95%B0%E7%9A%84%E6%B1%82%E5%92%8C%E7%BB%93%E6%9E%9C%20res%20%3D%20%7Bres%7D%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
@@ -1668,22 +1570,6 @@ comments: true
end
```
=== "Zig"
```zig title="recursion.zig"
// 斐波那契数列
fn fib(n: i32) i32 {
// 终止条件 f(1) = 0, f(2) = 1
if (n == 1 or n == 2) {
return n - 1;
}
// 递归调用 f(n) = f(n-1) + f(n-2)
const res: i32 = fib(n - 1) + fib(n - 2);
// 返回结果 f(n)
return res;
}
```
??? pythontutor "可视化运行"
<div style="height: 459px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20fib%28n%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E6%96%90%E6%B3%A2%E9%82%A3%E5%A5%91%E6%95%B0%E5%88%97%EF%BC%9A%E9%80%92%E5%BD%92%22%22%22%0A%20%20%20%20%23%20%E7%BB%88%E6%AD%A2%E6%9D%A1%E4%BB%B6%20f%281%29%20%3D%200,%20f%282%29%20%3D%201%0A%20%20%20%20if%20n%20%3D%3D%201%20or%20n%20%3D%3D%202%3A%0A%20%20%20%20%20%20%20%20return%20n%20-%201%0A%20%20%20%20%23%20%E9%80%92%E5%BD%92%E8%B0%83%E7%94%A8%20f%28n%29%20%3D%20f%28n-1%29%20%2B%20f%28n-2%29%0A%20%20%20%20res%20%3D%20fib%28n%20-%201%29%20%2B%20fib%28n%20-%202%29%0A%20%20%20%20%23%20%E8%BF%94%E5%9B%9E%E7%BB%93%E6%9E%9C%20f%28n%29%0A%20%20%20%20return%20res%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%205%0A%20%20%20%20res%20%3D%20fib%28n%29%0A%20%20%20%20print%28f%22%5Cn%E6%96%90%E6%B3%A2%E9%82%A3%E5%A5%91%E6%95%B0%E5%88%97%E7%9A%84%E7%AC%AC%20%7Bn%7D%20%E9%A1%B9%E4%B8%BA%20%7Bres%7D%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
@@ -2029,31 +1915,6 @@ comments: true
end
```
=== "Zig"
```zig title="recursion.zig"
// 使用迭代模拟递归
fn forLoopRecur(comptime n: i32) i32 {
// 使用一个显式的栈来模拟系统调用栈
var stack: [n]i32 = undefined;
var res: i32 = 0;
// 递：递归调用
var i: usize = n;
while (i > 0) {
stack[i - 1] = @intCast(i);
i -= 1;
}
// 归：返回结果
var index: usize = n;
while (index > 0) {
index -= 1;
res += stack[index];
}
// res = 1+2+3+...+n
return res;
}
```
??? pythontutor "可视化运行"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20for_loop_recur%28n%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E4%BD%BF%E7%94%A8%E8%BF%AD%E4%BB%A3%E6%A8%A1%E6%8B%9F%E9%80%92%E5%BD%92%22%22%22%0A%20%20%20%20%23%20%E4%BD%BF%E7%94%A8%E4%B8%80%E4%B8%AA%E6%98%BE%E5%BC%8F%E7%9A%84%E6%A0%88%E6%9D%A5%E6%A8%A1%E6%8B%9F%E7%B3%BB%E7%BB%9F%E8%B0%83%E7%94%A8%E6%A0%88%0A%20%20%20%20stack%20%3D%20%5B%5D%0A%20%20%20%20res%20%3D%200%0A%20%20%20%20%23%20%E9%80%92%EF%BC%9A%E9%80%92%E5%BD%92%E8%B0%83%E7%94%A8%0A%20%20%20%20for%20i%20in%20range%28n,%200,%20-1%29%3A%0A%20%20%20%20%20%20%20%20%23%20%E9%80%9A%E8%BF%87%E2%80%9C%E5%85%A5%E6%A0%88%E6%93%8D%E4%BD%9C%E2%80%9D%E6%A8%A1%E6%8B%9F%E2%80%9C%E9%80%92%E2%80%9D%0A%20%20%20%20%20%20%20%20stack.append%28i%29%0A%20%20%20%20%23%20%E5%BD%92%EF%BC%9A%E8%BF%94%E5%9B%9E%E7%BB%93%E6%9E%9C%0A%20%20%20%20while%20stack%3A%0A%20%20%20%20%20%20%20%20%23%20%E9%80%9A%E8%BF%87%E2%80%9C%E5%87%BA%E6%A0%88%E6%93%8D%E4%BD%9C%E2%80%9D%E6%A8%A1%E6%8B%9F%E2%80%9C%E5%BD%92%E2%80%9D%0A%20%20%20%20%20%20%20%20res%20%2B%3D%20stack.pop%28%29%0A%20%20%20%20%23%20res%20%3D%201%2B2%2B3%2B...%2Bn%0A%20%20%20%20return%20res%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%205%0A%20%20%20%20res%20%3D%20for_loop_recur%28n%29%0A%20%20%20%20print%28f%22%5Cn%E4%BD%BF%E7%94%A8%E8%BF%AD%E4%BB%A3%E6%A8%A1%E6%8B%9F%E9%80%92%E5%BD%92%E6%B1%82%E5%92%8C%E7%BB%93%E6%9E%9C%20res%20%3D%20%7Bres%7D%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
docs/chapter_computational_complexity/space_complexity.md
-137
View File
@@ -367,12 +367,6 @@ comments: true
end
```
=== "Zig"
```zig title=""
```
## 2.4.2 &nbsp; 推算方法
空间复杂度的推算方法与时间复杂度大致相同，只需将统计对象从“操作数量”转为“使用空间大小”。
@@ -535,12 +529,6 @@ comments: true
end
```
=== "Zig"
```zig title=""
```
**在递归函数中，需要注意统计栈帧空间**。观察以下代码：
=== "Python"
@@ -813,12 +801,6 @@ comments: true
end
```
=== "Zig"
```zig title=""
```
函数 `loop()` 和 `recur()` 的时间复杂度都为 $O(n)$ ，但空间复杂度不同。
- 函数 `loop()` 在循环中调用了 $n$ 次 `function()` ，每轮中的 `function()` 都返回并释放了栈帧空间，因此空间复杂度仍为 $O(1)$ 。
@@ -1195,40 +1177,6 @@ $$
end
```
=== "Zig"
```zig title="space_complexity.zig"
// 函数
fn function() i32 {
// 执行某些操作
return 0;
}
// 常数阶
fn constant(n: i32) void {
// 常量、变量、对象占用 O(1) 空间
const a: i32 = 0;
const b: i32 = 0;
const nums = [_]i32{0} ** 10000;
const node = ListNode(i32){ .val = 0 };
var i: i32 = 0;
// 循环中的变量占用 O(1) 空间
while (i < n) : (i += 1) {
const c: i32 = 0;
_ = c;
}
// 循环中的函数占用 O(1) 空间
i = 0;
while (i < n) : (i += 1) {
_ = function();
}
_ = a;
_ = b;
_ = nums;
_ = node;
}
```
??? pythontutor "可视化运行"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=class%20ListNode%3A%0A%20%20%20%20%22%22%22%E9%93%BE%E8%A1%A8%E8%8A%82%E7%82%B9%E7%B1%BB%22%22%22%0A%20%20%20%20def%20__init__%28self,%20val%3A%20int%29%3A%0A%20%20%20%20%20%20%20%20self.val%3A%20int%20%3D%20val%20%20%23%20%E8%8A%82%E7%82%B9%E5%80%BC%0A%20%20%20%20%20%20%20%20self.next%3A%20ListNode%20%7C%20None%20%3D%20None%20%20%23%20%E5%90%8E%E7%BB%A7%E8%8A%82%E7%82%B9%E5%BC%95%E7%94%A8%0A%0Adef%20function%28%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E5%87%BD%E6%95%B0%22%22%22%0A%20%20%20%20%23%20%E6%89%A7%E8%A1%8C%E6%9F%90%E4%BA%9B%E6%93%8D%E4%BD%9C%0A%20%20%20%20return%200%0A%0Adef%20constant%28n%3A%20int%29%3A%0A%20%20%20%20%22%22%22%E5%B8%B8%E6%95%B0%E9%98%B6%22%22%22%0A%20%20%20%20%23%20%E5%B8%B8%E9%87%8F%E3%80%81%E5%8F%98%E9%87%8F%E3%80%81%E5%AF%B9%E8%B1%A1%E5%8D%A0%E7%94%A8%20O%281%29%20%E7%A9%BA%E9%97%B4%0A%20%20%20%20a%20%3D%200%0A%20%20%20%20nums%20%3D%20%5B0%5D%20*%2010%0A%20%20%20%20node%20%3D%20ListNode%280%29%0A%20%20%20%20%23%20%E5%BE%AA%E7%8E%AF%E4%B8%AD%E7%9A%84%E5%8F%98%E9%87%8F%E5%8D%A0%E7%94%A8%20O%281%29%20%E7%A9%BA%E9%97%B4%0A%20%20%20%20for%20_%20in%20range%28n%29%3A%0A%20%20%20%20%20%20%20%20c%20%3D%200%0A%20%20%20%20%23%20%E5%BE%AA%E7%8E%AF%E4%B8%AD%E7%9A%84%E5%87%BD%E6%95%B0%E5%8D%A0%E7%94%A8%20O%281%29%20%E7%A9%BA%E9%97%B4%0A%20%20%20%20for%20_%20in%20range%28n%29%3A%0A%20%20%20%20%20%20%20%20function%28%29%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%205%0A%20%20%20%20print%28%22%E8%BE%93%E5%85%A5%E6%95%B0%E6%8D%AE%E5%A4%A7%E5%B0%8F%20n%20%3D%22,%20n%29%0A%0A%20%20%20%20%23%20%E5%B8%B8%E6%95%B0%E9%98%B6%0A%20%20%20%20constant%28n%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=6&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
@@ -1507,33 +1455,6 @@ $$
end
```
=== "Zig"
```zig title="space_complexity.zig"
// 线性阶
fn linear(comptime n: i32) !void {
// 长度为 n 的数组占用 O(n) 空间
const nums = [_]i32{0} ** n;
// 长度为 n 的列表占用 O(n) 空间
var nodes = std.ArrayList(i32).init(std.heap.page_allocator);
defer nodes.deinit();
var i: i32 = 0;
while (i < n) : (i += 1) {
try nodes.append(i);
}
// 长度为 n 的哈希表占用 O(n) 空间
var map = std.AutoArrayHashMap(i32, []const u8).init(std.heap.page_allocator);
defer map.deinit();
var j: i32 = 0;
while (j < n) : (j += 1) {
const string = try std.fmt.allocPrint(std.heap.page_allocator, "{d}", .{j});
defer std.heap.page_allocator.free(string);
try map.put(i, string);
}
_ = nums;
}
```
??? pythontutor "可视化运行"
<div style="height: 477px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20linear%28n%3A%20int%29%3A%0A%20%20%20%20%22%22%22%E7%BA%BF%E6%80%A7%E9%98%B6%22%22%22%0A%20%20%20%20%23%20%E9%95%BF%E5%BA%A6%E4%B8%BA%20n%20%E7%9A%84%E5%88%97%E8%A1%A8%E5%8D%A0%E7%94%A8%20O%28n%29%20%E7%A9%BA%E9%97%B4%0A%20%20%20%20nums%20%3D%20%5B0%5D%20*%20n%0A%20%20%20%20%23%20%E9%95%BF%E5%BA%A6%E4%B8%BA%20n%20%E7%9A%84%E5%93%88%E5%B8%8C%E8%A1%A8%E5%8D%A0%E7%94%A8%20O%28n%29%20%E7%A9%BA%E9%97%B4%0A%20%20%20%20hmap%20%3D%20dict%5Bint,%20str%5D%28%29%0A%20%20%20%20for%20i%20in%20range%28n%29%3A%0A%20%20%20%20%20%20%20%20hmap%5Bi%5D%20%3D%20str%28i%29%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%205%0A%20%20%20%20print%28%22%E8%BE%93%E5%85%A5%E6%95%B0%E6%8D%AE%E5%A4%A7%E5%B0%8F%20n%20%3D%22,%20n%29%0A%0A%20%20%20%20%23%20%E7%BA%BF%E6%80%A7%E9%98%B6%0A%20%20%20%20linear%28n%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=20&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
@@ -1694,17 +1615,6 @@ $$
end
```
=== "Zig"
```zig title="space_complexity.zig"
// 线性阶（递归实现）
fn linearRecur(comptime n: i32) void {
std.debug.print("递归 n = {}\n", .{n});
if (n == 1) return;
linearRecur(n - 1);
}
```
??? pythontutor "可视化运行"
<div style="height: 441px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20linear_recur%28n%3A%20int%29%3A%0A%20%20%20%20%22%22%22%E7%BA%BF%E6%80%A7%E9%98%B6%EF%BC%88%E9%80%92%E5%BD%92%E5%AE%9E%E7%8E%B0%EF%BC%89%22%22%22%0A%20%20%20%20print%28%22%E9%80%92%E5%BD%92%20n%20%3D%22,%20n%29%0A%20%20%20%20if%20n%20%3D%3D%201%3A%0A%20%20%20%20%20%20%20%20return%0A%20%20%20%20linear_recur%28n%20-%201%29%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%205%0A%20%20%20%20print%28%22%E8%BE%93%E5%85%A5%E6%95%B0%E6%8D%AE%E5%A4%A7%E5%B0%8F%20n%20%3D%22,%20n%29%0A%0A%20%20%20%20%23%20%E7%BA%BF%E6%80%A7%E9%98%B6%0A%20%20%20%20linear_recur%28n%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=25&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
@@ -1938,27 +1848,6 @@ $$
end
```
=== "Zig"
```zig title="space_complexity.zig"
// 平方阶
fn quadratic(n: i32) !void {
// 二维列表占用 O(n^2) 空间
var nodes = std.ArrayList(std.ArrayList(i32)).init(std.heap.page_allocator);
defer nodes.deinit();
var i: i32 = 0;
while (i < n) : (i += 1) {
var tmp = std.ArrayList(i32).init(std.heap.page_allocator);
defer tmp.deinit();
var j: i32 = 0;
while (j < n) : (j += 1) {
try tmp.append(0);
}
try nodes.append(tmp);
}
}
```
??? pythontutor "可视化运行"
<div style="height: 405px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20quadratic%28n%3A%20int%29%3A%0A%20%20%20%20%22%22%22%E5%B9%B3%E6%96%B9%E9%98%B6%22%22%22%0A%20%20%20%20%23%20%E4%BA%8C%E7%BB%B4%E5%88%97%E8%A1%A8%E5%8D%A0%E7%94%A8%20O%28n%5E2%29%20%E7%A9%BA%E9%97%B4%0A%20%20%20%20num_matrix%20%3D%20%5B%5B0%5D%20*%20n%20for%20_%20in%20range%28n%29%5D%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%205%0A%20%20%20%20print%28%22%E8%BE%93%E5%85%A5%E6%95%B0%E6%8D%AE%E5%A4%A7%E5%B0%8F%20n%20%3D%22,%20n%29%0A%0A%20%20%20%20%23%20%E5%B9%B3%E6%96%B9%E9%98%B6%0A%20%20%20%20quadratic%28n%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=16&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
@@ -2140,18 +2029,6 @@ $$
end
```
=== "Zig"
```zig title="space_complexity.zig"
// 平方阶（递归实现）
fn quadraticRecur(comptime n: i32) i32 {
if (n <= 0) return 0;
const nums = [_]i32{0} ** n;
std.debug.print("递归 n = {} 中的 nums 长度 = {}\n", .{ n, nums.len });
return quadraticRecur(n - 1);
}
```
??? pythontutor "可视化运行"
<div style="height: 459px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20quadratic_recur%28n%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E5%B9%B3%E6%96%B9%E9%98%B6%EF%BC%88%E9%80%92%E5%BD%92%E5%AE%9E%E7%8E%B0%EF%BC%89%22%22%22%0A%20%20%20%20if%20n%20%3C%3D%200%3A%0A%20%20%20%20%20%20%20%20return%200%0A%20%20%20%20%23%20%E6%95%B0%E7%BB%84%20nums%20%E9%95%BF%E5%BA%A6%E4%B8%BA%20n,%20n-1,%20...,%202,%201%0A%20%20%20%20nums%20%3D%20%5B0%5D%20*%20n%0A%20%20%20%20return%20quadratic_recur%28n%20-%201%29%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%205%0A%20%20%20%20print%28%22%E8%BE%93%E5%85%A5%E6%95%B0%E6%8D%AE%E5%A4%A7%E5%B0%8F%20n%20%3D%22,%20n%29%0A%0A%20%20%20%20%23%20%E5%B9%B3%E6%96%B9%E9%98%B6%0A%20%20%20%20quadratic_recur%28n%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=28&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
@@ -2346,20 +2223,6 @@ $$
end
```
=== "Zig"
```zig title="space_complexity.zig"
// 指数阶（建立满二叉树）
fn buildTree(allocator: std.mem.Allocator, n: i32) !?*TreeNode(i32) {
if (n == 0) return null;
const root = try allocator.create(TreeNode(i32));
root.init(0);
root.left = try buildTree(allocator, n - 1);
root.right = try buildTree(allocator, n - 1);
return root;
}
```
??? pythontutor "可视化运行"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=class%20TreeNode%3A%0A%20%20%20%20%22%22%22%E4%BA%8C%E5%8F%89%E6%A0%91%E8%8A%82%E7%82%B9%E7%B1%BB%22%22%22%0A%20%20%20%20def%20__init__%28self,%20val%3A%20int%20%3D%200%29%3A%0A%20%20%20%20%20%20%20%20self.val%3A%20int%20%3D%20val%20%20%23%20%E8%8A%82%E7%82%B9%E5%80%BC%0A%20%20%20%20%20%20%20%20self.left%3A%20TreeNode%20%7C%20None%20%3D%20None%20%20%23%20%E5%B7%A6%E5%AD%90%E8%8A%82%E7%82%B9%E5%BC%95%E7%94%A8%0A%20%20%20%20%20%20%20%20self.right%3A%20TreeNode%20%7C%20None%20%3D%20None%20%20%23%20%E5%8F%B3%E5%AD%90%E8%8A%82%E7%82%B9%E5%BC%95%E7%94%A8%0A%0Adef%20build_tree%28n%3A%20int%29%20-%3E%20TreeNode%20%7C%20None%3A%0A%20%20%20%20%22%22%22%E6%8C%87%E6%95%B0%E9%98%B6%EF%BC%88%E5%BB%BA%E7%AB%8B%E6%BB%A1%E4%BA%8C%E5%8F%89%E6%A0%91%EF%BC%89%22%22%22%0A%20%20%20%20if%20n%20%3D%3D%200%3A%0A%20%20%20%20%20%20%20%20return%20None%0A%20%20%20%20root%20%3D%20TreeNode%280%29%0A%20%20%20%20root.left%20%3D%20build_tree%28n%20-%201%29%0A%20%20%20%20root.right%20%3D%20build_tree%28n%20-%201%29%0A%20%20%20%20return%20root%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%205%0A%20%20%20%20print%28%22%E8%BE%93%E5%85%A5%E6%95%B0%E6%8D%AE%E5%A4%A7%E5%B0%8F%20n%20%3D%22,%20n%29%0A%0A%20%20%20%20%23%20%E6%8C%87%E6%95%B0%E9%98%B6%0A%20%20%20%20root%20%3D%20build_tree%28n%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=507&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
docs/chapter_computational_complexity/time_complexity.md
-273
View File
@@ -205,21 +205,6 @@ comments: true
end
```
=== "Zig"
```zig title=""
// 在某运行平台下
fn algorithm(n: usize) void {
var a: i32 = 2; // 1 ns
a += 1; // 1 ns
a *= 2; // 10 ns
// 循环 n 次
for (0..n) |_| { // 1 ns
std.debug.print("{}\n", .{0}); // 5 ns
}
}
```
根据以上方法，可以得到算法的运行时间为 $(6n + 12)$ ns ：
$$
@@ -503,29 +488,6 @@ $$
end
```
=== "Zig"
```zig title=""
// 算法 A 的时间复杂度：常数阶
fn algorithm_A(n: usize) void {
_ = n;
std.debug.print("{}\n", .{0});
}
// 算法 B 的时间复杂度：线性阶
fn algorithm_B(n: i32) void {
for (0..n) |_| {
std.debug.print("{}\n", .{0});
}
}
// 算法 C 的时间复杂度：常数阶
fn algorithm_C(n: i32) void {
_ = n;
for (0..1000000) |_| {
std.debug.print("{}\n", .{0});
}
}
```
图 2-7 展示了以上三个算法函数的时间复杂度。
- 算法 `A` 只有 $1$ 个打印操作，算法运行时间不随着 $n$ 增大而增长。我们称此算法的时间复杂度为“常数阶”。
@@ -727,20 +689,6 @@ $$
end
```
=== "Zig"
```zig title=""
fn algorithm(n: usize) void {
var a: i32 = 1; // +1
a += 1; // +1
a *= 2; // +1
// 循环 n 次
for (0..n) |_| { // +1（每轮都执行 i ++）
std.debug.print("{}\n", .{0}); // +1
}
}
```
设算法的操作数量是一个关于输入数据大小 $n$ 的函数，记为 $T(n)$ ，则以上函数的操作数量为：
$$
@@ -1020,27 +968,6 @@ $T(n)$ 是一次函数，说明其运行时间的增长趋势是线性的，因
end
```
=== "Zig"
```zig title=""
fn algorithm(n: usize) void {
var a: i32 = 1; // +0（技巧 1）
a = a + @as(i32, @intCast(n)); // +0（技巧 1）
// +n（技巧 2）
for(0..(5 * n + 1)) |_| {
std.debug.print("{}\n", .{0});
}
// +n*n（技巧 3）
for(0..(2 * n)) |_| {
for(0..(n + 1)) |_| {
std.debug.print("{}\n", .{0});
}
}
}
```
以下公式展示了使用上述技巧前后的统计结果，两者推算出的时间复杂度都为 $O(n^2)$ 。
$$
@@ -1266,22 +1193,6 @@ $$
end
```
=== "Zig"
```zig title="time_complexity.zig"
// 常数阶
fn constant(n: i32) i32 {
_ = n;
var count: i32 = 0;
const size: i32 = 100_000;
var i: i32 = 0;
while (i < size) : (i += 1) {
count += 1;
}
return count;
}
```
??? pythontutor "可视化运行"
<div style="height: 459px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20constant%28n%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E5%B8%B8%E6%95%B0%E9%98%B6%22%22%22%0A%20%20%20%20count%20%3D%200%0A%20%20%20%20size%20%3D%2010%0A%20%20%20%20for%20_%20in%20range%28size%29%3A%0A%20%20%20%20%20%20%20%20count%20%2B%3D%201%0A%20%20%20%20return%20count%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%208%0A%20%20%20%20print%28%22%E8%BE%93%E5%85%A5%E6%95%B0%E6%8D%AE%E5%A4%A7%E5%B0%8F%20n%20%3D%22,%20n%29%0A%0A%20%20%20%20count%20%3D%20constant%28n%29%0A%20%20%20%20print%28%22%E5%B8%B8%E6%95%B0%E9%98%B6%E7%9A%84%E6%93%8D%E4%BD%9C%E6%95%B0%E9%87%8F%20%3D%22,%20count%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
@@ -1448,20 +1359,6 @@ $$
end
```
=== "Zig"
```zig title="time_complexity.zig"
// 线性阶
fn linear(n: i32) i32 {
var count: i32 = 0;
var i: i32 = 0;
while (i < n) : (i += 1) {
count += 1;
}
return count;
}
```
??? pythontutor "可视化运行"
<div style="height: 441px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20linear%28n%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E7%BA%BF%E6%80%A7%E9%98%B6%22%22%22%0A%20%20%20%20count%20%3D%200%0A%20%20%20%20for%20_%20in%20range%28n%29%3A%0A%20%20%20%20%20%20%20%20count%20%2B%3D%201%0A%20%20%20%20return%20count%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%208%0A%20%20%20%20print%28%22%E8%BE%93%E5%85%A5%E6%95%B0%E6%8D%AE%E5%A4%A7%E5%B0%8F%20n%20%3D%22,%20n%29%0A%0A%20%20%20%20count%20%3D%20linear%28n%29%0A%20%20%20%20print%28%22%E7%BA%BF%E6%80%A7%E9%98%B6%E7%9A%84%E6%93%8D%E4%BD%9C%E6%95%B0%E9%87%8F%20%3D%22,%20count%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
@@ -1651,20 +1548,6 @@ $$
end
```
=== "Zig"
```zig title="time_complexity.zig"
// 线性阶（遍历数组）
fn arrayTraversal(nums: []i32) i32 {
var count: i32 = 0;
// 循环次数与数组长度成正比
for (nums) |_| {
count += 1;
}
return count;
}
```
??? pythontutor "可视化运行"
<div style="height: 459px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20array_traversal%28nums%3A%20list%5Bint%5D%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E7%BA%BF%E6%80%A7%E9%98%B6%EF%BC%88%E9%81%8D%E5%8E%86%E6%95%B0%E7%BB%84%EF%BC%89%22%22%22%0A%20%20%20%20count%20%3D%200%0A%20%20%20%20%23%20%E5%BE%AA%E7%8E%AF%E6%AC%A1%E6%95%B0%E4%B8%8E%E6%95%B0%E7%BB%84%E9%95%BF%E5%BA%A6%E6%88%90%E6%AD%A3%E6%AF%94%0A%20%20%20%20for%20num%20in%20nums%3A%0A%20%20%20%20%20%20%20%20count%20%2B%3D%201%0A%20%20%20%20return%20count%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%208%0A%20%20%20%20print%28%22%E8%BE%93%E5%85%A5%E6%95%B0%E6%8D%AE%E5%A4%A7%E5%B0%8F%20n%20%3D%22,%20n%29%0A%0A%20%20%20%20count%20%3D%20array_traversal%28%5B0%5D%20*%20n%29%0A%20%20%20%20print%28%22%E7%BA%BF%E6%80%A7%E9%98%B6%EF%BC%88%E9%81%8D%E5%8E%86%E6%95%B0%E7%BB%84%EF%BC%89%E7%9A%84%E6%93%8D%E4%BD%9C%E6%95%B0%E9%87%8F%20%3D%22,%20count%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
@@ -1883,24 +1766,6 @@ $$
end
```
=== "Zig"
```zig title="time_complexity.zig"
// 平方阶
fn quadratic(n: i32) i32 {
var count: i32 = 0;
var i: i32 = 0;
// 循环次数与数据大小 n 成平方关系
while (i < n) : (i += 1) {
var j: i32 = 0;
while (j < n) : (j += 1) {
count += 1;
}
}
return count;
}
```
??? pythontutor "可视化运行"
<div style="height: 477px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20quadratic%28n%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E5%B9%B3%E6%96%B9%E9%98%B6%22%22%22%0A%20%20%20%20count%20%3D%200%0A%20%20%20%20%23%20%E5%BE%AA%E7%8E%AF%E6%AC%A1%E6%95%B0%E4%B8%8E%E6%95%B0%E6%8D%AE%E5%A4%A7%E5%B0%8F%20n%20%E6%88%90%E5%B9%B3%E6%96%B9%E5%85%B3%E7%B3%BB%0A%20%20%20%20for%20i%20in%20range%28n%29%3A%0A%20%20%20%20%20%20%20%20for%20j%20in%20range%28n%29%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20count%20%2B%3D%201%0A%20%20%20%20return%20count%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%208%0A%20%20%20%20print%28%22%E8%BE%93%E5%85%A5%E6%95%B0%E6%8D%AE%E5%A4%A7%E5%B0%8F%20n%20%3D%22,%20n%29%0A%0A%20%20%20%20count%20%3D%20quadratic%28n%29%0A%20%20%20%20print%28%22%E5%B9%B3%E6%96%B9%E9%98%B6%E7%9A%84%E6%93%8D%E4%BD%9C%E6%95%B0%E9%87%8F%20%3D%22,%20count%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
@@ -2210,31 +2075,6 @@ $$
end
```
=== "Zig"
```zig title="time_complexity.zig"
// 平方阶（冒泡排序）
fn bubbleSort(nums: []i32) i32 {
var count: i32 = 0; // 计数器
// 外循环：未排序区间为 [0, i]
var i: i32 = @as(i32, @intCast(nums.len)) - 1;
while (i > 0) : (i -= 1) {
var j: usize = 0;
// 内循环：将未排序区间 [0, i] 中的最大元素交换至该区间的最右端
while (j < i) : (j += 1) {
if (nums[j] > nums[j + 1]) {
// 交换 nums[j] 与 nums[j + 1]
const tmp = nums[j];
nums[j] = nums[j + 1];
nums[j + 1] = tmp;
count += 3; // 元素交换包含 3 个单元操作
}
}
}
return count;
}
```
??? pythontutor "可视化运行"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20bubble_sort%28nums%3A%20list%5Bint%5D%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E5%B9%B3%E6%96%B9%E9%98%B6%EF%BC%88%E5%86%92%E6%B3%A1%E6%8E%92%E5%BA%8F%EF%BC%89%22%22%22%0A%20%20%20%20count%20%3D%200%20%20%23%20%E8%AE%A1%E6%95%B0%E5%99%A8%0A%20%20%20%20%23%20%E5%A4%96%E5%BE%AA%E7%8E%AF%EF%BC%9A%E6%9C%AA%E6%8E%92%E5%BA%8F%E5%8C%BA%E9%97%B4%E4%B8%BA%20%5B0,%20i%5D%0A%20%20%20%20for%20i%20in%20range%28len%28nums%29%20-%201,%200,%20-1%29%3A%0A%20%20%20%20%20%20%20%20%23%20%E5%86%85%E5%BE%AA%E7%8E%AF%EF%BC%9A%E5%B0%86%E6%9C%AA%E6%8E%92%E5%BA%8F%E5%8C%BA%E9%97%B4%20%5B0,%20i%5D%20%E4%B8%AD%E7%9A%84%E6%9C%80%E5%A4%A7%E5%85%83%E7%B4%A0%E4%BA%A4%E6%8D%A2%E8%87%B3%E8%AF%A5%E5%8C%BA%E9%97%B4%E7%9A%84%E6%9C%80%E5%8F%B3%E7%AB%AF%0A%20%20%20%20%20%20%20%20for%20j%20in%20range%28i%29%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20if%20nums%5Bj%5D%20%3E%20nums%5Bj%20%2B%201%5D%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%23%20%E4%BA%A4%E6%8D%A2%20nums%5Bj%5D%20%E4%B8%8E%20nums%5Bj%20%2B%201%5D%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20tmp%20%3D%20nums%5Bj%5D%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20nums%5Bj%5D%20%3D%20nums%5Bj%20%2B%201%5D%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20nums%5Bj%20%2B%201%5D%20%3D%20tmp%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20count%20%2B%3D%203%20%20%23%20%E5%85%83%E7%B4%A0%E4%BA%A4%E6%8D%A2%E5%8C%85%E5%90%AB%203%20%E4%B8%AA%E5%8D%95%E5%85%83%E6%93%8D%E4%BD%9C%0A%20%20%20%20return%20count%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%208%0A%20%20%20%20print%28%22%E8%BE%93%E5%85%A5%E6%95%B0%E6%8D%AE%E5%A4%A7%E5%B0%8F%20n%20%3D%22,%20n%29%0A%0A%20%20%20%20nums%20%3D%20%5Bi%20for%20i%20in%20range%28n,%200,%20-1%29%5D%20%20%23%20%5Bn,%20n-1,%20...,%202,%201%5D%0A%20%20%20%20count%20%3D%20bubble_sort%28nums%29%0A%20%20%20%20print%28%22%E5%B9%B3%E6%96%B9%E9%98%B6%EF%BC%88%E5%86%92%E6%B3%A1%E6%8E%92%E5%BA%8F%EF%BC%89%E7%9A%84%E6%93%8D%E4%BD%9C%E6%95%B0%E9%87%8F%20%3D%22,%20count%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
@@ -2484,27 +2324,6 @@ $$
end
```
=== "Zig"
```zig title="time_complexity.zig"
// 指数阶（循环实现）
fn exponential(n: i32) i32 {
var count: i32 = 0;
var bas: i32 = 1;
var i: i32 = 0;
// 细胞每轮一分为二，形成数列 1, 2, 4, 8, ..., 2^(n-1)
while (i < n) : (i += 1) {
var j: i32 = 0;
while (j < bas) : (j += 1) {
count += 1;
}
bas *= 2;
}
// count = 1 + 2 + 4 + 8 + .. + 2^(n-1) = 2^n - 1
return count;
}
```
??? pythontutor "可视化运行"
<div style="height: 531px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20exponential%28n%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E6%8C%87%E6%95%B0%E9%98%B6%EF%BC%88%E5%BE%AA%E7%8E%AF%E5%AE%9E%E7%8E%B0%EF%BC%89%22%22%22%0A%20%20%20%20count%20%3D%200%0A%20%20%20%20base%20%3D%201%0A%20%20%20%20%23%20%E7%BB%86%E8%83%9E%E6%AF%8F%E8%BD%AE%E4%B8%80%E5%88%86%E4%B8%BA%E4%BA%8C%EF%BC%8C%E5%BD%A2%E6%88%90%E6%95%B0%E5%88%97%201,%202,%204,%208,%20...,%202%5E%28n-1%29%0A%20%20%20%20for%20_%20in%20range%28n%29%3A%0A%20%20%20%20%20%20%20%20for%20_%20in%20range%28base%29%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20count%20%2B%3D%201%0A%20%20%20%20%20%20%20%20base%20*%3D%202%0A%20%20%20%20%23%20count%20%3D%201%20%2B%202%20%2B%204%20%2B%208%20%2B%20..%20%2B%202%5E%28n-1%29%20%3D%202%5En%20-%201%0A%20%20%20%20return%20count%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%208%0A%20%20%20%20print%28%22%E8%BE%93%E5%85%A5%E6%95%B0%E6%8D%AE%E5%A4%A7%E5%B0%8F%20n%20%3D%22,%20n%29%0A%0A%20%20%20%20count%20%3D%20exponential%28n%29%0A%20%20%20%20print%28%22%E6%8C%87%E6%95%B0%E9%98%B6%EF%BC%88%E5%BE%AA%E7%8E%AF%E5%AE%9E%E7%8E%B0%EF%BC%89%E7%9A%84%E6%93%8D%E4%BD%9C%E6%95%B0%E9%87%8F%20%3D%22,%20count%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
@@ -2657,16 +2476,6 @@ $$
end
```
=== "Zig"
```zig title="time_complexity.zig"
// 指数阶（递归实现）
fn expRecur(n: i32) i32 {
if (n == 1) return 1;
return expRecur(n - 1) + expRecur(n - 1) + 1;
}
```
??? pythontutor "可视化运行"
<div style="height: 423px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20exp_recur%28n%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E6%8C%87%E6%95%B0%E9%98%B6%EF%BC%88%E9%80%92%E5%BD%92%E5%AE%9E%E7%8E%B0%EF%BC%89%22%22%22%0A%20%20%20%20if%20n%20%3D%3D%201%3A%0A%20%20%20%20%20%20%20%20return%201%0A%20%20%20%20return%20exp_recur%28n%20-%201%29%20%2B%20exp_recur%28n%20-%201%29%20%2B%201%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%207%0A%20%20%20%20print%28%22%E8%BE%93%E5%85%A5%E6%95%B0%E6%8D%AE%E5%A4%A7%E5%B0%8F%20n%20%3D%22,%20n%29%0A%0A%20%20%20%20count%20%3D%20exp_recur%28n%29%0A%20%20%20%20print%28%22%E6%8C%87%E6%95%B0%E9%98%B6%EF%BC%88%E9%80%92%E5%BD%92%E5%AE%9E%E7%8E%B0%EF%BC%89%E7%9A%84%E6%93%8D%E4%BD%9C%E6%95%B0%E9%87%8F%20%3D%22,%20count%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
@@ -2864,20 +2673,6 @@ $$
end
```
=== "Zig"
```zig title="time_complexity.zig"
// 对数阶（循环实现）
fn logarithmic(n: i32) i32 {
var count: i32 = 0;
var n_var: i32 = n;
while (n_var > 1) : (n_var = @divTrunc(n_var, 2)) {
count += 1;
}
return count;
}
```
??? pythontutor "可视化运行"
<div style="height: 459px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20logarithmic%28n%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E5%AF%B9%E6%95%B0%E9%98%B6%EF%BC%88%E5%BE%AA%E7%8E%AF%E5%AE%9E%E7%8E%B0%EF%BC%89%22%22%22%0A%20%20%20%20count%20%3D%200%0A%20%20%20%20while%20n%20%3E%201%3A%0A%20%20%20%20%20%20%20%20n%20%3D%20n%20/%202%0A%20%20%20%20%20%20%20%20count%20%2B%3D%201%0A%20%20%20%20return%20count%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%208%0A%20%20%20%20print%28%22%E8%BE%93%E5%85%A5%E6%95%B0%E6%8D%AE%E5%A4%A7%E5%B0%8F%20n%20%3D%22,%20n%29%0A%0A%20%20%20%20count%20%3D%20logarithmic%28n%29%0A%20%20%20%20print%28%22%E5%AF%B9%E6%95%B0%E9%98%B6%EF%BC%88%E5%BE%AA%E7%8E%AF%E5%AE%9E%E7%8E%B0%EF%BC%89%E7%9A%84%E6%93%8D%E4%BD%9C%E6%95%B0%E9%87%8F%20%3D%22,%20count%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
@@ -3029,16 +2824,6 @@ $$
end
```
=== "Zig"
```zig title="time_complexity.zig"
// 对数阶（递归实现）
fn logRecur(n: i32) i32 {
if (n <= 1) return 0;
return logRecur(@divTrunc(n, 2)) + 1;
}
```
??? pythontutor "可视化运行"
<div style="height: 423px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20log_recur%28n%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E5%AF%B9%E6%95%B0%E9%98%B6%EF%BC%88%E9%80%92%E5%BD%92%E5%AE%9E%E7%8E%B0%EF%BC%89%22%22%22%0A%20%20%20%20if%20n%20%3C%3D%201%3A%0A%20%20%20%20%20%20%20%20return%200%0A%20%20%20%20return%20log_recur%28n%20/%202%29%20%2B%201%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%208%0A%20%20%20%20print%28%22%E8%BE%93%E5%85%A5%E6%95%B0%E6%8D%AE%E5%A4%A7%E5%B0%8F%20n%20%3D%22,%20n%29%0A%0A%20%20%20%20count%20%3D%20log_recur%28n%29%0A%20%20%20%20print%28%22%E5%AF%B9%E6%95%B0%E9%98%B6%EF%BC%88%E9%80%92%E5%BD%92%E5%AE%9E%E7%8E%B0%EF%BC%89%E7%9A%84%E6%93%8D%E4%BD%9C%E6%95%B0%E9%87%8F%20%3D%22,%20count%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=4&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
@@ -3253,21 +3038,6 @@ $$
end
```
=== "Zig"
```zig title="time_complexity.zig"
// 线性对数阶
fn linearLogRecur(n: i32) i32 {
if (n <= 1) return 1;
var count: i32 = linearLogRecur(@divTrunc(n, 2)) + linearLogRecur(@divTrunc(n, 2));
var i: i32 = 0;
while (i < n) : (i += 1) {
count += 1;
}
return count;
}
```
??? pythontutor "可视化运行"
<div style="height: 477px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20linear_log_recur%28n%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E7%BA%BF%E6%80%A7%E5%AF%B9%E6%95%B0%E9%98%B6%22%22%22%0A%20%20%20%20if%20n%20%3C%3D%201%3A%0A%20%20%20%20%20%20%20%20return%201%0A%20%20%20%20count%20%3D%20linear_log_recur%28n%20//%202%29%20%2B%20linear_log_recur%28n%20//%202%29%0A%20%20%20%20for%20_%20in%20range%28n%29%3A%0A%20%20%20%20%20%20%20%20count%20%2B%3D%201%0A%20%20%20%20return%20count%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%208%0A%20%20%20%20print%28%22%E8%BE%93%E5%85%A5%E6%95%B0%E6%8D%AE%E5%A4%A7%E5%B0%8F%20n%20%3D%22,%20n%29%0A%0A%20%20%20%20count%20%3D%20linear_log_recur%28n%29%0A%20%20%20%20print%28%22%E7%BA%BF%E6%80%A7%E5%AF%B9%E6%95%B0%E9%98%B6%EF%BC%88%E9%80%92%E5%BD%92%E5%AE%9E%E7%8E%B0%EF%BC%89%E7%9A%84%E6%93%8D%E4%BD%9C%E6%95%B0%E9%87%8F%20%3D%22,%20count%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=4&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
@@ -3494,22 +3264,6 @@ $$
end
```
=== "Zig"
```zig title="time_complexity.zig"
// 阶乘阶（递归实现）
fn factorialRecur(n: i32) i32 {
if (n == 0) return 1;
var count: i32 = 0;
var i: i32 = 0;
// 从 1 个分裂出 n 个
while (i < n) : (i += 1) {
count += factorialRecur(n - 1);
}
return count;
}
```
??? pythontutor "可视化运行"
<div style="height: 495px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20factorial_recur%28n%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E9%98%B6%E4%B9%98%E9%98%B6%EF%BC%88%E9%80%92%E5%BD%92%E5%AE%9E%E7%8E%B0%EF%BC%89%22%22%22%0A%20%20%20%20if%20n%20%3D%3D%200%3A%0A%20%20%20%20%20%20%20%20return%201%0A%20%20%20%20count%20%3D%200%0A%20%20%20%20%23%20%E4%BB%8E%201%20%E4%B8%AA%E5%88%86%E8%A3%82%E5%87%BA%20n%20%E4%B8%AA%0A%20%20%20%20for%20_%20in%20range%28n%29%3A%0A%20%20%20%20%20%20%20%20count%20%2B%3D%20factorial_recur%28n%20-%201%29%0A%20%20%20%20return%20count%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%204%0A%20%20%20%20print%28%22%E8%BE%93%E5%85%A5%E6%95%B0%E6%8D%AE%E5%A4%A7%E5%B0%8F%20n%20%3D%22,%20n%29%0A%0A%20%20%20%20count%20%3D%20factorial_recur%28n%29%0A%20%20%20%20print%28%22%E9%98%B6%E4%B9%98%E9%98%B6%EF%BC%88%E9%80%92%E5%BD%92%E5%AE%9E%E7%8E%B0%EF%BC%89%E7%9A%84%E6%93%8D%E4%BD%9C%E6%95%B0%E9%87%8F%20%3D%22,%20count%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
@@ -3904,33 +3658,6 @@ $$
end
```
=== "Zig"
```zig title="worst_best_time_complexity.zig"
// 生成一个数组，元素为 { 1, 2, ..., n }，顺序被打乱
fn randomNumbers(comptime n: usize) [n]i32 {
var nums: [n]i32 = undefined;
// 生成数组 nums = { 1, 2, 3, ..., n }
for (&nums, 0..) |*num, i| {
num.* = @as(i32, @intCast(i)) + 1;
}
// 随机打乱数组元素
const rand = std.crypto.random;
rand.shuffle(i32, &nums);
return nums;
}
// 查找数组 nums 中数字 1 所在索引
fn findOne(nums: []i32) i32 {
for (nums, 0..) |num, i| {
// 当元素 1 在数组头部时，达到最佳时间复杂度 O(1)
// 当元素 1 在数组尾部时，达到最差时间复杂度 O(n)
if (num == 1) return @intCast(i);
}
return -1;
}
```
??? pythontutor "可视化运行"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=import%20random%0A%0Adef%20random_numbers%28n%3A%20int%29%20-%3E%20list%5Bint%5D%3A%0A%20%20%20%20%22%22%22%E7%94%9F%E6%88%90%E4%B8%80%E4%B8%AA%E6%95%B0%E7%BB%84%EF%BC%8C%E5%85%83%E7%B4%A0%E4%B8%BA%3A%201,%202,%20...,%20n%20%EF%BC%8C%E9%A1%BA%E5%BA%8F%E8%A2%AB%E6%89%93%E4%B9%B1%22%22%22%0A%20%20%20%20%23%20%E7%94%9F%E6%88%90%E6%95%B0%E7%BB%84%20nums%20%3D%3A%201,%202,%203,%20...,%20n%0A%20%20%20%20nums%20%3D%20%5Bi%20for%20i%20in%20range%281,%20n%20%2B%201%29%5D%0A%20%20%20%20%23%20%E9%9A%8F%E6%9C%BA%E6%89%93%E4%B9%B1%E6%95%B0%E7%BB%84%E5%85%83%E7%B4%A0%0A%20%20%20%20random.shuffle%28nums%29%0A%20%20%20%20return%20nums%0A%0Adef%20find_one%28nums%3A%20list%5Bint%5D%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E6%9F%A5%E6%89%BE%E6%95%B0%E7%BB%84%20nums%20%E4%B8%AD%E6%95%B0%E5%AD%97%201%20%E6%89%80%E5%9C%A8%E7%B4%A2%E5%BC%95%22%22%22%0A%20%20%20%20for%20i%20in%20range%28len%28nums%29%29%3A%0A%20%20%20%20%20%20%20%20%23%20%E5%BD%93%E5%85%83%E7%B4%A0%201%20%E5%9C%A8%E6%95%B0%E7%BB%84%E5%A4%B4%E9%83%A8%E6%97%B6%EF%BC%8C%E8%BE%BE%E5%88%B0%E6%9C%80%E4%BD%B3%E6%97%B6%E9%97%B4%E5%A4%8D%E6%9D%82%E5%BA%A6%20O%281%29%0A%20%20%20%20%20%20%20%20%23%20%E5%BD%93%E5%85%83%E7%B4%A0%201%20%E5%9C%A8%E6%95%B0%E7%BB%84%E5%B0%BE%E9%83%A8%E6%97%B6%EF%BC%8C%E8%BE%BE%E5%88%B0%E6%9C%80%E5%B7%AE%E6%97%B6%E9%97%B4%E5%A4%8D%E6%9D%82%E5%BA%A6%20O%28n%29%0A%20%20%20%20%20%20%20%20if%20nums%5Bi%5D%20%3D%3D%201%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20return%20i%0A%20%20%20%20return%20-1%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%2010%0A%20%20%20%20nums%20%3D%20random_numbers%28n%29%0A%20%20%20%20index%20%3D%20find_one%28nums%29%0A%20%20%20%20print%28%22%5Cn%E6%95%B0%E7%BB%84%20%5B%201,%202,%20...,%20n%20%5D%20%E8%A2%AB%E6%89%93%E4%B9%B1%E5%90%8E%20%3D%22,%20nums%29%0A%20%20%20%20print%28%22%E6%95%B0%E5%AD%97%201%20%E7%9A%84%E7%B4%A2%E5%BC%95%E4%B8%BA%22,%20index%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=25&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
docs/chapter_data_structure/basic_data_types.md
-18
View File
@@ -178,24 +178,6 @@ comments: true
data = [0, 0.0, 'a', false, ListNode(0)]
```
=== "Zig"
```zig title=""
const hello = [5]u8{ 'h', 'e', 'l', 'l', 'o' };
// 以上代码展示了定义一个字面量数组的方式，其中你可以选择指明数组的大小或者使用 _ 代替。使用 _ 时，Zig 会尝试自动计算数组的长度
const matrix_4x4 = [4][4]f32{
[_]f32{ 1.0, 0.0, 0.0, 0.0 },
[_]f32{ 0.0, 1.0, 0.0, 1.0 },
[_]f32{ 0.0, 0.0, 1.0, 0.0 },
[_]f32{ 0.0, 0.0, 0.0, 1.0 },
};
// 多维数组（矩阵）实际上就是嵌套数组，我们很容易就可以创建一个多维数组出来
const array = [_:0]u8{ 1, 2, 3, 4 };
// 定义一个哨兵终止数组，本质上来说，这是为了兼容 C 中的规定的字符串结尾字符\0。我们使用语法 [N:x]T 来描述一个元素为类型 T，长度为 N 的数组，在它对应 N 的索引处的值应该是 x
```
??? pythontutor "可视化运行"
<div style="height: 477px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=class%20ListNode%3A%0A%20%20%20%20%22%22%22%E9%93%BE%E8%A1%A8%E8%8A%82%E7%82%B9%E7%B1%BB%22%22%22%0A%20%20%20%20def%20__init__%28self,%20val%3A%20int%29%3A%0A%20%20%20%20%20%20%20%20self.val%3A%20int%20%3D%20val%20%20%23%20%E8%8A%82%E7%82%B9%E5%80%BC%0A%20%20%20%20%20%20%20%20self.next%3A%20ListNode%20%7C%20None%20%3D%20None%20%20%23%20%E5%90%8E%E7%BB%A7%E8%8A%82%E7%82%B9%E5%BC%95%E7%94%A8%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20%23%20%E4%BD%BF%E7%94%A8%E5%A4%9A%E7%A7%8D%E5%9F%BA%E6%9C%AC%E6%95%B0%E6%8D%AE%E7%B1%BB%E5%9E%8B%E6%9D%A5%E5%88%9D%E5%A7%8B%E5%8C%96%E6%95%B0%E7%BB%84%0A%20%20%20%20numbers%20%3D%20%5B0%5D%20*%205%0A%20%20%20%20decimals%20%3D%20%5B0.0%5D%20*%205%0A%20%20%20%20%23%20Python%20%E7%9A%84%E5%AD%97%E7%AC%A6%E5%AE%9E%E9%99%85%E4%B8%8A%E6%98%AF%E9%95%BF%E5%BA%A6%E4%B8%BA%201%20%E7%9A%84%E5%AD%97%E7%AC%A6%E4%B8%B2%0A%20%20%20%20characters%20%3D%20%5B'0'%5D%20*%205%0A%20%20%20%20bools%20%3D%20%5BFalse%5D%20*%205%0A%20%20%20%20%23%20Python%20%E7%9A%84%E5%88%97%E8%A1%A8%E5%8F%AF%E4%BB%A5%E8%87%AA%E7%94%B1%E5%AD%98%E5%82%A8%E5%90%84%E7%A7%8D%E5%9F%BA%E6%9C%AC%E6%95%B0%E6%8D%AE%E7%B1%BB%E5%9E%8B%E5%92%8C%E5%AF%B9%E8%B1%A1%E5%BC%95%E7%94%A8%0A%20%20%20%20data%20%3D%20%5B0,%200.0,%20'a',%20False,%20ListNode%280%29%5D&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=12&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
docs/chapter_divide_and_conquer/binary_search_recur.md
-8
View File
@@ -450,14 +450,6 @@ comments: true
end
```
=== "Zig"
```zig title="binary_search_recur.zig"
[class]{}-[func]{dfs}
[class]{}-[func]{binarySearch}
```
??? pythontutor "可视化运行"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20dfs%28nums%3A%20list%5Bint%5D,%20target%3A%20int,%20i%3A%20int,%20j%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E4%BA%8C%E5%88%86%E6%9F%A5%E6%89%BE%EF%BC%9A%E9%97%AE%E9%A2%98%20f%28i,%20j%29%22%22%22%0A%20%20%20%20%23%20%E8%8B%A5%E5%8C%BA%E9%97%B4%E4%B8%BA%E7%A9%BA%EF%BC%8C%E4%BB%A3%E8%A1%A8%E6%97%A0%E7%9B%AE%E6%A0%87%E5%85%83%E7%B4%A0%EF%BC%8C%E5%88%99%E8%BF%94%E5%9B%9E%20-1%0A%20%20%20%20if%20i%20%3E%20j%3A%0A%20%20%20%20%20%20%20%20return%20-1%0A%20%20%20%20%23%20%E8%AE%A1%E7%AE%97%E4%B8%AD%E7%82%B9%E7%B4%A2%E5%BC%95%20m%0A%20%20%20%20m%20%3D%20%28i%20%2B%20j%29%20//%202%0A%20%20%20%20if%20nums%5Bm%5D%20%3C%20target%3A%0A%20%20%20%20%20%20%20%20%23%20%E9%80%92%E5%BD%92%E5%AD%90%E9%97%AE%E9%A2%98%20f%28m%2B1,%20j%29%0A%20%20%20%20%20%20%20%20return%20dfs%28nums,%20target,%20m%20%2B%201,%20j%29%0A%20%20%20%20elif%20nums%5Bm%5D%20%3E%20target%3A%0A%20%20%20%20%20%20%20%20%23%20%E9%80%92%E5%BD%92%E5%AD%90%E9%97%AE%E9%A2%98%20f%28i,%20m-1%29%0A%20%20%20%20%20%20%20%20return%20dfs%28nums,%20target,%20i,%20m%20-%201%29%0A%20%20%20%20else%3A%0A%20%20%20%20%20%20%20%20%23%20%E6%89%BE%E5%88%B0%E7%9B%AE%E6%A0%87%E5%85%83%E7%B4%A0%EF%BC%8C%E8%BF%94%E5%9B%9E%E5%85%B6%E7%B4%A2%E5%BC%95%0A%20%20%20%20%20%20%20%20return%20m%0A%0Adef%20binary_search%28nums%3A%20list%5Bint%5D,%20target%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E4%BA%8C%E5%88%86%E6%9F%A5%E6%89%BE%22%22%22%0A%20%20%20%20n%20%3D%20len%28nums%29%0A%20%20%20%20%23%20%E6%B1%82%E8%A7%A3%E9%97%AE%E9%A2%98%20f%280,%20n-1%29%0A%20%20%20%20return%20dfs%28nums,%20target,%200,%20n%20-%201%29%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20target%20%3D%206%0A%20%20%20%20nums%20%3D%20%5B1,%203,%206,%208,%2012,%2015,%2023,%2026,%2031,%2035%5D%0A%0A%20%20%20%20%23%20%E4%BA%8C%E5%88%86%E6%9F%A5%E6%89%BE%EF%BC%88%E5%8F%8C%E9%97%AD%E5%8C%BA%E9%97%B4%EF%BC%89%0A%20%20%20%20index%20%3D%20binary_search%28nums,%20target%29%0A%20%20%20%20print%28%22%E7%9B%AE%E6%A0%87%E5%85%83%E7%B4%A0%206%20%E7%9A%84%E7%B4%A2%E5%BC%95%20%3D%20%22,%20index%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=6&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
docs/chapter_divide_and_conquer/build_binary_tree_problem.md
-8
View File
@@ -512,14 +512,6 @@ comments: true
end
```
=== "Zig"
```zig title="build_tree.zig"
[class]{}-[func]{dfs}
[class]{}-[func]{buildTree}
```
??? pythontutor "可视化运行"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=class%20TreeNode%3A%0A%20%20%20%20%22%22%22%E4%BA%8C%E5%8F%89%E6%A0%91%E8%8A%82%E7%82%B9%E7%B1%BB%22%22%22%0A%20%20%20%20def%20__init__%28self,%20val%3A%20int%20%3D%200%29%3A%0A%20%20%20%20%20%20%20%20self.val%3A%20int%20%3D%20val%20%20%23%20%E8%8A%82%E7%82%B9%E5%80%BC%0A%20%20%20%20%20%20%20%20self.left%3A%20TreeNode%20%7C%20None%20%3D%20None%20%20%23%20%E5%B7%A6%E5%AD%90%E8%8A%82%E7%82%B9%E5%BC%95%E7%94%A8%0A%20%20%20%20%20%20%20%20self.right%3A%20TreeNode%20%7C%20None%20%3D%20None%20%20%23%20%E5%8F%B3%E5%AD%90%E8%8A%82%E7%82%B9%E5%BC%95%E7%94%A8%0A%0Adef%20dfs%28%0A%20%20%20%20preorder%3A%20list%5Bint%5D,%0A%20%20%20%20inorder_map%3A%20dict%5Bint,%20int%5D,%0A%20%20%20%20i%3A%20int,%0A%20%20%20%20l%3A%20int,%0A%20%20%20%20r%3A%20int,%0A%29%20-%3E%20TreeNode%20%7C%20None%3A%0A%20%20%20%20%22%22%22%E6%9E%84%E5%BB%BA%E4%BA%8C%E5%8F%89%E6%A0%91%EF%BC%9A%E5%88%86%E6%B2%BB%22%22%22%0A%20%20%20%20%23%20%E5%AD%90%E6%A0%91%E5%8C%BA%E9%97%B4%E4%B8%BA%E7%A9%BA%E6%97%B6%E7%BB%88%E6%AD%A2%0A%20%20%20%20if%20r%20-%20l%20%3C%200%3A%0A%20%20%20%20%20%20%20%20return%20None%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%E6%A0%B9%E8%8A%82%E7%82%B9%0A%20%20%20%20root%20%3D%20TreeNode%28preorder%5Bi%5D%29%0A%20%20%20%20%23%20%E6%9F%A5%E8%AF%A2%20m%20%EF%BC%8C%E4%BB%8E%E8%80%8C%E5%88%92%E5%88%86%E5%B7%A6%E5%8F%B3%E5%AD%90%E6%A0%91%0A%20%20%20%20m%20%3D%20inorder_map%5Bpreorder%5Bi%5D%5D%0A%20%20%20%20%23%20%E5%AD%90%E9%97%AE%E9%A2%98%EF%BC%9A%E6%9E%84%E5%BB%BA%E5%B7%A6%E5%AD%90%E6%A0%91%0A%20%20%20%20root.left%20%3D%20dfs%28preorder,%20inorder_map,%20i%20%2B%201,%20l,%20m%20-%201%29%0A%20%20%20%20%23%20%E5%AD%90%E9%97%AE%E9%A2%98%EF%BC%9A%E6%9E%84%E5%BB%BA%E5%8F%B3%E5%AD%90%E6%A0%91%0A%20%20%20%20root.right%20%3D%20dfs%28preorder,%20inorder_map,%20i%20%2B%201%20%2B%20m%20-%20l,%20m%20%2B%201,%20r%29%0A%20%20%20%20%23%20%E8%BF%94%E5%9B%9E%E6%A0%B9%E8%8A%82%E7%82%B9%0A%20%20%20%20return%20root%0A%0A%0Adef%20build_tree%28preorder%3A%20list%5Bint%5D,%20inorder%3A%20list%5Bint%5D%29%20-%3E%20TreeNode%20%7C%20None%3A%0A%20%20%20%20%22%22%22%E6%9E%84%E5%BB%BA%E4%BA%8C%E5%8F%89%E6%A0%91%22%22%22%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%E5%93%88%E5%B8%8C%E8%A1%A8%EF%BC%8C%E5%AD%98%E5%82%A8%20inorder%20%E5%85%83%E7%B4%A0%E5%88%B0%E7%B4%A2%E5%BC%95%E7%9A%84%E6%98%A0%E5%B0%84%0A%20%20%20%20inorder_map%20%3D%20%7Bval%3A%20i%20for%20i,%20val%20in%20enumerate%28inorder%29%7D%0A%20%20%20%20root%20%3D%20dfs%28preorder,%20inorder_map,%200,%200,%20len%28inorder%29%20-%201%29%0A%20%20%20%20return%20root%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20preorder%20%3D%20%5B3,%209,%202,%201,%207%5D%0A%20%20%20%20inorder%20%3D%20%5B9,%203,%201,%202,%207%5D%0A%20%20%20%20print%28f%22%E5%89%8D%E5%BA%8F%E9%81%8D%E5%8E%86%20%3D%20%7Bpreorder%7D%22%29%0A%20%20%20%20print%28f%22%E4%B8%AD%E5%BA%8F%E9%81%8D%E5%8E%86%20%3D%20%7Binorder%7D%22%29%0A%20%20%20%20root%20%3D%20build_tree%28preorder,%20inorder%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=21&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
docs/chapter_divide_and_conquer/hanota_problem.md
-10
View File
@@ -542,16 +542,6 @@ comments: true
end
```
=== "Zig"
```zig title="hanota.zig"
[class]{}-[func]{move}
[class]{}-[func]{dfs}
[class]{}-[func]{solveHanota}
```
??? pythontutor "可视化运行"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20move%28src%3A%20list%5Bint%5D,%20tar%3A%20list%5Bint%5D%29%3A%0A%20%20%20%20%22%22%22%E7%A7%BB%E5%8A%A8%E4%B8%80%E4%B8%AA%E5%9C%86%E7%9B%98%22%22%22%0A%20%20%20%20%23%20%E4%BB%8E%20src%20%E9%A1%B6%E9%83%A8%E6%8B%BF%E5%87%BA%E4%B8%80%E4%B8%AA%E5%9C%86%E7%9B%98%0A%20%20%20%20pan%20%3D%20src.pop%28%29%0A%20%20%20%20%23%20%E5%B0%86%E5%9C%86%E7%9B%98%E6%94%BE%E5%85%A5%20tar%20%E9%A1%B6%E9%83%A8%0A%20%20%20%20tar.append%28pan%29%0A%0A%0Adef%20dfs%28i%3A%20int,%20src%3A%20list%5Bint%5D,%20buf%3A%20list%5Bint%5D,%20tar%3A%20list%5Bint%5D%29%3A%0A%20%20%20%20%22%22%22%E6%B1%82%E8%A7%A3%E6%B1%89%E8%AF%BA%E5%A1%94%E9%97%AE%E9%A2%98%20f%28i%29%22%22%22%0A%20%20%20%20%23%20%E8%8B%A5%20src%20%E5%8F%AA%E5%89%A9%E4%B8%8B%E4%B8%80%E4%B8%AA%E5%9C%86%E7%9B%98%EF%BC%8C%E5%88%99%E7%9B%B4%E6%8E%A5%E5%B0%86%E5%85%B6%E7%A7%BB%E5%88%B0%20tar%0A%20%20%20%20if%20i%20%3D%3D%201%3A%0A%20%20%20%20%20%20%20%20move%28src,%20tar%29%0A%20%20%20%20%20%20%20%20return%0A%20%20%20%20%23%20%E5%AD%90%E9%97%AE%E9%A2%98%20f%28i-1%29%20%EF%BC%9A%E5%B0%86%20src%20%E9%A1%B6%E9%83%A8%20i-1%20%E4%B8%AA%E5%9C%86%E7%9B%98%E5%80%9F%E5%8A%A9%20tar%20%E7%A7%BB%E5%88%B0%20buf%0A%20%20%20%20dfs%28i%20-%201,%20src,%20tar,%20buf%29%0A%20%20%20%20%23%20%E5%AD%90%E9%97%AE%E9%A2%98%20f%281%29%20%EF%BC%9A%E5%B0%86%20src%20%E5%89%A9%E4%BD%99%E4%B8%80%E4%B8%AA%E5%9C%86%E7%9B%98%E7%A7%BB%E5%88%B0%20tar%0A%20%20%20%20move%28src,%20tar%29%0A%20%20%20%20%23%20%E5%AD%90%E9%97%AE%E9%A2%98%20f%28i-1%29%20%EF%BC%9A%E5%B0%86%20buf%20%E9%A1%B6%E9%83%A8%20i-1%20%E4%B8%AA%E5%9C%86%E7%9B%98%E5%80%9F%E5%8A%A9%20src%20%E7%A7%BB%E5%88%B0%20tar%0A%20%20%20%20dfs%28i%20-%201,%20buf,%20src,%20tar%29%0A%0A%0Adef%20solve_hanota%28A%3A%20list%5Bint%5D,%20B%3A%20list%5Bint%5D,%20C%3A%20list%5Bint%5D%29%3A%0A%20%20%20%20%22%22%22%E6%B1%82%E8%A7%A3%E6%B1%89%E8%AF%BA%E5%A1%94%E9%97%AE%E9%A2%98%22%22%22%0A%20%20%20%20n%20%3D%20len%28A%29%0A%20%20%20%20%23%20%E5%B0%86%20A%20%E9%A1%B6%E9%83%A8%20n%20%E4%B8%AA%E5%9C%86%E7%9B%98%E5%80%9F%E5%8A%A9%20B%20%E7%A7%BB%E5%88%B0%20C%0A%20%20%20%20dfs%28n,%20A,%20B,%20C%29%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20%23%20%E5%88%97%E8%A1%A8%E5%B0%BE%E9%83%A8%E6%98%AF%E6%9F%B1%E5%AD%90%E9%A1%B6%E9%83%A8%0A%20%20%20%20A%20%3D%20%5B5,%204,%203,%202,%201%5D%0A%20%20%20%20B%20%3D%20%5B%5D%0A%20%20%20%20C%20%3D%20%5B%5D%0A%20%20%20%20print%28%22%E5%88%9D%E5%A7%8B%E7%8A%B6%E6%80%81%E4%B8%8B%EF%BC%9A%22%29%0A%20%20%20%20print%28f%22A%20%3D%20%7BA%7D%22%29%0A%20%20%20%20print%28f%22B%20%3D%20%7BB%7D%22%29%0A%20%20%20%20print%28f%22C%20%3D%20%7BC%7D%22%29%0A%0A%20%20%20%20solve_hanota%28A,%20B,%20C%29%0A%0A%20%20%20%20print%28%22%E5%9C%86%E7%9B%98%E7%A7%BB%E5%8A%A8%E5%AE%8C%E6%88%90%E5%90%8E%EF%BC%9A%22%29%0A%20%20%20%20print%28f%22A%20%3D%20%7BA%7D%22%29%0A%20%20%20%20print%28f%22B%20%3D%20%7BB%7D%22%29%0A%20%20%20%20print%28f%22C%20%3D%20%7BC%7D%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=12&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
docs/chapter_divide_and_conquer/summary.md
+2
View File
@@ -4,6 +4,8 @@ comments: true
# 12.5 &nbsp; 小结
### 1. &nbsp; 重点回顾
- 分治是一种常见的算法设计策略，包括分（划分）和治（合并）两个阶段，通常基于递归实现。
- 判断是否是分治算法问题的依据包括：问题能否分解、子问题是否独立、子问题能否合并。
- 归并排序是分治策略的典型应用，其递归地将数组划分为等长的两个子数组，直到只剩一个元素时开始逐层合并，从而完成排序。
docs/chapter_dynamic_programming/dp_problem_features.md
-67
View File
@@ -318,28 +318,6 @@ $$
end
```
=== "Zig"
```zig title="min_cost_climbing_stairs_dp.zig"
// 爬楼梯最小代价：动态规划
fn minCostClimbingStairsDP(comptime cost: []i32) i32 {
comptime var n = cost.len - 1;
if (n == 1 or n == 2) {
return cost[n];
}
// 初始化 dp 表，用于存储子问题的解
var dp = [_]i32{-1} ** (n + 1);
// 初始状态：预设最小子问题的解
dp[1] = cost[1];
dp[2] = cost[2];
// 状态转移：从较小子问题逐步求解较大子问题
for (3..n + 1) |i| {
dp[i] = @min(dp[i - 1], dp[i - 2]) + cost[i];
}
return dp[n];
}
```
??? pythontutor "可视化运行"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20min_cost_climbing_stairs_dp%28cost%3A%20list%5Bint%5D%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E7%88%AC%E6%A5%BC%E6%A2%AF%E6%9C%80%E5%B0%8F%E4%BB%A3%E4%BB%B7%EF%BC%9A%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%22%22%22%0A%20%20%20%20n%20%3D%20len%28cost%29%20-%201%0A%20%20%20%20if%20n%20%3D%3D%201%20or%20n%20%3D%3D%202%3A%0A%20%20%20%20%20%20%20%20return%20cost%5Bn%5D%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%20dp%20%E8%A1%A8%EF%BC%8C%E7%94%A8%E4%BA%8E%E5%AD%98%E5%82%A8%E5%AD%90%E9%97%AE%E9%A2%98%E7%9A%84%E8%A7%A3%0A%20%20%20%20dp%20%3D%20%5B0%5D%20*%20%28n%20%2B%201%29%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E7%8A%B6%E6%80%81%EF%BC%9A%E9%A2%84%E8%AE%BE%E6%9C%80%E5%B0%8F%E5%AD%90%E9%97%AE%E9%A2%98%E7%9A%84%E8%A7%A3%0A%20%20%20%20dp%5B1%5D,%20dp%5B2%5D%20%3D%20cost%5B1%5D,%20cost%5B2%5D%0A%20%20%20%20%23%20%E7%8A%B6%E6%80%81%E8%BD%AC%E7%A7%BB%EF%BC%9A%E4%BB%8E%E8%BE%83%E5%B0%8F%E5%AD%90%E9%97%AE%E9%A2%98%E9%80%90%E6%AD%A5%E6%B1%82%E8%A7%A3%E8%BE%83%E5%A4%A7%E5%AD%90%E9%97%AE%E9%A2%98%0A%20%20%20%20for%20i%20in%20range%283,%20n%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20dp%5Bi%5D%20%3D%20min%28dp%5Bi%20-%201%5D,%20dp%5Bi%20-%202%5D%29%20%2B%20cost%5Bi%5D%0A%20%20%20%20return%20dp%5Bn%5D%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20cost%20%3D%20%5B0,%201,%2010,%201,%201,%201,%2010,%201,%201,%2010,%201%5D%0A%20%20%20%20print%28f%22%E8%BE%93%E5%85%A5%E6%A5%BC%E6%A2%AF%E7%9A%84%E4%BB%A3%E4%BB%B7%E5%88%97%E8%A1%A8%E4%B8%BA%20%7Bcost%7D%22%29%0A%0A%20%20%20%20res%20%3D%20min_cost_climbing_stairs_dp%28cost%29%0A%20%20%20%20print%28f%22%E7%88%AC%E5%AE%8C%E6%A5%BC%E6%A2%AF%E7%9A%84%E6%9C%80%E4%BD%8E%E4%BB%A3%E4%BB%B7%E4%B8%BA%20%7Bres%7D%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=4&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
@@ -603,27 +581,6 @@ $$
end
```
=== "Zig"
```zig title="min_cost_climbing_stairs_dp.zig"
// 爬楼梯最小代价：空间优化后的动态规划
fn minCostClimbingStairsDPComp(cost: []i32) i32 {
var n = cost.len - 1;
if (n == 1 or n == 2) {
return cost[n];
}
var a = cost[1];
var b = cost[2];
// 状态转移：从较小子问题逐步求解较大子问题
for (3..n + 1) |i| {
var tmp = b;
b = @min(a, tmp) + cost[i];
a = tmp;
}
return b;
}
```
??? pythontutor "可视化运行"
<div style="height: 513px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20min_cost_climbing_stairs_dp_comp%28cost%3A%20list%5Bint%5D%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E7%88%AC%E6%A5%BC%E6%A2%AF%E6%9C%80%E5%B0%8F%E4%BB%A3%E4%BB%B7%EF%BC%9A%E7%A9%BA%E9%97%B4%E4%BC%98%E5%8C%96%E5%90%8E%E7%9A%84%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%22%22%22%0A%20%20%20%20n%20%3D%20len%28cost%29%20-%201%0A%20%20%20%20if%20n%20%3D%3D%201%20or%20n%20%3D%3D%202%3A%0A%20%20%20%20%20%20%20%20return%20cost%5Bn%5D%0A%20%20%20%20a,%20b%20%3D%20cost%5B1%5D,%20cost%5B2%5D%0A%20%20%20%20for%20i%20in%20range%283,%20n%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20a,%20b%20%3D%20b,%20min%28a,%20b%29%20%2B%20cost%5Bi%5D%0A%20%20%20%20return%20b%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20cost%20%3D%20%5B0,%201,%2010,%201,%201,%201,%2010,%201,%201,%2010,%201%5D%0A%20%20%20%20print%28f%22%E8%BE%93%E5%85%A5%E6%A5%BC%E6%A2%AF%E7%9A%84%E4%BB%A3%E4%BB%B7%E5%88%97%E8%A1%A8%E4%B8%BA%20%7Bcost%7D%22%29%0A%0A%20%20%20%20res%20%3D%20min_cost_climbing_stairs_dp_comp%28cost%29%0A%20%20%20%20print%28f%22%E7%88%AC%E5%AE%8C%E6%A5%BC%E6%A2%AF%E7%9A%84%E6%9C%80%E4%BD%8E%E4%BB%A3%E4%BB%B7%E4%B8%BA%20%7Bres%7D%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=5&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
@@ -985,30 +942,6 @@ $$
end
```
=== "Zig"
```zig title="climbing_stairs_constraint_dp.zig"
// 带约束爬楼梯：动态规划
fn climbingStairsConstraintDP(comptime n: usize) i32 {
if (n == 1 or n == 2) {
return 1;
}
// 初始化 dp 表，用于存储子问题的解
var dp = [_][3]i32{ [_]i32{ -1, -1, -1 } } ** (n + 1);
// 初始状态：预设最小子问题的解
dp[1][1] = 1;
dp[1][2] = 0;
dp[2][1] = 0;
dp[2][2] = 1;
// 状态转移：从较小子问题逐步求解较大子问题
for (3..n + 1) |i| {
dp[i][1] = dp[i - 1][2];
dp[i][2] = dp[i - 2][1] + dp[i - 2][2];
}
return dp[n][1] + dp[n][2];
}
```
??? pythontutor "可视化运行"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20climbing_stairs_constraint_dp%28n%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E5%B8%A6%E7%BA%A6%E6%9D%9F%E7%88%AC%E6%A5%BC%E6%A2%AF%EF%BC%9A%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%22%22%22%0A%20%20%20%20if%20n%20%3D%3D%201%20or%20n%20%3D%3D%202%3A%0A%20%20%20%20%20%20%20%20return%201%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%20dp%20%E8%A1%A8%EF%BC%8C%E7%94%A8%E4%BA%8E%E5%AD%98%E5%82%A8%E5%AD%90%E9%97%AE%E9%A2%98%E7%9A%84%E8%A7%A3%0A%20%20%20%20dp%20%3D%20%5B%5B0%5D%20*%203%20for%20_%20in%20range%28n%20%2B%201%29%5D%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E7%8A%B6%E6%80%81%EF%BC%9A%E9%A2%84%E8%AE%BE%E6%9C%80%E5%B0%8F%E5%AD%90%E9%97%AE%E9%A2%98%E7%9A%84%E8%A7%A3%0A%20%20%20%20dp%5B1%5D%5B1%5D,%20dp%5B1%5D%5B2%5D%20%3D%201,%200%0A%20%20%20%20dp%5B2%5D%5B1%5D,%20dp%5B2%5D%5B2%5D%20%3D%200,%201%0A%20%20%20%20%23%20%E7%8A%B6%E6%80%81%E8%BD%AC%E7%A7%BB%EF%BC%9A%E4%BB%8E%E8%BE%83%E5%B0%8F%E5%AD%90%E9%97%AE%E9%A2%98%E9%80%90%E6%AD%A5%E6%B1%82%E8%A7%A3%E8%BE%83%E5%A4%A7%E5%AD%90%E9%97%AE%E9%A2%98%0A%20%20%20%20for%20i%20in%20range%283,%20n%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20dp%5Bi%5D%5B1%5D%20%3D%20dp%5Bi%20-%201%5D%5B2%5D%0A%20%20%20%20%20%20%20%20dp%5Bi%5D%5B2%5D%20%3D%20dp%5Bi%20-%202%5D%5B1%5D%20%2B%20dp%5Bi%20-%202%5D%5B2%5D%0A%20%20%20%20return%20dp%5Bn%5D%5B1%5D%20%2B%20dp%5Bn%5D%5B2%5D%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%209%0A%0A%20%20%20%20res%20%3D%20climbing_stairs_constraint_dp%28n%29%0A%20%20%20%20print%28f%22%E7%88%AC%20%7Bn%7D%20%E9%98%B6%E6%A5%BC%E6%A2%AF%E5%85%B1%E6%9C%89%20%7Bres%7D%20%E7%A7%8D%E6%96%B9%E6%A1%88%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=4&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
docs/chapter_dynamic_programming/dp_solution_pipeline.md
-102
View File
@@ -383,27 +383,6 @@ $$
end
```
=== "Zig"
```zig title="min_path_sum.zig"
// 最小路径和：暴力搜索
fn minPathSumDFS(grid: anytype, i: i32, j: i32) i32 {
// 若为左上角单元格，则终止搜索
if (i == 0 and j == 0) {
return grid[0][0];
}
// 若行列索引越界，则返回 +∞ 代价
if (i < 0 or j < 0) {
return std.math.maxInt(i32);
}
// 计算从左上角到 (i-1, j) 和 (i, j-1) 的最小路径代价
var up = minPathSumDFS(grid, i - 1, j);
var left = minPathSumDFS(grid, i, j - 1);
// 返回从左上角到 (i, j) 的最小路径代价
return @min(left, up) + grid[@as(usize, @intCast(i))][@as(usize, @intCast(j))];
}
```
??? pythontutor "可视化运行"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=from%20math%20import%20inf%0A%0Adef%20min_path_sum_dfs%28grid%3A%20list%5Blist%5Bint%5D%5D,%20i%3A%20int,%20j%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E6%9C%80%E5%B0%8F%E8%B7%AF%E5%BE%84%E5%92%8C%EF%BC%9A%E6%9A%B4%E5%8A%9B%E6%90%9C%E7%B4%A2%22%22%22%0A%20%20%20%20%23%20%E8%8B%A5%E4%B8%BA%E5%B7%A6%E4%B8%8A%E8%A7%92%E5%8D%95%E5%85%83%E6%A0%BC%EF%BC%8C%E5%88%99%E7%BB%88%E6%AD%A2%E6%90%9C%E7%B4%A2%0A%20%20%20%20if%20i%20%3D%3D%200%20and%20j%20%3D%3D%200%3A%0A%20%20%20%20%20%20%20%20return%20grid%5B0%5D%5B0%5D%0A%20%20%20%20%23%20%E8%8B%A5%E8%A1%8C%E5%88%97%E7%B4%A2%E5%BC%95%E8%B6%8A%E7%95%8C%EF%BC%8C%E5%88%99%E8%BF%94%E5%9B%9E%20%2B%E2%88%9E%20%E4%BB%A3%E4%BB%B7%0A%20%20%20%20if%20i%20%3C%200%20or%20j%20%3C%200%3A%0A%20%20%20%20%20%20%20%20return%20inf%0A%20%20%20%20%23%20%E8%AE%A1%E7%AE%97%E4%BB%8E%E5%B7%A6%E4%B8%8A%E8%A7%92%E5%88%B0%20%28i-1,%20j%29%20%E5%92%8C%20%28i,%20j-1%29%20%E7%9A%84%E6%9C%80%E5%B0%8F%E8%B7%AF%E5%BE%84%E4%BB%A3%E4%BB%B7%0A%20%20%20%20up%20%3D%20min_path_sum_dfs%28grid,%20i%20-%201,%20j%29%0A%20%20%20%20left%20%3D%20min_path_sum_dfs%28grid,%20i,%20j%20-%201%29%0A%20%20%20%20%23%20%E8%BF%94%E5%9B%9E%E4%BB%8E%E5%B7%A6%E4%B8%8A%E8%A7%92%E5%88%B0%20%28i,%20j%29%20%E7%9A%84%E6%9C%80%E5%B0%8F%E8%B7%AF%E5%BE%84%E4%BB%A3%E4%BB%B7%0A%20%20%20%20return%20min%28left,%20up%29%20%2B%20grid%5Bi%5D%5Bj%5D%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20grid%20%3D%20%5B%5B1,%203,%201,%205%5D,%20%5B2,%202,%204,%202%5D,%20%5B5,%203,%202,%201%5D,%20%5B4,%203,%205,%202%5D%5D%0A%20%20%20%20n,%20m%20%3D%20len%28grid%29,%20len%28grid%5B0%5D%29%0A%0A%20%20%20%20%23%20%E6%9A%B4%E5%8A%9B%E6%90%9C%E7%B4%A2%0A%20%20%20%20res%20%3D%20min_path_sum_dfs%28grid,%20n%20-%201,%20m%20-%201%29%0A%20%20%20%20print%28f%22%E4%BB%8E%E5%B7%A6%E4%B8%8A%E8%A7%92%E5%88%B0%E5%8F%B3%E4%B8%8B%E8%A7%92%E7%9A%84%E5%81%9A%E5%B0%8F%E8%B7%AF%E5%BE%84%E5%92%8C%E4%B8%BA%20%7Bres%7D%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=6&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
@@ -763,33 +742,6 @@ $$
end
```
=== "Zig"
```zig title="min_path_sum.zig"
// 最小路径和：记忆化搜索
fn minPathSumDFSMem(grid: anytype, mem: anytype, i: i32, j: i32) i32 {
// 若为左上角单元格，则终止搜索
if (i == 0 and j == 0) {
return grid[0][0];
}
// 若行列索引越界，则返回 +∞ 代价
if (i < 0 or j < 0) {
return std.math.maxInt(i32);
}
// 若已有记录，则直接返回
if (mem[@as(usize, @intCast(i))][@as(usize, @intCast(j))] != -1) {
return mem[@as(usize, @intCast(i))][@as(usize, @intCast(j))];
}
// 计算从左上角到 (i-1, j) 和 (i, j-1) 的最小路径代价
var up = minPathSumDFSMem(grid, mem, i - 1, j);
var left = minPathSumDFSMem(grid, mem, i, j - 1);
// 返回从左上角到 (i, j) 的最小路径代价
// 记录并返回左上角到 (i, j) 的最小路径代价
mem[@as(usize, @intCast(i))][@as(usize, @intCast(j))] = @min(left, up) + grid[@as(usize, @intCast(i))][@as(usize, @intCast(j))];
return mem[@as(usize, @intCast(i))][@as(usize, @intCast(j))];
}
```
??? pythontutor "可视化运行"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=from%20math%20import%20inf%0A%0Adef%20min_path_sum_dfs_mem%28%0A%20%20%20%20grid%3A%20list%5Blist%5Bint%5D%5D,%20mem%3A%20list%5Blist%5Bint%5D%5D,%20i%3A%20int,%20j%3A%20int%0A%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E6%9C%80%E5%B0%8F%E8%B7%AF%E5%BE%84%E5%92%8C%EF%BC%9A%E8%AE%B0%E5%BF%86%E5%8C%96%E6%90%9C%E7%B4%A2%22%22%22%0A%20%20%20%20%23%20%E8%8B%A5%E4%B8%BA%E5%B7%A6%E4%B8%8A%E8%A7%92%E5%8D%95%E5%85%83%E6%A0%BC%EF%BC%8C%E5%88%99%E7%BB%88%E6%AD%A2%E6%90%9C%E7%B4%A2%0A%20%20%20%20if%20i%20%3D%3D%200%20and%20j%20%3D%3D%200%3A%0A%20%20%20%20%20%20%20%20return%20grid%5B0%5D%5B0%5D%0A%20%20%20%20%23%20%E8%8B%A5%E8%A1%8C%E5%88%97%E7%B4%A2%E5%BC%95%E8%B6%8A%E7%95%8C%EF%BC%8C%E5%88%99%E8%BF%94%E5%9B%9E%20%2B%E2%88%9E%20%E4%BB%A3%E4%BB%B7%0A%20%20%20%20if%20i%20%3C%200%20or%20j%20%3C%200%3A%0A%20%20%20%20%20%20%20%20return%20inf%0A%20%20%20%20%23%20%E8%8B%A5%E5%B7%B2%E6%9C%89%E8%AE%B0%E5%BD%95%EF%BC%8C%E5%88%99%E7%9B%B4%E6%8E%A5%E8%BF%94%E5%9B%9E%0A%20%20%20%20if%20mem%5Bi%5D%5Bj%5D%20!%3D%20-1%3A%0A%20%20%20%20%20%20%20%20return%20mem%5Bi%5D%5Bj%5D%0A%20%20%20%20%23%20%E5%B7%A6%E8%BE%B9%E5%92%8C%E4%B8%8A%E8%BE%B9%E5%8D%95%E5%85%83%E6%A0%BC%E7%9A%84%E6%9C%80%E5%B0%8F%E8%B7%AF%E5%BE%84%E4%BB%A3%E4%BB%B7%0A%20%20%20%20up%20%3D%20min_path_sum_dfs_mem%28grid,%20mem,%20i%20-%201,%20j%29%0A%20%20%20%20left%20%3D%20min_path_sum_dfs_mem%28grid,%20mem,%20i,%20j%20-%201%29%0A%20%20%20%20%23%20%E8%AE%B0%E5%BD%95%E5%B9%B6%E8%BF%94%E5%9B%9E%E5%B7%A6%E4%B8%8A%E8%A7%92%E5%88%B0%20%28i,%20j%29%20%E7%9A%84%E6%9C%80%E5%B0%8F%E8%B7%AF%E5%BE%84%E4%BB%A3%E4%BB%B7%0A%20%20%20%20mem%5Bi%5D%5Bj%5D%20%3D%20min%28left,%20up%29%20%2B%20grid%5Bi%5D%5Bj%5D%0A%20%20%20%20return%20mem%5Bi%5D%5Bj%5D%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20grid%20%3D%20%5B%5B1,%203,%201,%205%5D,%20%5B2,%202,%204,%202%5D,%20%5B5,%203,%202,%201%5D,%20%5B4,%203,%205,%202%5D%5D%0A%20%20%20%20n,%20m%20%3D%20len%28grid%29,%20len%28grid%5B0%5D%29%0A%0A%20%20%20%23%20%E8%AE%B0%E5%BF%86%E5%8C%96%E6%90%9C%E7%B4%A2%0A%20%20%20%20mem%20%3D%20%5B%5B-1%5D%20*%20m%20for%20_%20in%20range%28n%29%5D%0A%20%20%20%20res%20%3D%20min_path_sum_dfs_mem%28grid,%20mem,%20n%20-%201,%20m%20-%201%29%0A%20%20%20%20print%28f%22%E4%BB%8E%E5%B7%A6%E4%B8%8A%E8%A7%92%E5%88%B0%E5%8F%B3%E4%B8%8B%E8%A7%92%E7%9A%84%E5%81%9A%E5%B0%8F%E8%B7%AF%E5%BE%84%E5%92%8C%E4%B8%BA%20%7Bres%7D%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=16&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
@@ -1165,34 +1117,6 @@ $$
end
```
=== "Zig"
```zig title="min_path_sum.zig"
// 最小路径和：动态规划
fn minPathSumDP(comptime grid: anytype) i32 {
comptime var n = grid.len;
comptime var m = grid[0].len;
// 初始化 dp 表
var dp = [_][m]i32{[_]i32{0} ** m} ** n;
dp[0][0] = grid[0][0];
// 状态转移：首行
for (1..m) |j| {
dp[0][j] = dp[0][j - 1] + grid[0][j];
}
// 状态转移：首列
for (1..n) |i| {
dp[i][0] = dp[i - 1][0] + grid[i][0];
}
// 状态转移：其余行和列
for (1..n) |i| {
for (1..m) |j| {
dp[i][j] = @min(dp[i][j - 1], dp[i - 1][j]) + grid[i][j];
}
}
return dp[n - 1][m - 1];
}
```
??? pythontutor "可视化运行"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=from%20math%20import%20inf%0A%0Adef%20min_path_sum_dp%28grid%3A%20list%5Blist%5Bint%5D%5D%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E6%9C%80%E5%B0%8F%E8%B7%AF%E5%BE%84%E5%92%8C%EF%BC%9A%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%22%22%22%0A%20%20%20%20n,%20m%20%3D%20len%28grid%29,%20len%28grid%5B0%5D%29%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%20dp%20%E8%A1%A8%0A%20%20%20%20dp%20%3D%20%5B%5B0%5D%20*%20m%20for%20_%20in%20range%28n%29%5D%0A%20%20%20%20dp%5B0%5D%5B0%5D%20%3D%20grid%5B0%5D%5B0%5D%0A%20%20%20%20%23%20%E7%8A%B6%E6%80%81%E8%BD%AC%E7%A7%BB%EF%BC%9A%E9%A6%96%E8%A1%8C%0A%20%20%20%20for%20j%20in%20range%281,%20m%29%3A%0A%20%20%20%20%20%20%20%20dp%5B0%5D%5Bj%5D%20%3D%20dp%5B0%5D%5Bj%20-%201%5D%20%2B%20grid%5B0%5D%5Bj%5D%0A%20%20%20%20%23%20%E7%8A%B6%E6%80%81%E8%BD%AC%E7%A7%BB%EF%BC%9A%E9%A6%96%E5%88%97%0A%20%20%20%20for%20i%20in%20range%281,%20n%29%3A%0A%20%20%20%20%20%20%20%20dp%5Bi%5D%5B0%5D%20%3D%20dp%5Bi%20-%201%5D%5B0%5D%20%2B%20grid%5Bi%5D%5B0%5D%0A%20%20%20%20%23%20%E7%8A%B6%E6%80%81%E8%BD%AC%E7%A7%BB%EF%BC%9A%E5%85%B6%E4%BD%99%E8%A1%8C%E5%92%8C%E5%88%97%0A%20%20%20%20for%20i%20in%20range%281,%20n%29%3A%0A%20%20%20%20%20%20%20%20for%20j%20in%20range%281,%20m%29%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20dp%5Bi%5D%5Bj%5D%20%3D%20min%28dp%5Bi%5D%5Bj%20-%201%5D,%20dp%5Bi%20-%201%5D%5Bj%5D%29%20%2B%20grid%5Bi%5D%5Bj%5D%0A%20%20%20%20return%20dp%5Bn%20-%201%5D%5Bm%20-%201%5D%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20grid%20%3D%20%5B%5B1,%203,%201,%205%5D,%20%5B2,%202,%204,%202%5D,%20%5B5,%203,%202,%201%5D,%20%5B4,%203,%205,%202%5D%5D%0A%20%20%20%20n,%20m%20%3D%20len%28grid%29,%20len%28grid%5B0%5D%29%0A%0A%20%20%20%20%23%20%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%0A%20%20%20%20res%20%3D%20min_path_sum_dp%28grid%29%0A%20%20%20%20print%28f%22%E4%BB%8E%E5%B7%A6%E4%B8%8A%E8%A7%92%E5%88%B0%E5%8F%B3%E4%B8%8B%E8%A7%92%E7%9A%84%E5%81%9A%E5%B0%8F%E8%B7%AF%E5%BE%84%E5%92%8C%E4%B8%BA%20%7Bres%7D%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=6&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
@@ -1581,32 +1505,6 @@ $$
end
```
=== "Zig"
```zig title="min_path_sum.zig"
// 最小路径和：空间优化后的动态规划
fn minPathSumDPComp(comptime grid: anytype) i32 {
comptime var n = grid.len;
comptime var m = grid[0].len;
// 初始化 dp 表
var dp = [_]i32{0} ** m;
// 状态转移：首行
dp[0] = grid[0][0];
for (1..m) |j| {
dp[j] = dp[j - 1] + grid[0][j];
}
// 状态转移：其余行
for (1..n) |i| {
// 状态转移：首列
dp[0] = dp[0] + grid[i][0];
for (1..m) |j| {
dp[j] = @min(dp[j - 1], dp[j]) + grid[i][j];
}
}
return dp[m - 1];
}
```
??? pythontutor "可视化运行"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=from%20math%20import%20inf%0A%0Adef%20min_path_sum_dp_comp%28grid%3A%20list%5Blist%5Bint%5D%5D%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E6%9C%80%E5%B0%8F%E8%B7%AF%E5%BE%84%E5%92%8C%EF%BC%9A%E7%A9%BA%E9%97%B4%E4%BC%98%E5%8C%96%E5%90%8E%E7%9A%84%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%22%22%22%0A%20%20%20%20n,%20m%20%3D%20len%28grid%29,%20len%28grid%5B0%5D%29%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%20dp%20%E8%A1%A8%0A%20%20%20%20dp%20%3D%20%5B0%5D%20*%20m%0A%20%20%20%20%23%20%E7%8A%B6%E6%80%81%E8%BD%AC%E7%A7%BB%EF%BC%9A%E9%A6%96%E8%A1%8C%0A%20%20%20%20dp%5B0%5D%20%3D%20grid%5B0%5D%5B0%5D%0A%20%20%20%20for%20j%20in%20range%281,%20m%29%3A%0A%20%20%20%20%20%20%20%20dp%5Bj%5D%20%3D%20dp%5Bj%20-%201%5D%20%2B%20grid%5B0%5D%5Bj%5D%0A%20%20%20%20%23%20%E7%8A%B6%E6%80%81%E8%BD%AC%E7%A7%BB%EF%BC%9A%E5%85%B6%E4%BD%99%E8%A1%8C%0A%20%20%20%20for%20i%20in%20range%281,%20n%29%3A%0A%20%20%20%20%20%20%20%20%23%20%E7%8A%B6%E6%80%81%E8%BD%AC%E7%A7%BB%EF%BC%9A%E9%A6%96%E5%88%97%0A%20%20%20%20%20%20%20%20dp%5B0%5D%20%3D%20dp%5B0%5D%20%2B%20grid%5Bi%5D%5B0%5D%0A%20%20%20%20%20%20%20%20%23%20%E7%8A%B6%E6%80%81%E8%BD%AC%E7%A7%BB%EF%BC%9A%E5%85%B6%E4%BD%99%E5%88%97%0A%20%20%20%20%20%20%20%20for%20j%20in%20range%281,%20m%29%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20dp%5Bj%5D%20%3D%20min%28dp%5Bj%20-%201%5D,%20dp%5Bj%5D%29%20%2B%20grid%5Bi%5D%5Bj%5D%0A%20%20%20%20return%20dp%5Bm%20-%201%5D%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20grid%20%3D%20%5B%5B1,%203,%201,%205%5D,%20%5B2,%202,%204,%202%5D,%20%5B5,%203,%202,%201%5D,%20%5B4,%203,%205,%202%5D%5D%0A%20%20%20%20n,%20m%20%3D%20len%28grid%29,%20len%28grid%5B0%5D%29%0A%0A%20%20%20%20%23%20%E7%A9%BA%E9%97%B4%E4%BC%98%E5%8C%96%E5%90%8E%E7%9A%84%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%0A%20%20%20%20res%20%3D%20min_path_sum_dp_comp%28grid%29%0A%20%20%20%20print%28f%22%E4%BB%8E%E5%B7%A6%E4%B8%8A%E8%A7%92%E5%88%B0%E5%8F%B3%E4%B8%8B%E8%A7%92%E7%9A%84%E5%81%9A%E5%B0%8F%E8%B7%AF%E5%BE%84%E5%92%8C%E4%B8%BA%20%7Bres%7D%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=6&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
docs/chapter_dynamic_programming/edit_distance_problem.md
-65
View File
@@ -477,37 +477,6 @@ $$
end
```
=== "Zig"
```zig title="edit_distance.zig"
// 编辑距离：动态规划
fn editDistanceDP(comptime s: []const u8, comptime t: []const u8) i32 {
comptime var n = s.len;
comptime var m = t.len;
var dp = [_][m + 1]i32{[_]i32{0} ** (m + 1)} ** (n + 1);
// 状态转移：首行首列
for (1..n + 1) |i| {
dp[i][0] = @intCast(i);
}
for (1..m + 1) |j| {
dp[0][j] = @intCast(j);
}
// 状态转移：其余行和列
for (1..n + 1) |i| {
for (1..m + 1) |j| {
if (s[i - 1] == t[j - 1]) {
// 若两字符相等，则直接跳过此两字符
dp[i][j] = dp[i - 1][j - 1];
} else {
// 最少编辑步数 = 插入、删除、替换这三种操作的最少编辑步数 + 1
dp[i][j] = @min(@min(dp[i][j - 1], dp[i - 1][j]), dp[i - 1][j - 1]) + 1;
}
}
}
return dp[n][m];
}
```
??? pythontutor "可视化运行"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20edit_distance_dp%28s%3A%20str,%20t%3A%20str%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E7%BC%96%E8%BE%91%E8%B7%9D%E7%A6%BB%EF%BC%9A%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%22%22%22%0A%20%20%20%20n,%20m%20%3D%20len%28s%29,%20len%28t%29%0A%20%20%20%20dp%20%3D%20%5B%5B0%5D%20*%20%28m%20%2B%201%29%20for%20_%20in%20range%28n%20%2B%201%29%5D%0A%20%20%20%20%23%20%E7%8A%B6%E6%80%81%E8%BD%AC%E7%A7%BB%EF%BC%9A%E9%A6%96%E8%A1%8C%E9%A6%96%E5%88%97%0A%20%20%20%20for%20i%20in%20range%281,%20n%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20dp%5Bi%5D%5B0%5D%20%3D%20i%0A%20%20%20%20for%20j%20in%20range%281,%20m%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20dp%5B0%5D%5Bj%5D%20%3D%20j%0A%20%20%20%20%23%20%E7%8A%B6%E6%80%81%E8%BD%AC%E7%A7%BB%EF%BC%9A%E5%85%B6%E4%BD%99%E8%A1%8C%E5%92%8C%E5%88%97%0A%20%20%20%20for%20i%20in%20range%281,%20n%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20for%20j%20in%20range%281,%20m%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20if%20s%5Bi%20-%201%5D%20%3D%3D%20t%5Bj%20-%201%5D%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%23%20%E8%8B%A5%E4%B8%A4%E5%AD%97%E7%AC%A6%E7%9B%B8%E7%AD%89%EF%BC%8C%E5%88%99%E7%9B%B4%E6%8E%A5%E8%B7%B3%E8%BF%87%E6%AD%A4%E4%B8%A4%E5%AD%97%E7%AC%A6%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20dp%5Bi%5D%5Bj%5D%20%3D%20dp%5Bi%20-%201%5D%5Bj%20-%201%5D%0A%20%20%20%20%20%20%20%20%20%20%20%20else%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%23%20%E6%9C%80%E5%B0%91%E7%BC%96%E8%BE%91%E6%AD%A5%E6%95%B0%20%3D%20%E6%8F%92%E5%85%A5%E3%80%81%E5%88%A0%E9%99%A4%E3%80%81%E6%9B%BF%E6%8D%A2%E8%BF%99%E4%B8%89%E7%A7%8D%E6%93%8D%E4%BD%9C%E7%9A%84%E6%9C%80%E5%B0%91%E7%BC%96%E8%BE%91%E6%AD%A5%E6%95%B0%20%2B%201%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20dp%5Bi%5D%5Bj%5D%20%3D%20min%28dp%5Bi%5D%5Bj%20-%201%5D,%20dp%5Bi%20-%201%5D%5Bj%5D,%20dp%5Bi%20-%201%5D%5Bj%20-%201%5D%29%20%2B%201%0A%20%20%20%20return%20dp%5Bn%5D%5Bm%5D%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20s%20%3D%20%22bag%22%0A%20%20%20%20t%20%3D%20%22pack%22%0A%20%20%20%20n,%20m%20%3D%20len%28s%29,%20len%28t%29%0A%0A%20%20%20%20%23%20%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%0A%20%20%20%20res%20%3D%20edit_distance_dp%28s,%20t%29%0A%20%20%20%20print%28f%22%E5%B0%86%20%7Bs%7D%20%E6%9B%B4%E6%94%B9%E4%B8%BA%20%7Bt%7D%20%E6%9C%80%E5%B0%91%E9%9C%80%E8%A6%81%E7%BC%96%E8%BE%91%20%7Bres%7D%20%E6%AD%A5%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=6&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
@@ -997,40 +966,6 @@ $$
end
```
=== "Zig"
```zig title="edit_distance.zig"
// 编辑距离：空间优化后的动态规划
fn editDistanceDPComp(comptime s: []const u8, comptime t: []const u8) i32 {
comptime var n = s.len;
comptime var m = t.len;
var dp = [_]i32{0} ** (m + 1);
// 状态转移：首行
for (1..m + 1) |j| {
dp[j] = @intCast(j);
}
// 状态转移：其余行
for (1..n + 1) |i| {
// 状态转移：首列
var leftup = dp[0]; // 暂存 dp[i-1, j-1]
dp[0] = @intCast(i);
// 状态转移：其余列
for (1..m + 1) |j| {
var temp = dp[j];
if (s[i - 1] == t[j - 1]) {
// 若两字符相等，则直接跳过此两字符
dp[j] = leftup;
} else {
// 最少编辑步数 = 插入、删除、替换这三种操作的最少编辑步数 + 1
dp[j] = @min(@min(dp[j - 1], dp[j]), leftup) + 1;
}
leftup = temp; // 更新为下一轮的 dp[i-1, j-1]
}
}
return dp[m];
}
```
??? pythontutor "可视化运行"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20edit_distance_dp_comp%28s%3A%20str,%20t%3A%20str%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E7%BC%96%E8%BE%91%E8%B7%9D%E7%A6%BB%EF%BC%9A%E7%A9%BA%E9%97%B4%E4%BC%98%E5%8C%96%E5%90%8E%E7%9A%84%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%22%22%22%0A%20%20%20%20n,%20m%20%3D%20len%28s%29,%20len%28t%29%0A%20%20%20%20dp%20%3D%20%5B0%5D%20*%20%28m%20%2B%201%29%0A%20%20%20%20%23%20%E7%8A%B6%E6%80%81%E8%BD%AC%E7%A7%BB%EF%BC%9A%E9%A6%96%E8%A1%8C%0A%20%20%20%20for%20j%20in%20range%281,%20m%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20dp%5Bj%5D%20%3D%20j%0A%20%20%20%20%23%20%E7%8A%B6%E6%80%81%E8%BD%AC%E7%A7%BB%EF%BC%9A%E5%85%B6%E4%BD%99%E8%A1%8C%0A%20%20%20%20for%20i%20in%20range%281,%20n%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20%23%20%E7%8A%B6%E6%80%81%E8%BD%AC%E7%A7%BB%EF%BC%9A%E9%A6%96%E5%88%97%0A%20%20%20%20%20%20%20%20leftup%20%3D%20dp%5B0%5D%20%20%23%20%E6%9A%82%E5%AD%98%20dp%5Bi-1,%20j-1%5D%0A%20%20%20%20%20%20%20%20dp%5B0%5D%20%2B%3D%201%0A%20%20%20%20%20%20%20%20%23%20%E7%8A%B6%E6%80%81%E8%BD%AC%E7%A7%BB%EF%BC%9A%E5%85%B6%E4%BD%99%E5%88%97%0A%20%20%20%20%20%20%20%20for%20j%20in%20range%281,%20m%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20temp%20%3D%20dp%5Bj%5D%0A%20%20%20%20%20%20%20%20%20%20%20%20if%20s%5Bi%20-%201%5D%20%3D%3D%20t%5Bj%20-%201%5D%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%23%20%E8%8B%A5%E4%B8%A4%E5%AD%97%E7%AC%A6%E7%9B%B8%E7%AD%89%EF%BC%8C%E5%88%99%E7%9B%B4%E6%8E%A5%E8%B7%B3%E8%BF%87%E6%AD%A4%E4%B8%A4%E5%AD%97%E7%AC%A6%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20dp%5Bj%5D%20%3D%20leftup%0A%20%20%20%20%20%20%20%20%20%20%20%20else%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%23%20%E6%9C%80%E5%B0%91%E7%BC%96%E8%BE%91%E6%AD%A5%E6%95%B0%20%3D%20%E6%8F%92%E5%85%A5%E3%80%81%E5%88%A0%E9%99%A4%E3%80%81%E6%9B%BF%E6%8D%A2%E8%BF%99%E4%B8%89%E7%A7%8D%E6%93%8D%E4%BD%9C%E7%9A%84%E6%9C%80%E5%B0%91%E7%BC%96%E8%BE%91%E6%AD%A5%E6%95%B0%20%2B%201%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20dp%5Bj%5D%20%3D%20min%28dp%5Bj%20-%201%5D,%20dp%5Bj%5D,%20leftup%29%20%2B%201%0A%20%20%20%20%20%20%20%20%20%20%20%20leftup%20%3D%20temp%20%20%23%20%E6%9B%B4%E6%96%B0%E4%B8%BA%E4%B8%8B%E4%B8%80%E8%BD%AE%E7%9A%84%20dp%5Bi-1,%20j-1%5D%0A%20%20%20%20return%20dp%5Bm%5D%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20s%20%3D%20%22bag%22%0A%20%20%20%20t%20%3D%20%22pack%22%0A%20%20%20%20n,%20m%20%3D%20len%28s%29,%20len%28t%29%0A%0A%20%20%20%20%23%20%E7%A9%BA%E9%97%B4%E4%BC%98%E5%8C%96%E5%90%8E%E7%9A%84%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%0A%20%20%20%20res%20%3D%20edit_distance_dp_comp%28s,%20t%29%0A%20%20%20%20print%28f%22%E5%B0%86%20%7Bs%7D%20%E6%9B%B4%E6%94%B9%E4%B8%BA%20%7Bt%7D%20%E6%9C%80%E5%B0%91%E9%9C%80%E8%A6%81%E7%BC%96%E8%BE%91%20%7Bres%7D%20%E6%AD%A5%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=6&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
docs/chapter_dynamic_programming/intro_to_dynamic_programming.md
-122
View File
@@ -420,39 +420,6 @@ comments: true
end
```
=== "Zig"
```zig title="climbing_stairs_backtrack.zig"
// 回溯
fn backtrack(choices: []i32, state: i32, n: i32, res: std.ArrayList(i32)) void {
// 当爬到第 n 阶时，方案数量加 1
if (state == n) {
res.items[0] = res.items[0] + 1;
}
// 遍历所有选择
for (choices) |choice| {
// 剪枝：不允许越过第 n 阶
if (state + choice > n) {
continue;
}
// 尝试：做出选择，更新状态
backtrack(choices, state + choice, n, res);
// 回退
}
}
// 爬楼梯：回溯
fn climbingStairsBacktrack(n: usize) !i32 {
var choices = [_]i32{ 1, 2 }; // 可选择向上爬 1 阶或 2 阶
var state: i32 = 0; // 从第 0 阶开始爬
var res = std.ArrayList(i32).init(std.heap.page_allocator);
defer res.deinit();
try res.append(0); // 使用 res[0] 记录方案数量
backtrack(&choices, state, @intCast(n), res);
return res.items[0];
}
```
??? pythontutor "可视化运行"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20backtrack%28choices%3A%20list%5Bint%5D,%20state%3A%20int,%20n%3A%20int,%20res%3A%20list%5Bint%5D%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E5%9B%9E%E6%BA%AF%22%22%22%0A%20%20%20%20%23%20%E5%BD%93%E7%88%AC%E5%88%B0%E7%AC%AC%20n%20%E9%98%B6%E6%97%B6%EF%BC%8C%E6%96%B9%E6%A1%88%E6%95%B0%E9%87%8F%E5%8A%A0%201%0A%20%20%20%20if%20state%20%3D%3D%20n%3A%0A%20%20%20%20%20%20%20%20res%5B0%5D%20%2B%3D%201%0A%20%20%20%20%23%20%E9%81%8D%E5%8E%86%E6%89%80%E6%9C%89%E9%80%89%E6%8B%A9%0A%20%20%20%20for%20choice%20in%20choices%3A%0A%20%20%20%20%20%20%20%20%23%20%E5%89%AA%E6%9E%9D%EF%BC%9A%E4%B8%8D%E5%85%81%E8%AE%B8%E8%B6%8A%E8%BF%87%E7%AC%AC%20n%20%E9%98%B6%0A%20%20%20%20%20%20%20%20if%20state%20%2B%20choice%20%3E%20n%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20continue%0A%20%20%20%20%20%20%20%20%23%20%E5%B0%9D%E8%AF%95%EF%BC%9A%E5%81%9A%E5%87%BA%E9%80%89%E6%8B%A9%EF%BC%8C%E6%9B%B4%E6%96%B0%E7%8A%B6%E6%80%81%0A%20%20%20%20%20%20%20%20backtrack%28choices,%20state%20%2B%20choice,%20n,%20res%29%0A%20%20%20%20%20%20%20%20%23%20%E5%9B%9E%E9%80%80%0A%0A%0Adef%20climbing_stairs_backtrack%28n%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E7%88%AC%E6%A5%BC%E6%A2%AF%EF%BC%9A%E5%9B%9E%E6%BA%AF%22%22%22%0A%20%20%20%20choices%20%3D%20%5B1,%202%5D%20%20%23%20%E5%8F%AF%E9%80%89%E6%8B%A9%E5%90%91%E4%B8%8A%E7%88%AC%201%20%E9%98%B6%E6%88%96%202%20%E9%98%B6%0A%20%20%20%20state%20%3D%200%20%20%23%20%E4%BB%8E%E7%AC%AC%200%20%E9%98%B6%E5%BC%80%E5%A7%8B%E7%88%AC%0A%20%20%20%20res%20%3D%20%5B0%5D%20%20%23%20%E4%BD%BF%E7%94%A8%20res%5B0%5D%20%E8%AE%B0%E5%BD%95%E6%96%B9%E6%A1%88%E6%95%B0%E9%87%8F%0A%20%20%20%20backtrack%28choices,%20state,%20n,%20res%29%0A%20%20%20%20return%20res%5B0%5D%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%204%0A%0A%20%20%20%20res%20%3D%20climbing_stairs_backtrack%28n%29%0A%20%20%20%20print%28f%22%E7%88%AC%20%7Bn%7D%20%E9%98%B6%E6%A5%BC%E6%A2%AF%E5%85%B1%E6%9C%89%20%7Bres%7D%20%E7%A7%8D%E6%96%B9%E6%A1%88%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=5&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
@@ -728,26 +695,6 @@ $$
end
```
=== "Zig"
```zig title="climbing_stairs_dfs.zig"
// 搜索
fn dfs(i: usize) i32 {
// 已知 dp[1] 和 dp[2] ，返回之
if (i == 1 or i == 2) {
return @intCast(i);
}
// dp[i] = dp[i-1] + dp[i-2]
var count = dfs(i - 1) + dfs(i - 2);
return count;
}
// 爬楼梯：搜索
fn climbingStairsDFS(comptime n: usize) i32 {
return dfs(n);
}
```
??? pythontutor "可视化运行"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20dfs%28i%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E6%90%9C%E7%B4%A2%22%22%22%0A%20%20%20%20%23%20%E5%B7%B2%E7%9F%A5%20dp%5B1%5D%20%E5%92%8C%20dp%5B2%5D%20%EF%BC%8C%E8%BF%94%E5%9B%9E%E4%B9%8B%0A%20%20%20%20if%20i%20%3D%3D%201%20or%20i%20%3D%3D%202%3A%0A%20%20%20%20%20%20%20%20return%20i%0A%20%20%20%20%23%20dp%5Bi%5D%20%3D%20dp%5Bi-1%5D%20%2B%20dp%5Bi-2%5D%0A%20%20%20%20count%20%3D%20dfs%28i%20-%201%29%20%2B%20dfs%28i%20-%202%29%0A%20%20%20%20return%20count%0A%0A%0Adef%20climbing_stairs_dfs%28n%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E7%88%AC%E6%A5%BC%E6%A2%AF%EF%BC%9A%E6%90%9C%E7%B4%A2%22%22%22%0A%20%20%20%20return%20dfs%28n%29%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%209%0A%0A%20%20%20%20res%20%3D%20climbing_stairs_dfs%28n%29%0A%20%20%20%20print%28f%22%E7%88%AC%20%7Bn%7D%20%E9%98%B6%E6%A5%BC%E6%A2%AF%E5%85%B1%E6%9C%89%20%7Bres%7D%20%E7%A7%8D%E6%96%B9%E6%A1%88%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=5&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
@@ -1115,34 +1062,6 @@ $$
end
```
=== "Zig"
```zig title="climbing_stairs_dfs_mem.zig"
// 记忆化搜索
fn dfs(i: usize, mem: []i32) i32 {
// 已知 dp[1] 和 dp[2] ，返回之
if (i == 1 or i == 2) {
return @intCast(i);
}
// 若存在记录 dp[i] ，则直接返回之
if (mem[i] != -1) {
return mem[i];
}
// dp[i] = dp[i-1] + dp[i-2]
var count = dfs(i - 1, mem) + dfs(i - 2, mem);
// 记录 dp[i]
mem[i] = count;
return count;
}
// 爬楼梯：记忆化搜索
fn climbingStairsDFSMem(comptime n: usize) i32 {
// mem[i] 记录爬到第 i 阶的方案总数，-1 代表无记录
var mem = [_]i32{ -1 } ** (n + 1);
return dfs(n, &mem);
}
```
??? pythontutor "可视化运行"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20dfs%28i%3A%20int,%20mem%3A%20list%5Bint%5D%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E8%AE%B0%E5%BF%86%E5%8C%96%E6%90%9C%E7%B4%A2%22%22%22%0A%20%20%20%20%23%20%E5%B7%B2%E7%9F%A5%20dp%5B1%5D%20%E5%92%8C%20dp%5B2%5D%20%EF%BC%8C%E8%BF%94%E5%9B%9E%E4%B9%8B%0A%20%20%20%20if%20i%20%3D%3D%201%20or%20i%20%3D%3D%202%3A%0A%20%20%20%20%20%20%20%20return%20i%0A%20%20%20%20%23%20%E8%8B%A5%E5%AD%98%E5%9C%A8%E8%AE%B0%E5%BD%95%20dp%5Bi%5D%20%EF%BC%8C%E5%88%99%E7%9B%B4%E6%8E%A5%E8%BF%94%E5%9B%9E%E4%B9%8B%0A%20%20%20%20if%20mem%5Bi%5D%20!%3D%20-1%3A%0A%20%20%20%20%20%20%20%20return%20mem%5Bi%5D%0A%20%20%20%20%23%20dp%5Bi%5D%20%3D%20dp%5Bi-1%5D%20%2B%20dp%5Bi-2%5D%0A%20%20%20%20count%20%3D%20dfs%28i%20-%201,%20mem%29%20%2B%20dfs%28i%20-%202,%20mem%29%0A%20%20%20%20%23%20%E8%AE%B0%E5%BD%95%20dp%5Bi%5D%0A%20%20%20%20mem%5Bi%5D%20%3D%20count%0A%20%20%20%20return%20count%0A%0A%0Adef%20climbing_stairs_dfs_mem%28n%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E7%88%AC%E6%A5%BC%E6%A2%AF%EF%BC%9A%E8%AE%B0%E5%BF%86%E5%8C%96%E6%90%9C%E7%B4%A2%22%22%22%0A%20%20%20%20%23%20mem%5Bi%5D%20%E8%AE%B0%E5%BD%95%E7%88%AC%E5%88%B0%E7%AC%AC%20i%20%E9%98%B6%E7%9A%84%E6%96%B9%E6%A1%88%E6%80%BB%E6%95%B0%EF%BC%8C-1%20%E4%BB%A3%E8%A1%A8%E6%97%A0%E8%AE%B0%E5%BD%95%0A%20%20%20%20mem%20%3D%20%5B-1%5D%20*%20%28n%20%2B%201%29%0A%20%20%20%20return%20dfs%28n,%20mem%29%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%209%0A%0A%20%20%20%20res%20%3D%20climbing_stairs_dfs_mem%28n%29%0A%20%20%20%20print%28f%22%E7%88%AC%20%7Bn%7D%20%E9%98%B6%E6%A5%BC%E6%A2%AF%E5%85%B1%E6%9C%89%20%7Bres%7D%20%E7%A7%8D%E6%96%B9%E6%A1%88%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=5&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
@@ -1419,28 +1338,6 @@ $$
end
```
=== "Zig"
```zig title="climbing_stairs_dp.zig"
// 爬楼梯：动态规划
fn climbingStairsDP(comptime n: usize) i32 {
// 已知 dp[1] 和 dp[2] ，返回之
if (n == 1 or n == 2) {
return @intCast(n);
}
// 初始化 dp 表，用于存储子问题的解
var dp = [_]i32{-1} ** (n + 1);
// 初始状态：预设最小子问题的解
dp[1] = 1;
dp[2] = 2;
// 状态转移：从较小子问题逐步求解较大子问题
for (3..n + 1) |i| {
dp[i] = dp[i - 1] + dp[i - 2];
}
return dp[n];
}
```
??? pythontutor "可视化运行"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20climbing_stairs_dp%28n%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E7%88%AC%E6%A5%BC%E6%A2%AF%EF%BC%9A%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%22%22%22%0A%20%20%20%20if%20n%20%3D%3D%201%20or%20n%20%3D%3D%202%3A%0A%20%20%20%20%20%20%20%20return%20n%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%20dp%20%E8%A1%A8%EF%BC%8C%E7%94%A8%E4%BA%8E%E5%AD%98%E5%82%A8%E5%AD%90%E9%97%AE%E9%A2%98%E7%9A%84%E8%A7%A3%0A%20%20%20%20dp%20%3D%20%5B0%5D%20*%20%28n%20%2B%201%29%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E7%8A%B6%E6%80%81%EF%BC%9A%E9%A2%84%E8%AE%BE%E6%9C%80%E5%B0%8F%E5%AD%90%E9%97%AE%E9%A2%98%E7%9A%84%E8%A7%A3%0A%20%20%20%20dp%5B1%5D,%20dp%5B2%5D%20%3D%201,%202%0A%20%20%20%20%23%20%E7%8A%B6%E6%80%81%E8%BD%AC%E7%A7%BB%EF%BC%9A%E4%BB%8E%E8%BE%83%E5%B0%8F%E5%AD%90%E9%97%AE%E9%A2%98%E9%80%90%E6%AD%A5%E6%B1%82%E8%A7%A3%E8%BE%83%E5%A4%A7%E5%AD%90%E9%97%AE%E9%A2%98%0A%20%20%20%20for%20i%20in%20range%283,%20n%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20dp%5Bi%5D%20%3D%20dp%5Bi%20-%201%5D%20%2B%20dp%5Bi%20-%202%5D%0A%20%20%20%20return%20dp%5Bn%5D%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%209%0A%0A%20%20%20%20res%20%3D%20climbing_stairs_dp%28n%29%0A%20%20%20%20print%28f%22%E7%88%AC%20%7Bn%7D%20%E9%98%B6%E6%A5%BC%E6%A2%AF%E5%85%B1%E6%9C%89%20%7Bres%7D%20%E7%A7%8D%E6%96%B9%E6%A1%88%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=4&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
@@ -1678,25 +1575,6 @@ $$
end
```
=== "Zig"
```zig title="climbing_stairs_dp.zig"
// 爬楼梯：空间优化后的动态规划
fn climbingStairsDPComp(comptime n: usize) i32 {
if (n == 1 or n == 2) {
return @intCast(n);
}
var a: i32 = 1;
var b: i32 = 2;
for (3..n + 1) |_| {
var tmp = b;
b = a + b;
a = tmp;
}
return b;
}
```
??? pythontutor "可视化运行"
<div style="height: 477px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20climbing_stairs_dp_comp%28n%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E7%88%AC%E6%A5%BC%E6%A2%AF%EF%BC%9A%E7%A9%BA%E9%97%B4%E4%BC%98%E5%8C%96%E5%90%8E%E7%9A%84%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%22%22%22%0A%20%20%20%20if%20n%20%3D%3D%201%20or%20n%20%3D%3D%202%3A%0A%20%20%20%20%20%20%20%20return%20n%0A%20%20%20%20a,%20b%20%3D%201,%202%0A%20%20%20%20for%20_%20in%20range%283,%20n%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20a,%20b%20%3D%20b,%20a%20%2B%20b%0A%20%20%20%20return%20b%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%209%0A%0A%20%20%20%20res%20%3D%20climbing_stairs_dp_comp%28n%29%0A%20%20%20%20print%28f%22%E7%88%AC%20%7Bn%7D%20%E9%98%B6%E6%A5%BC%E6%A2%AF%E5%85%B1%E6%9C%89%20%7Bres%7D%20%E7%A7%8D%E6%96%B9%E6%A1%88%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=4&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
docs/chapter_dynamic_programming/knapsack_problem.md
-94
View File
@@ -338,27 +338,6 @@ $$
end
```
=== "Zig"
```zig title="knapsack.zig"
// 0-1 背包：暴力搜索
fn knapsackDFS(wgt: []i32, val: []i32, i: usize, c: usize) i32 {
// 若已选完所有物品或背包无剩余容量，则返回价值 0
if (i == 0 or c == 0) {
return 0;
}
// 若超过背包容量，则只能选择不放入背包
if (wgt[i - 1] > c) {
return knapsackDFS(wgt, val, i - 1, c);
}
// 计算不放入和放入物品 i 的最大价值
var no = knapsackDFS(wgt, val, i - 1, c);
var yes = knapsackDFS(wgt, val, i - 1, c - @as(usize, @intCast(wgt[i - 1]))) + val[i - 1];
// 返回两种方案中价值更大的那一个
return @max(no, yes);
}
```
??? pythontutor "可视化运行"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20knapsack_dfs%28wgt%3A%20list%5Bint%5D,%20val%3A%20list%5Bint%5D,%20i%3A%20int,%20c%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%220-1%20%E8%83%8C%E5%8C%85%EF%BC%9A%E6%9A%B4%E5%8A%9B%E6%90%9C%E7%B4%A2%22%22%22%0A%20%20%20%20%23%20%E8%8B%A5%E5%B7%B2%E9%80%89%E5%AE%8C%E6%89%80%E6%9C%89%E7%89%A9%E5%93%81%E6%88%96%E8%83%8C%E5%8C%85%E6%97%A0%E5%89%A9%E4%BD%99%E5%AE%B9%E9%87%8F%EF%BC%8C%E5%88%99%E8%BF%94%E5%9B%9E%E4%BB%B7%E5%80%BC%200%0A%20%20%20%20if%20i%20%3D%3D%200%20or%20c%20%3D%3D%200%3A%0A%20%20%20%20%20%20%20%20return%200%0A%20%20%20%20%23%20%E8%8B%A5%E8%B6%85%E8%BF%87%E8%83%8C%E5%8C%85%E5%AE%B9%E9%87%8F%EF%BC%8C%E5%88%99%E5%8F%AA%E8%83%BD%E9%80%89%E6%8B%A9%E4%B8%8D%E6%94%BE%E5%85%A5%E8%83%8C%E5%8C%85%0A%20%20%20%20if%20wgt%5Bi%20-%201%5D%20%3E%20c%3A%0A%20%20%20%20%20%20%20%20return%20knapsack_dfs%28wgt,%20val,%20i%20-%201,%20c%29%0A%20%20%20%20%23%20%E8%AE%A1%E7%AE%97%E4%B8%8D%E6%94%BE%E5%85%A5%E5%92%8C%E6%94%BE%E5%85%A5%E7%89%A9%E5%93%81%20i%20%E7%9A%84%E6%9C%80%E5%A4%A7%E4%BB%B7%E5%80%BC%0A%20%20%20%20no%20%3D%20knapsack_dfs%28wgt,%20val,%20i%20-%201,%20c%29%0A%20%20%20%20yes%20%3D%20knapsack_dfs%28wgt,%20val,%20i%20-%201,%20c%20-%20wgt%5Bi%20-%201%5D%29%20%2B%20val%5Bi%20-%201%5D%0A%20%20%20%20%23%20%E8%BF%94%E5%9B%9E%E4%B8%A4%E7%A7%8D%E6%96%B9%E6%A1%88%E4%B8%AD%E4%BB%B7%E5%80%BC%E6%9B%B4%E5%A4%A7%E7%9A%84%E9%82%A3%E4%B8%80%E4%B8%AA%0A%20%20%20%20return%20max%28no,%20yes%29%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20wgt%20%3D%20%5B10,%2020,%2030,%2040,%2050%5D%0A%20%20%20%20val%20%3D%20%5B50,%20120,%20150,%20210,%20240%5D%0A%20%20%20%20cap%20%3D%2050%0A%20%20%20%20n%20%3D%20len%28wgt%29%0A%0A%20%20%20%20%23%20%E6%9A%B4%E5%8A%9B%E6%90%9C%E7%B4%A2%0A%20%20%20%20res%20%3D%20knapsack_dfs%28wgt,%20val,%20n,%20cap%29%0A%20%20%20%20print%28f%22%E4%B8%8D%E8%B6%85%E8%BF%87%E8%83%8C%E5%8C%85%E5%AE%B9%E9%87%8F%E7%9A%84%E6%9C%80%E5%A4%A7%E7%89%A9%E5%93%81%E4%BB%B7%E5%80%BC%E4%B8%BA%20%7Bres%7D%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=7&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
@@ -727,32 +706,6 @@ $$
end
```
=== "Zig"
```zig title="knapsack.zig"
// 0-1 背包：记忆化搜索
fn knapsackDFSMem(wgt: []i32, val: []i32, mem: anytype, i: usize, c: usize) i32 {
// 若已选完所有物品或背包无剩余容量，则返回价值 0
if (i == 0 or c == 0) {
return 0;
}
// 若已有记录，则直接返回
if (mem[i][c] != -1) {
return mem[i][c];
}
// 若超过背包容量，则只能选择不放入背包
if (wgt[i - 1] > c) {
return knapsackDFSMem(wgt, val, mem, i - 1, c);
}
// 计算不放入和放入物品 i 的最大价值
var no = knapsackDFSMem(wgt, val, mem, i - 1, c);
var yes = knapsackDFSMem(wgt, val, mem, i - 1, c - @as(usize, @intCast(wgt[i - 1]))) + val[i - 1];
// 记录并返回两种方案中价值更大的那一个
mem[i][c] = @max(no, yes);
return mem[i][c];
}
```
??? pythontutor "可视化运行"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20knapsack_dfs_mem%28%0A%20%20%20%20wgt%3A%20list%5Bint%5D,%20val%3A%20list%5Bint%5D,%20mem%3A%20list%5Blist%5Bint%5D%5D,%20i%3A%20int,%20c%3A%20int%0A%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%220-1%20%E8%83%8C%E5%8C%85%EF%BC%9A%E8%AE%B0%E5%BF%86%E5%8C%96%E6%90%9C%E7%B4%A2%22%22%22%0A%20%20%20%20%23%20%E8%8B%A5%E5%B7%B2%E9%80%89%E5%AE%8C%E6%89%80%E6%9C%89%E7%89%A9%E5%93%81%E6%88%96%E8%83%8C%E5%8C%85%E6%97%A0%E5%89%A9%E4%BD%99%E5%AE%B9%E9%87%8F%EF%BC%8C%E5%88%99%E8%BF%94%E5%9B%9E%E4%BB%B7%E5%80%BC%200%0A%20%20%20%20if%20i%20%3D%3D%200%20or%20c%20%3D%3D%200%3A%0A%20%20%20%20%20%20%20%20return%200%0A%20%20%20%20%23%20%E8%8B%A5%E5%B7%B2%E6%9C%89%E8%AE%B0%E5%BD%95%EF%BC%8C%E5%88%99%E7%9B%B4%E6%8E%A5%E8%BF%94%E5%9B%9E%0A%20%20%20%20if%20mem%5Bi%5D%5Bc%5D%20!%3D%20-1%3A%0A%20%20%20%20%20%20%20%20return%20mem%5Bi%5D%5Bc%5D%0A%20%20%20%20%23%20%E8%8B%A5%E8%B6%85%E8%BF%87%E8%83%8C%E5%8C%85%E5%AE%B9%E9%87%8F%EF%BC%8C%E5%88%99%E5%8F%AA%E8%83%BD%E9%80%89%E6%8B%A9%E4%B8%8D%E6%94%BE%E5%85%A5%E8%83%8C%E5%8C%85%0A%20%20%20%20if%20wgt%5Bi%20-%201%5D%20%3E%20c%3A%0A%20%20%20%20%20%20%20%20return%20knapsack_dfs_mem%28wgt,%20val,%20mem,%20i%20-%201,%20c%29%0A%20%20%20%20%23%20%E8%AE%A1%E7%AE%97%E4%B8%8D%E6%94%BE%E5%85%A5%E5%92%8C%E6%94%BE%E5%85%A5%E7%89%A9%E5%93%81%20i%20%E7%9A%84%E6%9C%80%E5%A4%A7%E4%BB%B7%E5%80%BC%0A%20%20%20%20no%20%3D%20knapsack_dfs_mem%28wgt,%20val,%20mem,%20i%20-%201,%20c%29%0A%20%20%20%20yes%20%3D%20knapsack_dfs_mem%28wgt,%20val,%20mem,%20i%20-%201,%20c%20-%20wgt%5Bi%20-%201%5D%29%20%2B%20val%5Bi%20-%201%5D%0A%20%20%20%20%23%20%E8%AE%B0%E5%BD%95%E5%B9%B6%E8%BF%94%E5%9B%9E%E4%B8%A4%E7%A7%8D%E6%96%B9%E6%A1%88%E4%B8%AD%E4%BB%B7%E5%80%BC%E6%9B%B4%E5%A4%A7%E7%9A%84%E9%82%A3%E4%B8%80%E4%B8%AA%0A%20%20%20%20mem%5Bi%5D%5Bc%5D%20%3D%20max%28no,%20yes%29%0A%20%20%20%20return%20mem%5Bi%5D%5Bc%5D%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20wgt%20%3D%20%5B10,%2020,%2030,%2040,%2050%5D%0A%20%20%20%20val%20%3D%20%5B50,%20120,%20150,%20210,%20240%5D%0A%20%20%20%20cap%20%3D%2050%0A%20%20%20%20n%20%3D%20len%28wgt%29%0A%0A%20%20%20%20%23%20%E8%AE%B0%E5%BF%86%E5%8C%96%E6%90%9C%E7%B4%A2%0A%20%20%20%20mem%20%3D%20%5B%5B-1%5D%20*%20%28cap%20%2B%201%29%20for%20_%20in%20range%28n%20%2B%201%29%5D%0A%20%20%20%20res%20%3D%20knapsack_dfs_mem%28wgt,%20val,%20mem,%20n,%20cap%29%0A%20%20%20%20print%28f%22%E4%B8%8D%E8%B6%85%E8%BF%87%E8%83%8C%E5%8C%85%E5%AE%B9%E9%87%8F%E7%9A%84%E6%9C%80%E5%A4%A7%E7%89%A9%E5%93%81%E4%BB%B7%E5%80%BC%E4%B8%BA%20%7Bres%7D%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=20&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
@@ -1104,30 +1057,6 @@ $$
end
```
=== "Zig"
```zig title="knapsack.zig"
// 0-1 背包：动态规划
fn knapsackDP(comptime wgt: []i32, val: []i32, comptime cap: usize) i32 {
comptime var n = wgt.len;
// 初始化 dp 表
var dp = [_][cap + 1]i32{[_]i32{0} ** (cap + 1)} ** (n + 1);
// 状态转移
for (1..n + 1) |i| {
for (1..cap + 1) |c| {
if (wgt[i - 1] > c) {
// 若超过背包容量，则不选物品 i
dp[i][c] = dp[i - 1][c];
} else {
// 不选和选物品 i 这两种方案的较大值
dp[i][c] = @max(dp[i - 1][c], dp[i - 1][c - @as(usize, @intCast(wgt[i - 1]))] + val[i - 1]);
}
}
}
return dp[n][cap];
}
```
??? pythontutor "可视化运行"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20knapsack_dp%28wgt%3A%20list%5Bint%5D,%20val%3A%20list%5Bint%5D,%20cap%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%220-1%20%E8%83%8C%E5%8C%85%EF%BC%9A%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%22%22%22%0A%20%20%20%20n%20%3D%20len%28wgt%29%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%20dp%20%E8%A1%A8%0A%20%20%20%20dp%20%3D%20%5B%5B0%5D%20*%20%28cap%20%2B%201%29%20for%20_%20in%20range%28n%20%2B%201%29%5D%0A%20%20%20%20%23%20%E7%8A%B6%E6%80%81%E8%BD%AC%E7%A7%BB%0A%20%20%20%20for%20i%20in%20range%281,%20n%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20for%20c%20in%20range%281,%20cap%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20if%20wgt%5Bi%20-%201%5D%20%3E%20c%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%23%20%E8%8B%A5%E8%B6%85%E8%BF%87%E8%83%8C%E5%8C%85%E5%AE%B9%E9%87%8F%EF%BC%8C%E5%88%99%E4%B8%8D%E9%80%89%E7%89%A9%E5%93%81%20i%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20dp%5Bi%5D%5Bc%5D%20%3D%20dp%5Bi%20-%201%5D%5Bc%5D%0A%20%20%20%20%20%20%20%20%20%20%20%20else%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%23%20%E4%B8%8D%E9%80%89%E5%92%8C%E9%80%89%E7%89%A9%E5%93%81%20i%20%E8%BF%99%E4%B8%A4%E7%A7%8D%E6%96%B9%E6%A1%88%E7%9A%84%E8%BE%83%E5%A4%A7%E5%80%BC%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20dp%5Bi%5D%5Bc%5D%20%3D%20max%28dp%5Bi%20-%201%5D%5Bc%5D,%20dp%5Bi%20-%201%5D%5Bc%20-%20wgt%5Bi%20-%201%5D%5D%20%2B%20val%5Bi%20-%201%5D%29%0A%20%20%20%20return%20dp%5Bn%5D%5Bcap%5D%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20wgt%20%3D%20%5B10,%2020,%2030,%2040,%2050%5D%0A%20%20%20%20val%20%3D%20%5B50,%20120,%20150,%20210,%20240%5D%0A%20%20%20%20cap%20%3D%2050%0A%20%20%20%20n%20%3D%20len%28wgt%29%0A%0A%20%20%20%20%23%20%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%0A%20%20%20%20res%20%3D%20knapsack_dp%28wgt,%20val,%20cap%29%0A%20%20%20%20print%28f%22%E4%B8%8D%E8%B6%85%E8%BF%87%E8%83%8C%E5%8C%85%E5%AE%B9%E9%87%8F%E7%9A%84%E6%9C%80%E5%A4%A7%E7%89%A9%E5%93%81%E4%BB%B7%E5%80%BC%E4%B8%BA%20%7Bres%7D%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=7&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
@@ -1510,29 +1439,6 @@ $$
end
```
=== "Zig"
```zig title="knapsack.zig"
// 0-1 背包：空间优化后的动态规划
fn knapsackDPComp(wgt: []i32, val: []i32, comptime cap: usize) i32 {
var n = wgt.len;
// 初始化 dp 表
var dp = [_]i32{0} ** (cap + 1);
// 状态转移
for (1..n + 1) |i| {
// 倒序遍历
var c = cap;
while (c > 0) : (c -= 1) {
if (wgt[i - 1] < c) {
// 不选和选物品 i 这两种方案的较大值
dp[c] = @max(dp[c], dp[c - @as(usize, @intCast(wgt[i - 1]))] + val[i - 1]);
}
}
}
return dp[cap];
}
```
??? pythontutor "可视化运行"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20knapsack_dp_comp%28wgt%3A%20list%5Bint%5D,%20val%3A%20list%5Bint%5D,%20cap%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%220-1%20%E8%83%8C%E5%8C%85%EF%BC%9A%E7%A9%BA%E9%97%B4%E4%BC%98%E5%8C%96%E5%90%8E%E7%9A%84%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%22%22%22%0A%20%20%20%20n%20%3D%20len%28wgt%29%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%20dp%20%E8%A1%A8%0A%20%20%20%20dp%20%3D%20%5B0%5D%20*%20%28cap%20%2B%201%29%0A%20%20%20%20%23%20%E7%8A%B6%E6%80%81%E8%BD%AC%E7%A7%BB%0A%20%20%20%20for%20i%20in%20range%281,%20n%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20%23%20%E5%80%92%E5%BA%8F%E9%81%8D%E5%8E%86%0A%20%20%20%20%20%20%20%20for%20c%20in%20range%28cap,%200,%20-1%29%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20if%20wgt%5Bi%20-%201%5D%20%3E%20c%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%23%20%E8%8B%A5%E8%B6%85%E8%BF%87%E8%83%8C%E5%8C%85%E5%AE%B9%E9%87%8F%EF%BC%8C%E5%88%99%E4%B8%8D%E9%80%89%E7%89%A9%E5%93%81%20i%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20dp%5Bc%5D%20%3D%20dp%5Bc%5D%0A%20%20%20%20%20%20%20%20%20%20%20%20else%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%23%20%E4%B8%8D%E9%80%89%E5%92%8C%E9%80%89%E7%89%A9%E5%93%81%20i%20%E8%BF%99%E4%B8%A4%E7%A7%8D%E6%96%B9%E6%A1%88%E7%9A%84%E8%BE%83%E5%A4%A7%E5%80%BC%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20dp%5Bc%5D%20%3D%20max%28dp%5Bc%5D,%20dp%5Bc%20-%20wgt%5Bi%20-%201%5D%5D%20%2B%20val%5Bi%20-%201%5D%29%0A%20%20%20%20return%20dp%5Bcap%5D%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20wgt%20%3D%20%5B10,%2020,%2030,%2040,%2050%5D%0A%20%20%20%20val%20%3D%20%5B50,%20120,%20150,%20210,%20240%5D%0A%20%20%20%20cap%20%3D%2050%0A%20%20%20%20n%20%3D%20len%28wgt%29%0A%0A%20%20%20%20%23%20%E7%A9%BA%E9%97%B4%E4%BC%98%E5%8C%96%E5%90%8E%E7%9A%84%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%0A%20%20%20%20res%20%3D%20knapsack_dp_comp%28wgt,%20val,%20cap%29%0A%20%20%20%20print%28f%22%E4%B8%8D%E8%B6%85%E8%BF%87%E8%83%8C%E5%8C%85%E5%AE%B9%E9%87%8F%E7%9A%84%E6%9C%80%E5%A4%A7%E7%89%A9%E5%93%81%E4%BB%B7%E5%80%BC%E4%B8%BA%20%7Bres%7D%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=7&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
docs/chapter_dynamic_programming/summary.md
+2
View File
@@ -4,6 +4,8 @@ comments: true
# 14.7 &nbsp; 小结
### 1. &nbsp; 重点回顾
- 动态规划对问题进行分解，并通过存储子问题的解来规避重复计算，提高计算效率。
- 不考虑时间的前提下，所有动态规划问题都可以用回溯（暴力搜索）进行求解，但递归树中存在大量的重叠子问题，效率极低。通过引入记忆化列表，可以存储所有计算过的子问题的解，从而保证重叠子问题只被计算一次。
- 记忆化搜索是一种从顶至底的递归式解法，而与之对应的动态规划是一种从底至顶的递推式解法，其如同“填写表格”一样。由于当前状态仅依赖某些局部状态，因此我们可以消除 $dp$ 表的一个维度，从而降低空间复杂度。
docs/chapter_dynamic_programming/unbounded_knapsack_problem.md
-165
View File
@@ -371,30 +371,6 @@ $$
end
```
=== "Zig"
```zig title="unbounded_knapsack.zig"
// 完全背包：动态规划
fn unboundedKnapsackDP(comptime wgt: []i32, val: []i32, comptime cap: usize) i32 {
comptime var n = wgt.len;
// 初始化 dp 表
var dp = [_][cap + 1]i32{[_]i32{0} ** (cap + 1)} ** (n + 1);
// 状态转移
for (1..n + 1) |i| {
for (1..cap + 1) |c| {
if (wgt[i - 1] > c) {
// 若超过背包容量，则不选物品 i
dp[i][c] = dp[i - 1][c];
} else {
// 不选和选物品 i 这两种方案的较大值
dp[i][c] = @max(dp[i - 1][c], dp[i][c - @as(usize, @intCast(wgt[i - 1]))] + val[i - 1]);
}
}
}
return dp[n][cap];
}
```
??? pythontutor "可视化运行"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20unbounded_knapsack_dp%28wgt%3A%20list%5Bint%5D,%20val%3A%20list%5Bint%5D,%20cap%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E5%AE%8C%E5%85%A8%E8%83%8C%E5%8C%85%EF%BC%9A%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%22%22%22%0A%20%20%20%20n%20%3D%20len%28wgt%29%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%20dp%20%E8%A1%A8%0A%20%20%20%20dp%20%3D%20%5B%5B0%5D%20*%20%28cap%20%2B%201%29%20for%20_%20in%20range%28n%20%2B%201%29%5D%0A%20%20%20%20%23%20%E7%8A%B6%E6%80%81%E8%BD%AC%E7%A7%BB%0A%20%20%20%20for%20i%20in%20range%281,%20n%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20for%20c%20in%20range%281,%20cap%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20if%20wgt%5Bi%20-%201%5D%20%3E%20c%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%23%20%E8%8B%A5%E8%B6%85%E8%BF%87%E8%83%8C%E5%8C%85%E5%AE%B9%E9%87%8F%EF%BC%8C%E5%88%99%E4%B8%8D%E9%80%89%E7%89%A9%E5%93%81%20i%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20dp%5Bi%5D%5Bc%5D%20%3D%20dp%5Bi%20-%201%5D%5Bc%5D%0A%20%20%20%20%20%20%20%20%20%20%20%20else%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%23%20%E4%B8%8D%E9%80%89%E5%92%8C%E9%80%89%E7%89%A9%E5%93%81%20i%20%E8%BF%99%E4%B8%A4%E7%A7%8D%E6%96%B9%E6%A1%88%E7%9A%84%E8%BE%83%E5%A4%A7%E5%80%BC%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20dp%5Bi%5D%5Bc%5D%20%3D%20max%28dp%5Bi%20-%201%5D%5Bc%5D,%20dp%5Bi%5D%5Bc%20-%20wgt%5Bi%20-%201%5D%5D%20%2B%20val%5Bi%20-%201%5D%29%0A%20%20%20%20return%20dp%5Bn%5D%5Bcap%5D%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20wgt%20%3D%20%5B1,%202,%203%5D%0A%20%20%20%20val%20%3D%20%5B5,%2011,%2015%5D%0A%20%20%20%20cap%20%3D%204%0A%0A%20%20%20%20%23%20%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%0A%20%20%20%20res%20%3D%20unbounded_knapsack_dp%28wgt,%20val,%20cap%29%0A%20%20%20%20print%28f%22%E4%B8%8D%E8%B6%85%E8%BF%87%E8%83%8C%E5%8C%85%E5%AE%B9%E9%87%8F%E7%9A%84%E6%9C%80%E5%A4%A7%E7%89%A9%E5%93%81%E4%BB%B7%E5%80%BC%E4%B8%BA%20%7Bres%7D%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=6&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
@@ -769,30 +745,6 @@ $$
end
```
=== "Zig"
```zig title="unbounded_knapsack.zig"
// 完全背包：空间优化后的动态规划
fn unboundedKnapsackDPComp(comptime wgt: []i32, val: []i32, comptime cap: usize) i32 {
comptime var n = wgt.len;
// 初始化 dp 表
var dp = [_]i32{0} ** (cap + 1);
// 状态转移
for (1..n + 1) |i| {
for (1..cap + 1) |c| {
if (wgt[i - 1] > c) {
// 若超过背包容量，则不选物品 i
dp[c] = dp[c];
} else {
// 不选和选物品 i 这两种方案的较大值
dp[c] = @max(dp[c], dp[c - @as(usize, @intCast(wgt[i - 1]))] + val[i - 1]);
}
}
}
return dp[cap];
}
```
??? pythontutor "可视化运行"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20unbounded_knapsack_dp_comp%28wgt%3A%20list%5Bint%5D,%20val%3A%20list%5Bint%5D,%20cap%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E5%AE%8C%E5%85%A8%E8%83%8C%E5%8C%85%EF%BC%9A%E7%A9%BA%E9%97%B4%E4%BC%98%E5%8C%96%E5%90%8E%E7%9A%84%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%22%22%22%0A%20%20%20%20n%20%3D%20len%28wgt%29%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%20dp%20%E8%A1%A8%0A%20%20%20%20dp%20%3D%20%5B0%5D%20*%20%28cap%20%2B%201%29%0A%20%20%20%20%23%20%E7%8A%B6%E6%80%81%E8%BD%AC%E7%A7%BB%0A%20%20%20%20for%20i%20in%20range%281,%20n%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20%23%20%E6%AD%A3%E5%BA%8F%E9%81%8D%E5%8E%86%0A%20%20%20%20%20%20%20%20for%20c%20in%20range%281,%20cap%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20if%20wgt%5Bi%20-%201%5D%20%3E%20c%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%23%20%E8%8B%A5%E8%B6%85%E8%BF%87%E8%83%8C%E5%8C%85%E5%AE%B9%E9%87%8F%EF%BC%8C%E5%88%99%E4%B8%8D%E9%80%89%E7%89%A9%E5%93%81%20i%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20dp%5Bc%5D%20%3D%20dp%5Bc%5D%0A%20%20%20%20%20%20%20%20%20%20%20%20else%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%23%20%E4%B8%8D%E9%80%89%E5%92%8C%E9%80%89%E7%89%A9%E5%93%81%20i%20%E8%BF%99%E4%B8%A4%E7%A7%8D%E6%96%B9%E6%A1%88%E7%9A%84%E8%BE%83%E5%A4%A7%E5%80%BC%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20dp%5Bc%5D%20%3D%20max%28dp%5Bc%5D,%20dp%5Bc%20-%20wgt%5Bi%20-%201%5D%5D%20%2B%20val%5Bi%20-%201%5D%29%0A%20%20%20%20return%20dp%5Bcap%5D%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20wgt%20%3D%20%5B1,%202,%203%5D%0A%20%20%20%20val%20%3D%20%5B5,%2011,%2015%5D%0A%20%20%20%20cap%20%3D%204%0A%0A%20%20%20%20%23%20%E7%A9%BA%E9%97%B4%E4%BC%98%E5%8C%96%E5%90%8E%E7%9A%84%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%0A%20%20%20%20res%20%3D%20unbounded_knapsack_dp_comp%28wgt,%20val,%20cap%29%0A%20%20%20%20print%28f%22%E4%B8%8D%E8%B6%85%E8%BF%87%E8%83%8C%E5%8C%85%E5%AE%B9%E9%87%8F%E7%9A%84%E6%9C%80%E5%A4%A7%E7%89%A9%E5%93%81%E4%BB%B7%E5%80%BC%E4%B8%BA%20%7Bres%7D%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=6&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
@@ -1240,39 +1192,6 @@ $$
end
```
=== "Zig"
```zig title="coin_change.zig"
// 零钱兑换：动态规划
fn coinChangeDP(comptime coins: []i32, comptime amt: usize) i32 {
comptime var n = coins.len;
comptime var max = amt + 1;
// 初始化 dp 表
var dp = [_][amt + 1]i32{[_]i32{0} ** (amt + 1)} ** (n + 1);
// 状态转移：首行首列
for (1..amt + 1) |a| {
dp[0][a] = max;
}
// 状态转移：其余行和列
for (1..n + 1) |i| {
for (1..amt + 1) |a| {
if (coins[i - 1] > @as(i32, @intCast(a))) {
// 若超过目标金额，则不选硬币 i
dp[i][a] = dp[i - 1][a];
} else {
// 不选和选硬币 i 这两种方案的较小值
dp[i][a] = @min(dp[i - 1][a], dp[i][a - @as(usize, @intCast(coins[i - 1]))] + 1);
}
}
}
if (dp[n][amt] != max) {
return @intCast(dp[n][amt]);
} else {
return -1;
}
}
```
??? pythontutor "可视化运行"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20coin_change_dp%28coins%3A%20list%5Bint%5D,%20amt%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E9%9B%B6%E9%92%B1%E5%85%91%E6%8D%A2%EF%BC%9A%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%22%22%22%0A%20%20%20%20n%20%3D%20len%28coins%29%0A%20%20%20%20MAX%20%3D%20amt%20%2B%201%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%20dp%20%E8%A1%A8%0A%20%20%20%20dp%20%3D%20%5B%5B0%5D%20*%20%28amt%20%2B%201%29%20for%20_%20in%20range%28n%20%2B%201%29%5D%0A%20%20%20%20%23%20%E7%8A%B6%E6%80%81%E8%BD%AC%E7%A7%BB%EF%BC%9A%E9%A6%96%E8%A1%8C%E9%A6%96%E5%88%97%0A%20%20%20%20for%20a%20in%20range%281,%20amt%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20dp%5B0%5D%5Ba%5D%20%3D%20MAX%0A%20%20%20%20%23%20%E7%8A%B6%E6%80%81%E8%BD%AC%E7%A7%BB%EF%BC%9A%E5%85%B6%E4%BD%99%E8%A1%8C%E5%92%8C%E5%88%97%0A%20%20%20%20for%20i%20in%20range%281,%20n%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20for%20a%20in%20range%281,%20amt%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20if%20coins%5Bi%20-%201%5D%20%3E%20a%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%23%20%E8%8B%A5%E8%B6%85%E8%BF%87%E7%9B%AE%E6%A0%87%E9%87%91%E9%A2%9D%EF%BC%8C%E5%88%99%E4%B8%8D%E9%80%89%E7%A1%AC%E5%B8%81%20i%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20dp%5Bi%5D%5Ba%5D%20%3D%20dp%5Bi%20-%201%5D%5Ba%5D%0A%20%20%20%20%20%20%20%20%20%20%20%20else%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%23%20%E4%B8%8D%E9%80%89%E5%92%8C%E9%80%89%E7%A1%AC%E5%B8%81%20i%20%E8%BF%99%E4%B8%A4%E7%A7%8D%E6%96%B9%E6%A1%88%E7%9A%84%E8%BE%83%E5%B0%8F%E5%80%BC%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20dp%5Bi%5D%5Ba%5D%20%3D%20min%28dp%5Bi%20-%201%5D%5Ba%5D,%20dp%5Bi%5D%5Ba%20-%20coins%5Bi%20-%201%5D%5D%20%2B%201%29%0A%20%20%20%20return%20dp%5Bn%5D%5Bamt%5D%20if%20dp%5Bn%5D%5Bamt%5D%20!%3D%20MAX%20else%20-1%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20coins%20%3D%20%5B1,%202,%205%5D%0A%20%20%20%20amt%20%3D%204%0A%0A%20%20%20%20%23%20%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%0A%20%20%20%20res%20%3D%20coin_change_dp%28coins,%20amt%29%0A%20%20%20%20print%28f%22%E5%87%91%E5%88%B0%E7%9B%AE%E6%A0%87%E9%87%91%E9%A2%9D%E6%89%80%E9%9C%80%E7%9A%84%E6%9C%80%E5%B0%91%E7%A1%AC%E5%B8%81%E6%95%B0%E9%87%8F%E4%B8%BA%20%7Bres%7D%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=5&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
@@ -1688,37 +1607,6 @@ $$
end
```
=== "Zig"
```zig title="coin_change.zig"
// 零钱兑换：空间优化后的动态规划
fn coinChangeDPComp(comptime coins: []i32, comptime amt: usize) i32 {
comptime var n = coins.len;
comptime var max = amt + 1;
// 初始化 dp 表
var dp = [_]i32{0} ** (amt + 1);
@memset(&dp, max);
dp[0] = 0;
// 状态转移
for (1..n + 1) |i| {
for (1..amt + 1) |a| {
if (coins[i - 1] > @as(i32, @intCast(a))) {
// 若超过目标金额，则不选硬币 i
dp[a] = dp[a];
} else {
// 不选和选硬币 i 这两种方案的较小值
dp[a] = @min(dp[a], dp[a - @as(usize, @intCast(coins[i - 1]))] + 1);
}
}
}
if (dp[amt] != max) {
return @intCast(dp[amt]);
} else {
return -1;
}
}
```
??? pythontutor "可视化运行"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20coin_change_dp_comp%28coins%3A%20list%5Bint%5D,%20amt%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E9%9B%B6%E9%92%B1%E5%85%91%E6%8D%A2%EF%BC%9A%E7%A9%BA%E9%97%B4%E4%BC%98%E5%8C%96%E5%90%8E%E7%9A%84%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%22%22%22%0A%20%20%20%20n%20%3D%20len%28coins%29%0A%20%20%20%20MAX%20%3D%20amt%20%2B%201%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%20dp%20%E8%A1%A8%0A%20%20%20%20dp%20%3D%20%5BMAX%5D%20*%20%28amt%20%2B%201%29%0A%20%20%20%20dp%5B0%5D%20%3D%200%0A%20%20%20%20%23%20%E7%8A%B6%E6%80%81%E8%BD%AC%E7%A7%BB%0A%20%20%20%20for%20i%20in%20range%281,%20n%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20%23%20%E6%AD%A3%E5%BA%8F%E9%81%8D%E5%8E%86%0A%20%20%20%20%20%20%20%20for%20a%20in%20range%281,%20amt%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20if%20coins%5Bi%20-%201%5D%20%3E%20a%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%23%20%E8%8B%A5%E8%B6%85%E8%BF%87%E7%9B%AE%E6%A0%87%E9%87%91%E9%A2%9D%EF%BC%8C%E5%88%99%E4%B8%8D%E9%80%89%E7%A1%AC%E5%B8%81%20i%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20dp%5Ba%5D%20%3D%20dp%5Ba%5D%0A%20%20%20%20%20%20%20%20%20%20%20%20else%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%23%20%E4%B8%8D%E9%80%89%E5%92%8C%E9%80%89%E7%A1%AC%E5%B8%81%20i%20%E8%BF%99%E4%B8%A4%E7%A7%8D%E6%96%B9%E6%A1%88%E7%9A%84%E8%BE%83%E5%B0%8F%E5%80%BC%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20dp%5Ba%5D%20%3D%20min%28dp%5Ba%5D,%20dp%5Ba%20-%20coins%5Bi%20-%201%5D%5D%20%2B%201%29%0A%20%20%20%20return%20dp%5Bamt%5D%20if%20dp%5Bamt%5D%20!%3D%20MAX%20else%20-1%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20coins%20%3D%20%5B1,%202,%205%5D%0A%20%20%20%20amt%20%3D%204%0A%0A%20%20%20%20%23%20%E7%A9%BA%E9%97%B4%E4%BC%98%E5%8C%96%E5%90%8E%E7%9A%84%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%0A%20%20%20%20res%20%3D%20coin_change_dp_comp%28coins,%20amt%29%0A%20%20%20%20print%28f%22%E5%87%91%E5%88%B0%E7%9B%AE%E6%A0%87%E9%87%91%E9%A2%9D%E6%89%80%E9%9C%80%E7%9A%84%E6%9C%80%E5%B0%91%E7%A1%AC%E5%B8%81%E6%95%B0%E9%87%8F%E4%B8%BA%20%7Bres%7D%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=5&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
@@ -2121,34 +2009,6 @@ $$
end
```
=== "Zig"
```zig title="coin_change_ii.zig"
// 零钱兑换 II：动态规划
fn coinChangeIIDP(comptime coins: []i32, comptime amt: usize) i32 {
comptime var n = coins.len;
// 初始化 dp 表
var dp = [_][amt + 1]i32{[_]i32{0} ** (amt + 1)} ** (n + 1);
// 初始化首列
for (0..n + 1) |i| {
dp[i][0] = 1;
}
// 状态转移
for (1..n + 1) |i| {
for (1..amt + 1) |a| {
if (coins[i - 1] > @as(i32, @intCast(a))) {
// 若超过目标金额，则不选硬币 i
dp[i][a] = dp[i - 1][a];
} else {
// 不选和选硬币 i 这两种方案的较小值
dp[i][a] = dp[i - 1][a] + dp[i][a - @as(usize, @intCast(coins[i - 1]))];
}
}
}
return dp[n][amt];
}
```
??? pythontutor "可视化运行"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20coin_change_ii_dp%28coins%3A%20list%5Bint%5D,%20amt%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E9%9B%B6%E9%92%B1%E5%85%91%E6%8D%A2%20II%EF%BC%9A%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%22%22%22%0A%20%20%20%20n%20%3D%20len%28coins%29%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%20dp%20%E8%A1%A8%0A%20%20%20%20dp%20%3D%20%5B%5B0%5D%20*%20%28amt%20%2B%201%29%20for%20_%20in%20range%28n%20%2B%201%29%5D%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%E9%A6%96%E5%88%97%0A%20%20%20%20for%20i%20in%20range%28n%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20dp%5Bi%5D%5B0%5D%20%3D%201%0A%20%20%20%20%23%20%E7%8A%B6%E6%80%81%E8%BD%AC%E7%A7%BB%0A%20%20%20%20for%20i%20in%20range%281,%20n%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20for%20a%20in%20range%281,%20amt%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20if%20coins%5Bi%20-%201%5D%20%3E%20a%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%23%20%E8%8B%A5%E8%B6%85%E8%BF%87%E7%9B%AE%E6%A0%87%E9%87%91%E9%A2%9D%EF%BC%8C%E5%88%99%E4%B8%8D%E9%80%89%E7%A1%AC%E5%B8%81%20i%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20dp%5Bi%5D%5Ba%5D%20%3D%20dp%5Bi%20-%201%5D%5Ba%5D%0A%20%20%20%20%20%20%20%20%20%20%20%20else%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%23%20%E4%B8%8D%E9%80%89%E5%92%8C%E9%80%89%E7%A1%AC%E5%B8%81%20i%20%E8%BF%99%E4%B8%A4%E7%A7%8D%E6%96%B9%E6%A1%88%E4%B9%8B%E5%92%8C%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20dp%5Bi%5D%5Ba%5D%20%3D%20dp%5Bi%20-%201%5D%5Ba%5D%20%2B%20dp%5Bi%5D%5Ba%20-%20coins%5Bi%20-%201%5D%5D%0A%20%20%20%20return%20dp%5Bn%5D%5Bamt%5D%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20coins%20%3D%20%5B1,%202,%205%5D%0A%20%20%20%20amt%20%3D%205%0A%0A%20%20%20%20%23%20%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%0A%20%20%20%20res%20%3D%20coin_change_ii_dp%28coins,%20amt%29%0A%20%20%20%20print%28f%22%E5%87%91%E5%87%BA%E7%9B%AE%E6%A0%87%E9%87%91%E9%A2%9D%E7%9A%84%E7%A1%AC%E5%B8%81%E7%BB%84%E5%90%88%E6%95%B0%E9%87%8F%E4%B8%BA%20%7Bres%7D%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=5&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
@@ -2485,31 +2345,6 @@ $$
end
```
=== "Zig"
```zig title="coin_change_ii.zig"
// 零钱兑换 II：空间优化后的动态规划
fn coinChangeIIDPComp(comptime coins: []i32, comptime amt: usize) i32 {
comptime var n = coins.len;
// 初始化 dp 表
var dp = [_]i32{0} ** (amt + 1);
dp[0] = 1;
// 状态转移
for (1..n + 1) |i| {
for (1..amt + 1) |a| {
if (coins[i - 1] > @as(i32, @intCast(a))) {
// 若超过目标金额，则不选硬币 i
dp[a] = dp[a];
} else {
// 不选和选硬币 i 这两种方案的较小值
dp[a] = dp[a] + dp[a - @as(usize, @intCast(coins[i - 1]))];
}
}
}
return dp[amt];
}
```
??? pythontutor "可视化运行"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20coin_change_ii_dp_comp%28coins%3A%20list%5Bint%5D,%20amt%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E9%9B%B6%E9%92%B1%E5%85%91%E6%8D%A2%20II%EF%BC%9A%E7%A9%BA%E9%97%B4%E4%BC%98%E5%8C%96%E5%90%8E%E7%9A%84%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%22%22%22%0A%20%20%20%20n%20%3D%20len%28coins%29%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%20dp%20%E8%A1%A8%0A%20%20%20%20dp%20%3D%20%5B0%5D%20*%20%28amt%20%2B%201%29%0A%20%20%20%20dp%5B0%5D%20%3D%201%0A%20%20%20%20%23%20%E7%8A%B6%E6%80%81%E8%BD%AC%E7%A7%BB%0A%20%20%20%20for%20i%20in%20range%281,%20n%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20%23%20%E6%AD%A3%E5%BA%8F%E9%81%8D%E5%8E%86%0A%20%20%20%20%20%20%20%20for%20a%20in%20range%281,%20amt%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20if%20coins%5Bi%20-%201%5D%20%3E%20a%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%23%20%E8%8B%A5%E8%B6%85%E8%BF%87%E7%9B%AE%E6%A0%87%E9%87%91%E9%A2%9D%EF%BC%8C%E5%88%99%E4%B8%8D%E9%80%89%E7%A1%AC%E5%B8%81%20i%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20dp%5Ba%5D%20%3D%20dp%5Ba%5D%0A%20%20%20%20%20%20%20%20%20%20%20%20else%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%23%20%E4%B8%8D%E9%80%89%E5%92%8C%E9%80%89%E7%A1%AC%E5%B8%81%20i%20%E8%BF%99%E4%B8%A4%E7%A7%8D%E6%96%B9%E6%A1%88%E4%B9%8B%E5%92%8C%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20dp%5Ba%5D%20%3D%20dp%5Ba%5D%20%2B%20dp%5Ba%20-%20coins%5Bi%20-%201%5D%5D%0A%20%20%20%20return%20dp%5Bamt%5D%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20coins%20%3D%20%5B1,%202,%205%5D%0A%20%20%20%20amt%20%3D%205%0A%0A%20%20%20%20%23%20%E7%A9%BA%E9%97%B4%E4%BC%98%E5%8C%96%E5%90%8E%E7%9A%84%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%0A%20%20%20%20res%20%3D%20coin_change_ii_dp_comp%28coins,%20amt%29%0A%20%20%20%20print%28f%22%E5%87%91%E5%87%BA%E7%9B%AE%E6%A0%87%E9%87%91%E9%A2%9D%E7%9A%84%E7%A1%AC%E5%B8%81%E7%BB%84%E5%90%88%E6%95%B0%E9%87%8F%E4%B8%BA%20%7Bres%7D%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=5&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
docs/chapter_graph/graph_operations.md
-12
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docs/chapter_graph/graph_traversal.md
-14
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docs/chapter_greedy/fractional_knapsack_problem.md
-8
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@@ -531,14 +531,6 @@ comments: true
end
```
=== "Zig"
```zig title="fractional_knapsack.zig"
[class]{Item}-[func]{}
[class]{}-[func]{fractionalKnapsack}
```
??? pythontutor "可视化运行"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=class%20Item%3A%0A%20%20%20%20%22%22%22%E7%89%A9%E5%93%81%22%22%22%0A%20%20%20%20def%20__init__%28self,%20w%3A%20int,%20v%3A%20int%29%3A%0A%20%20%20%20%20%20%20%20self.w%20%3D%20w%20%20%23%20%E7%89%A9%E5%93%81%E9%87%8D%E9%87%8F%0A%20%20%20%20%20%20%20%20self.v%20%3D%20v%20%20%23%20%E7%89%A9%E5%93%81%E4%BB%B7%E5%80%BC%0A%0Adef%20fractional_knapsack%28wgt%3A%20list%5Bint%5D,%20val%3A%20list%5Bint%5D,%20cap%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E5%88%86%E6%95%B0%E8%83%8C%E5%8C%85%EF%BC%9A%E8%B4%AA%E5%BF%83%22%22%22%0A%20%20%20%20%23%20%E5%88%9B%E5%BB%BA%E7%89%A9%E5%93%81%E5%88%97%E8%A1%A8%EF%BC%8C%E5%8C%85%E5%90%AB%E4%B8%A4%E4%B8%AA%E5%B1%9E%E6%80%A7%EF%BC%9A%E9%87%8D%E9%87%8F%E3%80%81%E4%BB%B7%E5%80%BC%0A%20%20%20%20items%20%3D%20%5BItem%28w,%20v%29%20for%20w,%20v%20in%20zip%28wgt,%20val%29%5D%0A%20%20%20%20%23%20%E6%8C%89%E7%85%A7%E5%8D%95%E4%BD%8D%E4%BB%B7%E5%80%BC%20item.v%20/%20item.w%20%E4%BB%8E%E9%AB%98%E5%88%B0%E4%BD%8E%E8%BF%9B%E8%A1%8C%E6%8E%92%E5%BA%8F%0A%20%20%20%20items.sort%28key%3Dlambda%20item%3A%20item.v%20/%20item.w,%20reverse%3DTrue%29%0A%20%20%20%20%23%20%E5%BE%AA%E7%8E%AF%E8%B4%AA%E5%BF%83%E9%80%89%E6%8B%A9%0A%20%20%20%20res%20%3D%200%0A%20%20%20%20for%20item%20in%20items%3A%0A%20%20%20%20%20%20%20%20if%20item.w%20%3C%3D%20cap%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20%23%20%E8%8B%A5%E5%89%A9%E4%BD%99%E5%AE%B9%E9%87%8F%E5%85%85%E8%B6%B3%EF%BC%8C%E5%88%99%E5%B0%86%E5%BD%93%E5%89%8D%E7%89%A9%E5%93%81%E6%95%B4%E4%B8%AA%E8%A3%85%E8%BF%9B%E8%83%8C%E5%8C%85%0A%20%20%20%20%20%20%20%20%20%20%20%20res%20%2B%3D%20item.v%0A%20%20%20%20%20%20%20%20%20%20%20%20cap%20-%3D%20item.w%0A%20%20%20%20%20%20%20%20else%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20%23%20%E8%8B%A5%E5%89%A9%E4%BD%99%E5%AE%B9%E9%87%8F%E4%B8%8D%E8%B6%B3%EF%BC%8C%E5%88%99%E5%B0%86%E5%BD%93%E5%89%8D%E7%89%A9%E5%93%81%E7%9A%84%E4%B8%80%E9%83%A8%E5%88%86%E8%A3%85%E8%BF%9B%E8%83%8C%E5%8C%85%0A%20%20%20%20%20%20%20%20%20%20%20%20res%20%2B%3D%20%28item.v%20/%20item.w%29%20*%20cap%0A%20%20%20%20%20%20%20%20%20%20%20%20%23%20%E5%B7%B2%E6%97%A0%E5%89%A9%E4%BD%99%E5%AE%B9%E9%87%8F%EF%BC%8C%E5%9B%A0%E6%AD%A4%E8%B7%B3%E5%87%BA%E5%BE%AA%E7%8E%AF%0A%20%20%20%20%20%20%20%20%20%20%20%20break%0A%20%20%20%20return%20res%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20wgt%20%3D%20%5B10,%2020,%2030,%2040,%2050%5D%0A%20%20%20%20val%20%3D%20%5B50,%20120,%20150,%20210,%20240%5D%0A%20%20%20%20cap%20%3D%2050%0A%20%20%20%20n%20%3D%20len%28wgt%29%0A%0A%20%20%20%20%23%20%E8%B4%AA%E5%BF%83%E7%AE%97%E6%B3%95%0A%20%20%20%20res%20%3D%20fractional_knapsack%28wgt,%20val,%20cap%29%0A%20%20%20%20print%28f%22%E4%B8%8D%E8%B6%85%E8%BF%87%E8%83%8C%E5%8C%85%E5%AE%B9%E9%87%8F%E7%9A%84%E6%9C%80%E5%A4%A7%E7%89%A9%E5%93%81%E4%BB%B7%E5%80%BC%E4%B8%BA%20%7Bres%7D%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=8&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
docs/chapter_greedy/greedy_algorithm.md
-6
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@@ -330,12 +330,6 @@ comments: true
end
```
=== "Zig"
```zig title="coin_change_greedy.zig"
[class]{}-[func]{coinChangeGreedy}
```
??? pythontutor "可视化运行"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20coin_change_greedy%28coins%3A%20list%5Bint%5D,%20amt%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E9%9B%B6%E9%92%B1%E5%85%91%E6%8D%A2%EF%BC%9A%E8%B4%AA%E5%BF%83%22%22%22%0A%20%20%20%20%23%20%E5%81%87%E8%AE%BE%20coins%20%E5%88%97%E8%A1%A8%E6%9C%89%E5%BA%8F%0A%20%20%20%20i%20%3D%20len%28coins%29%20-%201%0A%20%20%20%20count%20%3D%200%0A%20%20%20%20%23%20%E5%BE%AA%E7%8E%AF%E8%BF%9B%E8%A1%8C%E8%B4%AA%E5%BF%83%E9%80%89%E6%8B%A9%EF%BC%8C%E7%9B%B4%E5%88%B0%E6%97%A0%E5%89%A9%E4%BD%99%E9%87%91%E9%A2%9D%0A%20%20%20%20while%20amt%20%3E%200%3A%0A%20%20%20%20%20%20%20%20%23%20%E6%89%BE%E5%88%B0%E5%B0%8F%E4%BA%8E%E4%B8%94%E6%9C%80%E6%8E%A5%E8%BF%91%E5%89%A9%E4%BD%99%E9%87%91%E9%A2%9D%E7%9A%84%E7%A1%AC%E5%B8%81%0A%20%20%20%20%20%20%20%20while%20i%20%3E%200%20and%20coins%5Bi%5D%20%3E%20amt%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20i%20-%3D%201%0A%20%20%20%20%20%20%20%20%23%20%E9%80%89%E6%8B%A9%20coins%5Bi%5D%0A%20%20%20%20%20%20%20%20amt%20-%3D%20coins%5Bi%5D%0A%20%20%20%20%20%20%20%20count%20%2B%3D%201%0A%20%20%20%20%23%20%E8%8B%A5%E6%9C%AA%E6%89%BE%E5%88%B0%E5%8F%AF%E8%A1%8C%E6%96%B9%E6%A1%88%EF%BC%8C%E5%88%99%E8%BF%94%E5%9B%9E%20-1%0A%20%20%20%20return%20count%20if%20amt%20%3D%3D%200%20else%20-1%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20%23%20%E8%B4%AA%E5%BF%83%EF%BC%9A%E8%83%BD%E5%A4%9F%E4%BF%9D%E8%AF%81%E6%89%BE%E5%88%B0%E5%85%A8%E5%B1%80%E6%9C%80%E4%BC%98%E8%A7%A3%0A%20%20%20%20coins%20%3D%20%5B1,%205,%2010,%2020,%2050,%20100%5D%0A%20%20%20%20amt%20%3D%20186%0A%20%20%20%20res%20%3D%20coin_change_greedy%28coins,%20amt%29%0A%20%20%20%20print%28f%22%5Cncoins%20%3D%20%7Bcoins%7D,%20amt%20%3D%20%7Bamt%7D%22%29%0A%20%20%20%20print%28f%22%E5%87%91%E5%88%B0%20%7Bamt%7D%20%E6%89%80%E9%9C%80%E7%9A%84%E6%9C%80%E5%B0%91%E7%A1%AC%E5%B8%81%E6%95%B0%E9%87%8F%E4%B8%BA%20%7Bres%7D%22%29%0A%0A%20%20%20%20%23%20%E8%B4%AA%E5%BF%83%EF%BC%9A%E6%97%A0%E6%B3%95%E4%BF%9D%E8%AF%81%E6%89%BE%E5%88%B0%E5%85%A8%E5%B1%80%E6%9C%80%E4%BC%98%E8%A7%A3%0A%20%20%20%20coins%20%3D%20%5B1,%2020,%2050%5D%0A%20%20%20%20amt%20%3D%2060%0A%20%20%20%20res%20%3D%20coin_change_greedy%28coins,%20amt%29%0A%20%20%20%20print%28f%22%5Cncoins%20%3D%20%7Bcoins%7D,%20amt%20%3D%20%7Bamt%7D%22%29%0A%20%20%20%20print%28f%22%E5%87%91%E5%88%B0%20%7Bamt%7D%20%E6%89%80%E9%9C%80%E7%9A%84%E6%9C%80%E5%B0%91%E7%A1%AC%E5%B8%81%E6%95%B0%E9%87%8F%E4%B8%BA%20%7Bres%7D%22%29%0A%20%20%20%20print%28f%22%E5%AE%9E%E9%99%85%E4%B8%8A%E9%9C%80%E8%A6%81%E7%9A%84%E6%9C%80%E5%B0%91%E6%95%B0%E9%87%8F%E4%B8%BA%203%20%EF%BC%8C%E5%8D%B3%2020%20%2B%2020%20%2B%2020%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=5&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
docs/chapter_greedy/max_capacity_problem.md
-6
View File
@@ -421,12 +421,6 @@ $$
end
```
=== "Zig"
```zig title="max_capacity.zig"
[class]{}-[func]{maxCapacity}
```
??? pythontutor "可视化运行"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20max_capacity%28ht%3A%20list%5Bint%5D%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E6%9C%80%E5%A4%A7%E5%AE%B9%E9%87%8F%EF%BC%9A%E8%B4%AA%E5%BF%83%22%22%22%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%20i,%20j%EF%BC%8C%E4%BD%BF%E5%85%B6%E5%88%86%E5%88%97%E6%95%B0%E7%BB%84%E4%B8%A4%E7%AB%AF%0A%20%20%20%20i,%20j%20%3D%200,%20len%28ht%29%20-%201%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E6%9C%80%E5%A4%A7%E5%AE%B9%E9%87%8F%E4%B8%BA%200%0A%20%20%20%20res%20%3D%200%0A%20%20%20%20%23%20%E5%BE%AA%E7%8E%AF%E8%B4%AA%E5%BF%83%E9%80%89%E6%8B%A9%EF%BC%8C%E7%9B%B4%E8%87%B3%E4%B8%A4%E6%9D%BF%E7%9B%B8%E9%81%87%0A%20%20%20%20while%20i%20%3C%20j%3A%0A%20%20%20%20%20%20%20%20%23%20%E6%9B%B4%E6%96%B0%E6%9C%80%E5%A4%A7%E5%AE%B9%E9%87%8F%0A%20%20%20%20%20%20%20%20cap%20%3D%20min%28ht%5Bi%5D,%20ht%5Bj%5D%29%20*%20%28j%20-%20i%29%0A%20%20%20%20%20%20%20%20res%20%3D%20max%28res,%20cap%29%0A%20%20%20%20%20%20%20%20%23%20%E5%90%91%E5%86%85%E7%A7%BB%E5%8A%A8%E7%9F%AD%E6%9D%BF%0A%20%20%20%20%20%20%20%20if%20ht%5Bi%5D%20%3C%20ht%5Bj%5D%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20i%20%2B%3D%201%0A%20%20%20%20%20%20%20%20else%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20j%20-%3D%201%0A%20%20%20%20return%20res%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20ht%20%3D%20%5B3,%208,%205,%202,%207,%207,%203,%204%5D%0A%0A%20%20%20%20%23%20%E8%B4%AA%E5%BF%83%E7%AE%97%E6%B3%95%0A%20%20%20%20res%20%3D%20max_capacity%28ht%29%0A%20%20%20%20print%28f%22%E6%9C%80%E5%A4%A7%E5%AE%B9%E9%87%8F%E4%B8%BA%20%7Bres%7D%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=4&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
docs/chapter_greedy/max_product_cutting_problem.md
-6
View File
@@ -386,12 +386,6 @@ $$
end
```
=== "Zig"
```zig title="max_product_cutting.zig"
[class]{}-[func]{maxProductCutting}
```
??? pythontutor "可视化运行"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=import%20math%0A%0Adef%20max_product_cutting%28n%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E6%9C%80%E5%A4%A7%E5%88%87%E5%88%86%E4%B9%98%E7%A7%AF%EF%BC%9A%E8%B4%AA%E5%BF%83%22%22%22%0A%20%20%20%20%23%20%E5%BD%93%20n%20%3C%3D%203%20%E6%97%B6%EF%BC%8C%E5%BF%85%E9%A1%BB%E5%88%87%E5%88%86%E5%87%BA%E4%B8%80%E4%B8%AA%201%0A%20%20%20%20if%20n%20%3C%3D%203%3A%0A%20%20%20%20%20%20%20%20return%201%20*%20%28n%20-%201%29%0A%20%20%20%20%23%20%E8%B4%AA%E5%BF%83%E5%9C%B0%E5%88%87%E5%88%86%E5%87%BA%203%20%EF%BC%8Ca%20%E4%B8%BA%203%20%E7%9A%84%E4%B8%AA%E6%95%B0%EF%BC%8Cb%20%E4%B8%BA%E4%BD%99%E6%95%B0%0A%20%20%20%20a,%20b%20%3D%20n%20//%203,%20n%20%25%203%0A%20%20%20%20if%20b%20%3D%3D%201%3A%0A%20%20%20%20%20%20%20%20%23%20%E5%BD%93%E4%BD%99%E6%95%B0%E4%B8%BA%201%20%E6%97%B6%EF%BC%8C%E5%B0%86%E4%B8%80%E5%AF%B9%201%20*%203%20%E8%BD%AC%E5%8C%96%E4%B8%BA%202%20*%202%0A%20%20%20%20%20%20%20%20return%20int%28math.pow%283,%20a%20-%201%29%29%20*%202%20*%202%0A%20%20%20%20if%20b%20%3D%3D%202%3A%0A%20%20%20%20%20%20%20%20%23%20%E5%BD%93%E4%BD%99%E6%95%B0%E4%B8%BA%202%20%E6%97%B6%EF%BC%8C%E4%B8%8D%E5%81%9A%E5%A4%84%E7%90%86%0A%20%20%20%20%20%20%20%20return%20int%28math.pow%283,%20a%29%29%20*%202%0A%20%20%20%20%23%20%E5%BD%93%E4%BD%99%E6%95%B0%E4%B8%BA%200%20%E6%97%B6%EF%BC%8C%E4%B8%8D%E5%81%9A%E5%A4%84%E7%90%86%0A%20%20%20%20return%20int%28math.pow%283,%20a%29%29%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%2058%0A%0A%20%20%20%20%23%20%E8%B4%AA%E5%BF%83%E7%AE%97%E6%B3%95%0A%20%20%20%20res%20%3D%20max_product_cutting%28n%29%0A%20%20%20%20print%28f%22%E6%9C%80%E5%A4%A7%E5%88%87%E5%88%86%E4%B9%98%E7%A7%AF%E4%B8%BA%20%7Bres%7D%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=5&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
docs/chapter_greedy/summary.md
+2
View File
@@ -4,6 +4,8 @@ comments: true
# 15.5 &nbsp; 小结
### 1. &nbsp; 重点回顾
- 贪心算法通常用于解决最优化问题，其原理是在每个决策阶段都做出局部最优的决策，以期获得全局最优解。
- 贪心算法会迭代地做出一个又一个的贪心选择，每轮都将问题转化成一个规模更小的子问题，直到问题被解决。
- 贪心算法不仅实现简单，还具有很高的解题效率。相比于动态规划，贪心算法的时间复杂度通常更低。
docs/chapter_hashing/hash_algorithm.md
-18
View File
@@ -642,18 +642,6 @@ index = hash(key) % capacity
end
```
=== "Zig"
```zig title="simple_hash.zig"
[class]{}-[func]{addHash}
[class]{}-[func]{mulHash}
[class]{}-[func]{xorHash}
[class]{}-[func]{rotHash}
```
??? pythontutor "可视化运行"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20add_hash%28key%3A%20str%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E5%8A%A0%E6%B3%95%E5%93%88%E5%B8%8C%22%22%22%0A%20%20%20%20hash%20%3D%200%0A%20%20%20%20modulus%20%3D%201000000007%0A%20%20%20%20for%20c%20in%20key%3A%0A%20%20%20%20%20%20%20%20hash%20%2B%3D%20ord%28c%29%0A%20%20%20%20return%20hash%20%25%20modulus%0A%0A%0Adef%20mul_hash%28key%3A%20str%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E4%B9%98%E6%B3%95%E5%93%88%E5%B8%8C%22%22%22%0A%20%20%20%20hash%20%3D%200%0A%20%20%20%20modulus%20%3D%201000000007%0A%20%20%20%20for%20c%20in%20key%3A%0A%20%20%20%20%20%20%20%20hash%20%3D%2031%20*%20hash%20%2B%20ord%28c%29%0A%20%20%20%20return%20hash%20%25%20modulus%0A%0A%0Adef%20xor_hash%28key%3A%20str%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E5%BC%82%E6%88%96%E5%93%88%E5%B8%8C%22%22%22%0A%20%20%20%20hash%20%3D%200%0A%20%20%20%20modulus%20%3D%201000000007%0A%20%20%20%20for%20c%20in%20key%3A%0A%20%20%20%20%20%20%20%20hash%20%5E%3D%20ord%28c%29%0A%20%20%20%20return%20hash%20%25%20modulus%0A%0A%0Adef%20rot_hash%28key%3A%20str%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E6%97%8B%E8%BD%AC%E5%93%88%E5%B8%8C%22%22%22%0A%20%20%20%20hash%20%3D%200%0A%20%20%20%20modulus%20%3D%201000000007%0A%20%20%20%20for%20c%20in%20key%3A%0A%20%20%20%20%20%20%20%20hash%20%3D%20%28hash%20%3C%3C%204%29%20%5E%20%28hash%20%3E%3E%2028%29%20%5E%20ord%28c%29%0A%20%20%20%20return%20hash%20%25%20modulus%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20key%20%3D%20%22Hello%20%E7%AE%97%E6%B3%95%22%0A%0A%20%20%20%20hash%20%3D%20add_hash%28key%29%0A%20%20%20%20print%28f%22%E5%8A%A0%E6%B3%95%E5%93%88%E5%B8%8C%E5%80%BC%E4%B8%BA%20%7Bhash%7D%22%29%0A%0A%20%20%20%20hash%20%3D%20mul_hash%28key%29%0A%20%20%20%20print%28f%22%E4%B9%98%E6%B3%95%E5%93%88%E5%B8%8C%E5%80%BC%E4%B8%BA%20%7Bhash%7D%22%29%0A%0A%20%20%20%20hash%20%3D%20xor_hash%28key%29%0A%20%20%20%20print%28f%22%E5%BC%82%E6%88%96%E5%93%88%E5%B8%8C%E5%80%BC%E4%B8%BA%20%7Bhash%7D%22%29%0A%0A%20%20%20%20hash%20%3D%20rot_hash%28key%29%0A%20%20%20%20print%28f%22%E6%97%8B%E8%BD%AC%E5%93%88%E5%B8%8C%E5%80%BC%E4%B8%BA%20%7Bhash%7D%22%29%0A&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=6&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
@@ -1012,12 +1000,6 @@ $$
# 节点对象 #<ListNode:0x000078133140ab70> 的哈希值为 4302940560806366381
```
=== "Zig"
```zig title="built_in_hash.zig"
```
??? pythontutor "可视化运行"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=class%20ListNode%3A%0A%20%20%20%20%22%22%22%E9%93%BE%E8%A1%A8%E8%8A%82%E7%82%B9%E7%B1%BB%22%22%22%0A%20%20%20%20def%20__init__%28self,%20val%3A%20int%29%3A%0A%20%20%20%20%20%20%20%20self.val%3A%20int%20%3D%20val%20%20%23%20%E8%8A%82%E7%82%B9%E5%80%BC%0A%20%20%20%20%20%20%20%20self.next%3A%20ListNode%20%7C%20None%20%3D%20None%20%20%23%20%E5%90%8E%E7%BB%A7%E8%8A%82%E7%82%B9%E5%BC%95%E7%94%A8%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20num%20%3D%203%0A%20%20%20%20hash_num%20%3D%20hash%28num%29%0A%20%20%20%20%23%20%E6%95%B4%E6%95%B0%203%20%E7%9A%84%E5%93%88%E5%B8%8C%E5%80%BC%E4%B8%BA%203%0A%0A%20%20%20%20bol%20%3D%20True%0A%20%20%20%20hash_bol%20%3D%20hash%28bol%29%0A%20%20%20%20%23%20%E5%B8%83%E5%B0%94%E9%87%8F%20True%20%E7%9A%84%E5%93%88%E5%B8%8C%E5%80%BC%E4%B8%BA%201%0A%0A%20%20%20%20dec%20%3D%203.14159%0A%20%20%20%20hash_dec%20%3D%20hash%28dec%29%0A%20%20%20%20%23%20%E5%B0%8F%E6%95%B0%203.14159%20%E7%9A%84%E5%93%88%E5%B8%8C%E5%80%BC%E4%B8%BA%20326484311674566659%0A%0A%20%20%20%20str%20%3D%20%22Hello%20%E7%AE%97%E6%B3%95%22%0A%20%20%20%20hash_str%20%3D%20hash%28str%29%0A%20%20%20%20%23%20%E5%AD%97%E7%AC%A6%E4%B8%B2%E2%80%9CHello%20%E7%AE%97%E6%B3%95%E2%80%9D%E7%9A%84%E5%93%88%E5%B8%8C%E5%80%BC%E4%B8%BA%204617003410720528961%0A%0A%20%20%20%20tup%20%3D%20%2812836,%20%22%E5%B0%8F%E5%93%88%22%29%0A%20%20%20%20hash_tup%20%3D%20hash%28tup%29%0A%20%20%20%20%23%20%E5%85%83%E7%BB%84%20%2812836,%20'%E5%B0%8F%E5%93%88'%29%20%E7%9A%84%E5%93%88%E5%B8%8C%E5%80%BC%E4%B8%BA%201029005403108185979%0A%0A%20%20%20%20obj%20%3D%20ListNode%280%29%0A%20%20%20%20hash_obj%20%3D%20hash%28obj%29%0A%20%20%20%20%23%20%E8%8A%82%E7%82%B9%E5%AF%B9%E8%B1%A1%20%3CListNode%20object%20at%200x1058fd810%3E%20%E7%9A%84%E5%93%88%E5%B8%8C%E5%80%BC%E4%B8%BA%20274267521&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=19&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
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docs/chapter_hashing/hash_map.md
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docs/chapter_heap/build_heap.md
-17
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@@ -322,23 +322,6 @@ comments: true
end
```
=== "Zig"
```zig title="my_heap.zig"
// 构造方法，根据输入列表建堆
fn init(self: *Self, allocator: std.mem.Allocator, nums: []const T) !void {
if (self.max_heap != null) return;
self.max_heap = std.ArrayList(T).init(allocator);
// 将列表元素原封不动添加进堆
try self.max_heap.?.appendSlice(nums);
// 堆化除叶节点以外的其他所有节点
var i: usize = parent(self.size() - 1) + 1;
while (i > 0) : (i -= 1) {
try self.siftDown(i - 1);
}
}
```
??? pythontutor "可视化运行"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=class%20MaxHeap%3A%0A%20%20%20%20%22%22%22%E5%A4%A7%E9%A1%B6%E5%A0%86%22%22%22%0A%0A%20%20%20%20def%20__init__%28self,%20nums%3A%20list%5Bint%5D%29%3A%0A%20%20%20%20%20%20%20%20%22%22%22%E6%9E%84%E9%80%A0%E6%96%B9%E6%B3%95%EF%BC%8C%E6%A0%B9%E6%8D%AE%E8%BE%93%E5%85%A5%E5%88%97%E8%A1%A8%E5%BB%BA%E5%A0%86%22%22%22%0A%20%20%20%20%20%20%20%20%23%20%E5%B0%86%E5%88%97%E8%A1%A8%E5%85%83%E7%B4%A0%E5%8E%9F%E5%B0%81%E4%B8%8D%E5%8A%A8%E6%B7%BB%E5%8A%A0%E8%BF%9B%E5%A0%86%0A%20%20%20%20%20%20%20%20self.max_heap%20%3D%20nums%0A%20%20%20%20%20%20%20%20%23%20%E5%A0%86%E5%8C%96%E9%99%A4%E5%8F%B6%E8%8A%82%E7%82%B9%E4%BB%A5%E5%A4%96%E7%9A%84%E5%85%B6%E4%BB%96%E6%89%80%E6%9C%89%E8%8A%82%E7%82%B9%0A%20%20%20%20%20%20%20%20for%20i%20in%20range%28self.parent%28self.size%28%29%20-%201%29,%20-1,%20-1%29%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20self.sift_down%28i%29%0A%0A%20%20%20%20def%20left%28self,%20i%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%20%20%20%20%22%22%22%E8%8E%B7%E5%8F%96%E5%B7%A6%E5%AD%90%E8%8A%82%E7%82%B9%E7%9A%84%E7%B4%A2%E5%BC%95%22%22%22%0A%20%20%20%20%20%20%20%20return%202%20*%20i%20%2B%201%0A%0A%20%20%20%20def%20right%28self,%20i%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%20%20%20%20%22%22%22%E8%8E%B7%E5%8F%96%E5%8F%B3%E5%AD%90%E8%8A%82%E7%82%B9%E7%9A%84%E7%B4%A2%E5%BC%95%22%22%22%0A%20%20%20%20%20%20%20%20return%202%20*%20i%20%2B%202%0A%0A%20%20%20%20def%20parent%28self,%20i%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%20%20%20%20%22%22%22%E8%8E%B7%E5%8F%96%E7%88%B6%E8%8A%82%E7%82%B9%E7%9A%84%E7%B4%A2%E5%BC%95%22%22%22%0A%20%20%20%20%20%20%20%20return%20%28i%20-%201%29%20//%202%20%20%23%20%E5%90%91%E4%B8%8B%E6%95%B4%E9%99%A4%0A%0A%20%20%20%20def%20swap%28self,%20i%3A%20int,%20j%3A%20int%29%3A%0A%20%20%20%20%20%20%20%20%22%22%22%E4%BA%A4%E6%8D%A2%E5%85%83%E7%B4%A0%22%22%22%0A%20%20%20%20%20%20%20%20self.max_heap%5Bi%5D,%20self.max_heap%5Bj%5D%20%3D%20self.max_heap%5Bj%5D,%20self.max_heap%5Bi%5D%0A%0A%20%20%20%20def%20size%28self%29%20-%3E%20int%3A%0A%20%20%20%20%20%20%20%20%22%22%22%E8%8E%B7%E5%8F%96%E5%A0%86%E5%A4%A7%E5%B0%8F%22%22%22%0A%20%20%20%20%20%20%20%20return%20len%28self.max_heap%29%0A%0A%20%20%20%20def%20sift_down%28self,%20i%3A%20int%29%3A%0A%20%20%20%20%20%20%20%20%22%22%22%E4%BB%8E%E8%8A%82%E7%82%B9%20i%20%E5%BC%80%E5%A7%8B%EF%BC%8C%E4%BB%8E%E9%A1%B6%E8%87%B3%E5%BA%95%E5%A0%86%E5%8C%96%22%22%22%0A%20%20%20%20%20%20%20%20while%20True%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20%23%20%E5%88%A4%E6%96%AD%E8%8A%82%E7%82%B9%20i,%20l,%20r%20%E4%B8%AD%E5%80%BC%E6%9C%80%E5%A4%A7%E7%9A%84%E8%8A%82%E7%82%B9%EF%BC%8C%E8%AE%B0%E4%B8%BA%20ma%0A%20%20%20%20%20%20%20%20%20%20%20%20l,%20r,%20ma%20%3D%20self.left%28i%29,%20self.right%28i%29,%20i%0A%20%20%20%20%20%20%20%20%20%20%20%20if%20l%20%3C%20self.size%28%29%20and%20self.max_heap%5Bl%5D%20%3E%20self.max_heap%5Bma%5D%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20ma%20%3D%20l%0A%20%20%20%20%20%20%20%20%20%20%20%20if%20r%20%3C%20self.size%28%29%20and%20self.max_heap%5Br%5D%20%3E%20self.max_heap%5Bma%5D%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20ma%20%3D%20r%0A%20%20%20%20%20%20%20%20%20%20%20%20%23%20%E8%8B%A5%E8%8A%82%E7%82%B9%20i%20%E6%9C%80%E5%A4%A7%E6%88%96%E7%B4%A2%E5%BC%95%20l,%20r%20%E8%B6%8A%E7%95%8C%EF%BC%8C%E5%88%99%E6%97%A0%E9%A1%BB%E7%BB%A7%E7%BB%AD%E5%A0%86%E5%8C%96%EF%BC%8C%E8%B7%B3%E5%87%BA%0A%20%20%20%20%20%20%20%20%20%20%20%20if%20ma%20%3D%3D%20i%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20break%0A%20%20%20%20%20%20%20%20%20%20%20%20%23%20%E4%BA%A4%E6%8D%A2%E4%B8%A4%E8%8A%82%E7%82%B9%0A%20%20%20%20%20%20%20%20%20%20%20%20self.swap%28i,%20ma%29%0A%20%20%20%20%20%20%20%20%20%20%20%20%23%20%E5%BE%AA%E7%8E%AF%E5%90%91%E4%B8%8B%E5%A0%86%E5%8C%96%0A%20%20%20%20%20%20%20%20%20%20%20%20i%20%3D%20ma%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%E5%A4%A7%E9%A1%B6%E5%A0%86%0A%20%20%20%20max_heap%20%3D%20MaxHeap%28%5B1,%202,%203,%204,%205%5D%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=4&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
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docs/chapter_heap/top_k.md
-6
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@@ -461,12 +461,6 @@ comments: true
end
```
=== "Zig"
```zig title="top_k.zig"
[class]{}-[func]{topKHeap}
```
??? pythontutor "可视化运行"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=import%20heapq%0A%0Adef%20top_k_heap%28nums%3A%20list%5Bint%5D,%20k%3A%20int%29%20-%3E%20list%5Bint%5D%3A%0A%20%20%20%20%22%22%22%E5%9F%BA%E4%BA%8E%E5%A0%86%E6%9F%A5%E6%89%BE%E6%95%B0%E7%BB%84%E4%B8%AD%E6%9C%80%E5%A4%A7%E7%9A%84%20k%20%E4%B8%AA%E5%85%83%E7%B4%A0%22%22%22%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%E5%B0%8F%E9%A1%B6%E5%A0%86%0A%20%20%20%20heap%20%3D%20%5B%5D%0A%20%20%20%20%23%20%E5%B0%86%E6%95%B0%E7%BB%84%E7%9A%84%E5%89%8D%20k%20%E4%B8%AA%E5%85%83%E7%B4%A0%E5%85%A5%E5%A0%86%0A%20%20%20%20for%20i%20in%20range%28k%29%3A%0A%20%20%20%20%20%20%20%20heapq.heappush%28heap,%20nums%5Bi%5D%29%0A%20%20%20%20%23%20%E4%BB%8E%E7%AC%AC%20k%2B1%20%E4%B8%AA%E5%85%83%E7%B4%A0%E5%BC%80%E5%A7%8B%EF%BC%8C%E4%BF%9D%E6%8C%81%E5%A0%86%E7%9A%84%E9%95%BF%E5%BA%A6%E4%B8%BA%20k%0A%20%20%20%20for%20i%20in%20range%28k,%20len%28nums%29%29%3A%0A%20%20%20%20%20%20%20%20%23%20%E8%8B%A5%E5%BD%93%E5%89%8D%E5%85%83%E7%B4%A0%E5%A4%A7%E4%BA%8E%E5%A0%86%E9%A1%B6%E5%85%83%E7%B4%A0%EF%BC%8C%E5%88%99%E5%B0%86%E5%A0%86%E9%A1%B6%E5%85%83%E7%B4%A0%E5%87%BA%E5%A0%86%E3%80%81%E5%BD%93%E5%89%8D%E5%85%83%E7%B4%A0%E5%85%A5%E5%A0%86%0A%20%20%20%20%20%20%20%20if%20nums%5Bi%5D%20%3E%20heap%5B0%5D%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20heapq.heappop%28heap%29%0A%20%20%20%20%20%20%20%20%20%20%20%20heapq.heappush%28heap,%20nums%5Bi%5D%29%0A%20%20%20%20return%20heap%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20nums%20%3D%20%5B1,%207,%206,%203,%202%5D%0A%20%20%20%20k%20%3D%203%0A%0A%20%20%20%20res%20%3D%20top_k_heap%28nums,%20k%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=6&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
docs/chapter_introduction/summary.md
+3 -1
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@@ -4,6 +4,8 @@ comments: true
# 1.3 &nbsp; 小结
### 1. &nbsp; 重点回顾
- 算法在日常生活中无处不在，并不是遥不可及的高深知识。实际上，我们已经在不知不觉中学会了许多算法，用以解决生活中的大小问题。
- 查字典的原理与二分查找算法相一致。二分查找算法体现了分而治之的重要算法思想。
- 整理扑克的过程与插入排序算法非常类似。插入排序算法适合排序小型数据集。
@@ -12,7 +14,7 @@ comments: true
- 数据结构与算法紧密相连。数据结构是算法的基石，而算法为数据结构注入生命力。
- 我们可以将数据结构与算法类比为拼装积木，积木代表数据，积木的形状和连接方式等代表数据结构，拼装积木的步骤则对应算法。
### 1. &nbsp; Q & A
### 2. &nbsp; Q & A
**Q**：作为一名程序员，我在日常工作中从未用算法解决过问题，常用算法都被编程语言封装好了，直接用就可以了；这是否意味着我们工作中的问题还没有到达需要算法的程度？
docs/chapter_preface/suggestions.md
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@@ -184,17 +184,6 @@ comments: true
# 注释
```
=== "Zig"
```zig title=""
// 标题注释，用于标注函数、类、测试样例等
// 内容注释，用于详解代码
// 多行
// 注释
```
## 0.2.2 &nbsp; 在动画图解中高效学习
相较于文字，视频和图片具有更高的信息密度和结构化程度，更易于理解。在本书中，**重点和难点知识将主要通过动画以图解形式展示**，而文字则作为解释与补充。
docs/chapter_preface/summary.md
+2
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@@ -4,6 +4,8 @@ comments: true
# 0.3 &nbsp; 小结
### 1. &nbsp; 重点回顾
- 本书的主要受众是算法初学者。如果你已有一定基础，本书能帮助你系统回顾算法知识，书中源代码也可作为“刷题工具库”使用。
- 书中内容主要包括复杂度分析、数据结构和算法三部分，涵盖了该领域的大部分主题。
- 对于算法新手，在初学阶段阅读一本入门书至关重要，可以少走许多弯路。
docs/chapter_searching/binary_search.md
-48
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@@ -362,30 +362,6 @@ comments: true
end
```
=== "Zig"
```zig title="binary_search.zig"
// 二分查找（双闭区间）
fn binarySearch(comptime T: type, nums: std.ArrayList(T), target: T) T {
// 初始化双闭区间 [0, n-1] ，即 i, j 分别指向数组首元素、尾元素
var i: usize = 0;
var j: usize = nums.items.len - 1;
// 循环，当搜索区间为空时跳出（当 i > j 时为空）
while (i <= j) {
var m = i + (j - i) / 2; // 计算中点索引 m
if (nums.items[m] < target) { // 此情况说明 target 在区间 [m+1, j] 中
i = m + 1;
} else if (nums.items[m] > target) { // 此情况说明 target 在区间 [i, m-1] 中
j = m - 1;
} else { // 找到目标元素，返回其索引
return @intCast(m);
}
}
// 未找到目标元素，返回 -1
return -1;
}
```
??? pythontutor "可视化运行"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20binary_search%28nums%3A%20list%5Bint%5D,%20target%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E4%BA%8C%E5%88%86%E6%9F%A5%E6%89%BE%EF%BC%88%E5%8F%8C%E9%97%AD%E5%8C%BA%E9%97%B4%EF%BC%89%22%22%22%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%E5%8F%8C%E9%97%AD%E5%8C%BA%E9%97%B4%20%5B0,%20n-1%5D%20%EF%BC%8C%E5%8D%B3%20i,%20j%20%E5%88%86%E5%88%AB%E6%8C%87%E5%90%91%E6%95%B0%E7%BB%84%E9%A6%96%E5%85%83%E7%B4%A0%E3%80%81%E5%B0%BE%E5%85%83%E7%B4%A0%0A%20%20%20%20i,%20j%20%3D%200,%20len%28nums%29%20-%201%0A%20%20%20%20%23%20%E5%BE%AA%E7%8E%AF%EF%BC%8C%E5%BD%93%E6%90%9C%E7%B4%A2%E5%8C%BA%E9%97%B4%E4%B8%BA%E7%A9%BA%E6%97%B6%E8%B7%B3%E5%87%BA%EF%BC%88%E5%BD%93%20i%20%3E%20j%20%E6%97%B6%E4%B8%BA%E7%A9%BA%EF%BC%89%0A%20%20%20%20while%20i%20%3C%3D%20j%3A%0A%20%20%20%20%20%20%20%20%23%20%E7%90%86%E8%AE%BA%E4%B8%8A%20Python%20%E7%9A%84%E6%95%B0%E5%AD%97%E5%8F%AF%E4%BB%A5%E6%97%A0%E9%99%90%E5%A4%A7%EF%BC%88%E5%8F%96%E5%86%B3%E4%BA%8E%E5%86%85%E5%AD%98%E5%A4%A7%E5%B0%8F%EF%BC%89%EF%BC%8C%E6%97%A0%E9%A1%BB%E8%80%83%E8%99%91%E5%A4%A7%E6%95%B0%E8%B6%8A%E7%95%8C%E9%97%AE%E9%A2%98%0A%20%20%20%20%20%20%20%20m%20%3D%20%28i%20%2B%20j%29%20//%202%20%20%23%20%E8%AE%A1%E7%AE%97%E4%B8%AD%E7%82%B9%E7%B4%A2%E5%BC%95%20m%0A%20%20%20%20%20%20%20%20if%20nums%5Bm%5D%20%3C%20target%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20i%20%3D%20m%20%2B%201%20%20%23%20%E6%AD%A4%E6%83%85%E5%86%B5%E8%AF%B4%E6%98%8E%20target%20%E5%9C%A8%E5%8C%BA%E9%97%B4%20%5Bm%2B1,%20j%5D%20%E4%B8%AD%0A%20%20%20%20%20%20%20%20elif%20nums%5Bm%5D%20%3E%20target%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20j%20%3D%20m%20-%201%20%20%23%20%E6%AD%A4%E6%83%85%E5%86%B5%E8%AF%B4%E6%98%8E%20target%20%E5%9C%A8%E5%8C%BA%E9%97%B4%20%5Bi,%20m-1%5D%20%E4%B8%AD%0A%20%20%20%20%20%20%20%20else%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20return%20m%20%20%23%20%E6%89%BE%E5%88%B0%E7%9B%AE%E6%A0%87%E5%85%83%E7%B4%A0%EF%BC%8C%E8%BF%94%E5%9B%9E%E5%85%B6%E7%B4%A2%E5%BC%95%0A%20%20%20%20return%20-1%20%20%23%20%E6%9C%AA%E6%89%BE%E5%88%B0%E7%9B%AE%E6%A0%87%E5%85%83%E7%B4%A0%EF%BC%8C%E8%BF%94%E5%9B%9E%20-1%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20target%20%3D%206%0A%20%20%20%20nums%20%3D%20%5B1,%203,%206,%208,%2012,%2015,%2023,%2026,%2031,%2035%5D%0A%0A%20%20%20%20%23%20%E4%BA%8C%E5%88%86%E6%9F%A5%E6%89%BE%EF%BC%88%E5%8F%8C%E9%97%AD%E5%8C%BA%E9%97%B4%EF%BC%89%0A%20%20%20%20index%20%3D%20binary_search%28nums,%20target%29%0A%20%20%20%20print%28%22%E7%9B%AE%E6%A0%87%E5%85%83%E7%B4%A0%206%20%E7%9A%84%E7%B4%A2%E5%BC%95%20%3D%20%22,%20index%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=5&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
@@ -710,30 +686,6 @@ comments: true
end
```
=== "Zig"
```zig title="binary_search.zig"
// 二分查找（左闭右开区间）
fn binarySearchLCRO(comptime T: type, nums: std.ArrayList(T), target: T) T {
// 初始化左闭右开区间 [0, n) ，即 i, j 分别指向数组首元素、尾元素+1
var i: usize = 0;
var j: usize = nums.items.len;
// 循环，当搜索区间为空时跳出（当 i = j 时为空）
while (i <= j) {
var m = i + (j - i) / 2; // 计算中点索引 m
if (nums.items[m] < target) { // 此情况说明 target 在区间 [m+1, j) 中
i = m + 1;
} else if (nums.items[m] > target) { // 此情况说明 target 在区间 [i, m) 中
j = m;
} else { // 找到目标元素，返回其索引
return @intCast(m);
}
}
// 未找到目标元素，返回 -1
return -1;
}
```
??? pythontutor "可视化运行"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20binary_search_lcro%28nums%3A%20list%5Bint%5D,%20target%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E4%BA%8C%E5%88%86%E6%9F%A5%E6%89%BE%EF%BC%88%E5%B7%A6%E9%97%AD%E5%8F%B3%E5%BC%80%E5%8C%BA%E9%97%B4%EF%BC%89%22%22%22%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%E5%B7%A6%E9%97%AD%E5%8F%B3%E5%BC%80%E5%8C%BA%E9%97%B4%20%5B0,%20n%29%20%EF%BC%8C%E5%8D%B3%20i,%20j%20%E5%88%86%E5%88%AB%E6%8C%87%E5%90%91%E6%95%B0%E7%BB%84%E9%A6%96%E5%85%83%E7%B4%A0%E3%80%81%E5%B0%BE%E5%85%83%E7%B4%A0%2B1%0A%20%20%20%20i,%20j%20%3D%200,%20len%28nums%29%0A%20%20%20%20%23%20%E5%BE%AA%E7%8E%AF%EF%BC%8C%E5%BD%93%E6%90%9C%E7%B4%A2%E5%8C%BA%E9%97%B4%E4%B8%BA%E7%A9%BA%E6%97%B6%E8%B7%B3%E5%87%BA%EF%BC%88%E5%BD%93%20i%20%3D%20j%20%E6%97%B6%E4%B8%BA%E7%A9%BA%EF%BC%89%0A%20%20%20%20while%20i%20%3C%20j%3A%0A%20%20%20%20%20%20%20%20m%20%3D%20%28i%20%2B%20j%29%20//%202%20%20%23%20%E8%AE%A1%E7%AE%97%E4%B8%AD%E7%82%B9%E7%B4%A2%E5%BC%95%20m%0A%20%20%20%20%20%20%20%20if%20nums%5Bm%5D%20%3C%20target%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20i%20%3D%20m%20%2B%201%20%20%23%20%E6%AD%A4%E6%83%85%E5%86%B5%E8%AF%B4%E6%98%8E%20target%20%E5%9C%A8%E5%8C%BA%E9%97%B4%20%5Bm%2B1,%20j%29%20%E4%B8%AD%0A%20%20%20%20%20%20%20%20elif%20nums%5Bm%5D%20%3E%20target%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20j%20%3D%20m%20%20%23%20%E6%AD%A4%E6%83%85%E5%86%B5%E8%AF%B4%E6%98%8E%20target%20%E5%9C%A8%E5%8C%BA%E9%97%B4%20%5Bi,%20m%29%20%E4%B8%AD%0A%20%20%20%20%20%20%20%20else%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20return%20m%20%20%23%20%E6%89%BE%E5%88%B0%E7%9B%AE%E6%A0%87%E5%85%83%E7%B4%A0%EF%BC%8C%E8%BF%94%E5%9B%9E%E5%85%B6%E7%B4%A2%E5%BC%95%0A%20%20%20%20return%20-1%20%20%23%20%E6%9C%AA%E6%89%BE%E5%88%B0%E7%9B%AE%E6%A0%87%E5%85%83%E7%B4%A0%EF%BC%8C%E8%BF%94%E5%9B%9E%20-1%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20target%20%3D%206%0A%20%20%20%20nums%20%3D%20%5B1,%203,%206,%208,%2012,%2015,%2023,%2026,%2031,%2035%5D%0A%0A%20%20%20%20%23%20%E4%BA%8C%E5%88%86%E6%9F%A5%E6%89%BE%EF%BC%88%E5%B7%A6%E9%97%AD%E5%8F%B3%E5%BC%80%E5%8C%BA%E9%97%B4%EF%BC%89%0A%20%20%20%20index%20%3D%20binary_search_lcro%28nums,%20target%29%0A%20%20%20%20print%28%22%E7%9B%AE%E6%A0%87%E5%85%83%E7%B4%A0%206%20%E7%9A%84%E7%B4%A2%E5%BC%95%20%3D%20%22,%20index%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=5&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
docs/chapter_searching/binary_search_edge.md
-12
View File
@@ -224,12 +224,6 @@ comments: true
end
```
=== "Zig"
```zig title="binary_search_edge.zig"
[class]{}-[func]{binarySearchLeftEdge}
```
??? pythontutor "可视化运行"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20binary_search_insertion%28nums%3A%20list%5Bint%5D,%20target%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E4%BA%8C%E5%88%86%E6%9F%A5%E6%89%BE%E6%8F%92%E5%85%A5%E7%82%B9%EF%BC%88%E5%AD%98%E5%9C%A8%E9%87%8D%E5%A4%8D%E5%85%83%E7%B4%A0%EF%BC%89%22%22%22%0A%20%20%20%20i,%20j%20%3D%200,%20len%28nums%29%20-%201%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%E5%8F%8C%E9%97%AD%E5%8C%BA%E9%97%B4%20%5B0,%20n-1%5D%0A%20%20%20%20while%20i%20%3C%3D%20j%3A%0A%20%20%20%20%20%20%20%20m%20%3D%20%28i%20%2B%20j%29%20//%202%20%20%23%20%E8%AE%A1%E7%AE%97%E4%B8%AD%E7%82%B9%E7%B4%A2%E5%BC%95%20m%0A%20%20%20%20%20%20%20%20if%20nums%5Bm%5D%20%3C%20target%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20i%20%3D%20m%20%2B%201%20%20%23%20target%20%E5%9C%A8%E5%8C%BA%E9%97%B4%20%5Bm%2B1,%20j%5D%20%E4%B8%AD%0A%20%20%20%20%20%20%20%20elif%20nums%5Bm%5D%20%3E%20target%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20j%20%3D%20m%20-%201%20%20%23%20target%20%E5%9C%A8%E5%8C%BA%E9%97%B4%20%5Bi,%20m-1%5D%20%E4%B8%AD%0A%20%20%20%20%20%20%20%20else%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20j%20%3D%20m%20-%201%20%20%23%20%E9%A6%96%E4%B8%AA%E5%B0%8F%E4%BA%8E%20target%20%E7%9A%84%E5%85%83%E7%B4%A0%E5%9C%A8%E5%8C%BA%E9%97%B4%20%5Bi,%20m-1%5D%20%E4%B8%AD%0A%20%20%20%20%23%20%E8%BF%94%E5%9B%9E%E6%8F%92%E5%85%A5%E7%82%B9%20i%0A%20%20%20%20return%20i%0A%0Adef%20binary_search_left_edge%28nums%3A%20list%5Bint%5D,%20target%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E4%BA%8C%E5%88%86%E6%9F%A5%E6%89%BE%E6%9C%80%E5%B7%A6%E4%B8%80%E4%B8%AA%20target%22%22%22%0A%20%20%20%20%23%20%E7%AD%89%E4%BB%B7%E4%BA%8E%E6%9F%A5%E6%89%BE%20target%20%E7%9A%84%E6%8F%92%E5%85%A5%E7%82%B9%0A%20%20%20%20i%20%3D%20binary_search_insertion%28nums,%20target%29%0A%20%20%20%20%23%20%E6%9C%AA%E6%89%BE%E5%88%B0%20target%20%EF%BC%8C%E8%BF%94%E5%9B%9E%20-1%0A%20%20%20%20if%20i%20%3D%3D%20len%28nums%29%20or%20nums%5Bi%5D%20!%3D%20target%3A%0A%20%20%20%20%20%20%20%20return%20-1%0A%20%20%20%20%23%20%E6%89%BE%E5%88%B0%20target%20%EF%BC%8C%E8%BF%94%E5%9B%9E%E7%B4%A2%E5%BC%95%20i%0A%20%20%20%20return%20i%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20%23%20%E5%8C%85%E5%90%AB%E9%87%8D%E5%A4%8D%E5%85%83%E7%B4%A0%E7%9A%84%E6%95%B0%E7%BB%84%0A%20%20%20%20nums%20%3D%20%5B1,%203,%206,%206,%206,%206,%206,%2010,%2012,%2015%5D%0A%20%20%20%20%23%20%E4%BA%8C%E5%88%86%E6%9F%A5%E6%89%BE%E5%B7%A6%E8%BE%B9%E7%95%8C%E5%92%8C%E5%8F%B3%E8%BE%B9%E7%95%8C%0A%20%20%20%20target%20%3D%206%0A%20%20%20%20index%20%3D%20binary_search_left_edge%28nums,%20target%29%0A%20%20%20%20print%28f%22%E6%9C%80%E5%B7%A6%E4%B8%80%E4%B8%AA%E5%85%83%E7%B4%A0%20%7Btarget%7D%20%E7%9A%84%E7%B4%A2%E5%BC%95%E4%B8%BA%20%7Bindex%7D%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=6&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
@@ -485,12 +479,6 @@ comments: true
end
```
=== "Zig"
```zig title="binary_search_edge.zig"
[class]{}-[func]{binarySearchRightEdge}
```
??? pythontutor "可视化运行"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20binary_search_insertion%28nums%3A%20list%5Bint%5D,%20target%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E4%BA%8C%E5%88%86%E6%9F%A5%E6%89%BE%E6%8F%92%E5%85%A5%E7%82%B9%EF%BC%88%E5%AD%98%E5%9C%A8%E9%87%8D%E5%A4%8D%E5%85%83%E7%B4%A0%EF%BC%89%22%22%22%0A%20%20%20%20i,%20j%20%3D%200,%20len%28nums%29%20-%201%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%E5%8F%8C%E9%97%AD%E5%8C%BA%E9%97%B4%20%5B0,%20n-1%5D%0A%20%20%20%20while%20i%20%3C%3D%20j%3A%0A%20%20%20%20%20%20%20%20m%20%3D%20%28i%20%2B%20j%29%20//%202%20%20%23%20%E8%AE%A1%E7%AE%97%E4%B8%AD%E7%82%B9%E7%B4%A2%E5%BC%95%20m%0A%20%20%20%20%20%20%20%20if%20nums%5Bm%5D%20%3C%20target%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20i%20%3D%20m%20%2B%201%20%20%23%20target%20%E5%9C%A8%E5%8C%BA%E9%97%B4%20%5Bm%2B1,%20j%5D%20%E4%B8%AD%0A%20%20%20%20%20%20%20%20elif%20nums%5Bm%5D%20%3E%20target%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20j%20%3D%20m%20-%201%20%20%23%20target%20%E5%9C%A8%E5%8C%BA%E9%97%B4%20%5Bi,%20m-1%5D%20%E4%B8%AD%0A%20%20%20%20%20%20%20%20else%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20j%20%3D%20m%20-%201%20%20%23%20%E9%A6%96%E4%B8%AA%E5%B0%8F%E4%BA%8E%20target%20%E7%9A%84%E5%85%83%E7%B4%A0%E5%9C%A8%E5%8C%BA%E9%97%B4%20%5Bi,%20m-1%5D%20%E4%B8%AD%0A%20%20%20%20%23%20%E8%BF%94%E5%9B%9E%E6%8F%92%E5%85%A5%E7%82%B9%20i%0A%20%20%20%20return%20i%0A%0Adef%20binary_search_right_edge%28nums%3A%20list%5Bint%5D,%20target%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E4%BA%8C%E5%88%86%E6%9F%A5%E6%89%BE%E6%9C%80%E5%8F%B3%E4%B8%80%E4%B8%AA%20target%22%22%22%0A%20%20%20%20%23%20%E8%BD%AC%E5%8C%96%E4%B8%BA%E6%9F%A5%E6%89%BE%E6%9C%80%E5%B7%A6%E4%B8%80%E4%B8%AA%20target%20%2B%201%0A%20%20%20%20i%20%3D%20binary_search_insertion%28nums,%20target%20%2B%201%29%0A%20%20%20%20%23%20j%20%E6%8C%87%E5%90%91%E6%9C%80%E5%8F%B3%E4%B8%80%E4%B8%AA%20target%20%EF%BC%8Ci%20%E6%8C%87%E5%90%91%E9%A6%96%E4%B8%AA%E5%A4%A7%E4%BA%8E%20target%20%E7%9A%84%E5%85%83%E7%B4%A0%0A%20%20%20%20j%20%3D%20i%20-%201%0A%20%20%20%20%23%20%E6%9C%AA%E6%89%BE%E5%88%B0%20target%20%EF%BC%8C%E8%BF%94%E5%9B%9E%20-1%0A%20%20%20%20if%20j%20%3D%3D%20-1%20or%20nums%5Bj%5D%20!%3D%20target%3A%0A%20%20%20%20%20%20%20%20return%20-1%0A%20%20%20%20%23%20%E6%89%BE%E5%88%B0%20target%20%EF%BC%8C%E8%BF%94%E5%9B%9E%E7%B4%A2%E5%BC%95%20j%0A%20%20%20%20return%20j%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20%23%20%E5%8C%85%E5%90%AB%E9%87%8D%E5%A4%8D%E5%85%83%E7%B4%A0%E7%9A%84%E6%95%B0%E7%BB%84%0A%20%20%20%20nums%20%3D%20%5B1,%203,%206,%206,%206,%206,%206,%2010,%2012,%2015%5D%0A%20%20%20%20%23%20%E4%BA%8C%E5%88%86%E6%9F%A5%E6%89%BE%E5%B7%A6%E8%BE%B9%E7%95%8C%E5%92%8C%E5%8F%B3%E8%BE%B9%E7%95%8C%0A%20%20%20%20target%20%3D%206%0A%20%20%20%20index%20%3D%20binary_search_right_edge%28nums,%20target%29%0A%20%20%20%20print%28f%22%E6%9C%80%E5%8F%B3%E4%B8%80%E4%B8%AA%E5%85%83%E7%B4%A0%20%7Btarget%7D%20%E7%9A%84%E7%B4%A2%E5%BC%95%E4%B8%BA%20%7Bindex%7D%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=6&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
docs/chapter_searching/binary_search_insertion.md
-12
View File
@@ -315,12 +315,6 @@ comments: true
end
```
=== "Zig"
```zig title="binary_search_insertion.zig"
[class]{}-[func]{binarySearchInsertionSimple}
```
??? pythontutor "可视化运行"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20binary_search_insertion_simple%28nums%3A%20list%5Bint%5D,%20target%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E4%BA%8C%E5%88%86%E6%9F%A5%E6%89%BE%E6%8F%92%E5%85%A5%E7%82%B9%EF%BC%88%E6%97%A0%E9%87%8D%E5%A4%8D%E5%85%83%E7%B4%A0%EF%BC%89%22%22%22%0A%20%20%20%20i,%20j%20%3D%200,%20len%28nums%29%20-%201%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%E5%8F%8C%E9%97%AD%E5%8C%BA%E9%97%B4%20%5B0,%20n-1%5D%0A%20%20%20%20while%20i%20%3C%3D%20j%3A%0A%20%20%20%20%20%20%20%20m%20%3D%20%28i%20%2B%20j%29%20//%202%20%20%23%20%E8%AE%A1%E7%AE%97%E4%B8%AD%E7%82%B9%E7%B4%A2%E5%BC%95%20m%0A%20%20%20%20%20%20%20%20if%20nums%5Bm%5D%20%3C%20target%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20i%20%3D%20m%20%2B%201%20%20%23%20target%20%E5%9C%A8%E5%8C%BA%E9%97%B4%20%5Bm%2B1,%20j%5D%20%E4%B8%AD%0A%20%20%20%20%20%20%20%20elif%20nums%5Bm%5D%20%3E%20target%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20j%20%3D%20m%20-%201%20%20%23%20target%20%E5%9C%A8%E5%8C%BA%E9%97%B4%20%5Bi,%20m-1%5D%20%E4%B8%AD%0A%20%20%20%20%20%20%20%20else%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20return%20m%20%20%23%20%E6%89%BE%E5%88%B0%20target%20%EF%BC%8C%E8%BF%94%E5%9B%9E%E6%8F%92%E5%85%A5%E7%82%B9%20m%0A%20%20%20%20%23%20%E6%9C%AA%E6%89%BE%E5%88%B0%20target%20%EF%BC%8C%E8%BF%94%E5%9B%9E%E6%8F%92%E5%85%A5%E7%82%B9%20i%0A%20%20%20%20return%20i%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20%23%20%E6%97%A0%E9%87%8D%E5%A4%8D%E5%85%83%E7%B4%A0%E7%9A%84%E6%95%B0%E7%BB%84%0A%20%20%20%20nums%20%3D%20%5B1,%203,%206,%208,%2012,%2015,%2023,%2026,%2031,%2035%5D%0A%20%20%20%20%23%20%E4%BA%8C%E5%88%86%E6%9F%A5%E6%89%BE%E6%8F%92%E5%85%A5%E7%82%B9%0A%20%20%20%20target%20%3D%206%0A%20%20%20%20index%20%3D%20binary_search_insertion_simple%28nums,%20target%29%0A%20%20%20%20print%28f%22%E5%85%83%E7%B4%A0%20%7Btarget%7D%20%E7%9A%84%E6%8F%92%E5%85%A5%E7%82%B9%E7%9A%84%E7%B4%A2%E5%BC%95%E4%B8%BA%20%7Bindex%7D%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=5&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
@@ -666,12 +660,6 @@ comments: true
end
```
=== "Zig"
```zig title="binary_search_insertion.zig"
[class]{}-[func]{binarySearchInsertion}
```
??? pythontutor "可视化运行"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20binary_search_insertion%28nums%3A%20list%5Bint%5D,%20target%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E4%BA%8C%E5%88%86%E6%9F%A5%E6%89%BE%E6%8F%92%E5%85%A5%E7%82%B9%EF%BC%88%E5%AD%98%E5%9C%A8%E9%87%8D%E5%A4%8D%E5%85%83%E7%B4%A0%EF%BC%89%22%22%22%0A%20%20%20%20i,%20j%20%3D%200,%20len%28nums%29%20-%201%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%E5%8F%8C%E9%97%AD%E5%8C%BA%E9%97%B4%20%5B0,%20n-1%5D%0A%20%20%20%20while%20i%20%3C%3D%20j%3A%0A%20%20%20%20%20%20%20%20m%20%3D%20%28i%20%2B%20j%29%20//%202%20%20%23%20%E8%AE%A1%E7%AE%97%E4%B8%AD%E7%82%B9%E7%B4%A2%E5%BC%95%20m%0A%20%20%20%20%20%20%20%20if%20nums%5Bm%5D%20%3C%20target%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20i%20%3D%20m%20%2B%201%20%20%23%20target%20%E5%9C%A8%E5%8C%BA%E9%97%B4%20%5Bm%2B1,%20j%5D%20%E4%B8%AD%0A%20%20%20%20%20%20%20%20elif%20nums%5Bm%5D%20%3E%20target%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20j%20%3D%20m%20-%201%20%20%23%20target%20%E5%9C%A8%E5%8C%BA%E9%97%B4%20%5Bi,%20m-1%5D%20%E4%B8%AD%0A%20%20%20%20%20%20%20%20else%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20j%20%3D%20m%20-%201%20%20%23%20%E9%A6%96%E4%B8%AA%E5%B0%8F%E4%BA%8E%20target%20%E7%9A%84%E5%85%83%E7%B4%A0%E5%9C%A8%E5%8C%BA%E9%97%B4%20%5Bi,%20m-1%5D%20%E4%B8%AD%0A%20%20%20%20%23%20%E8%BF%94%E5%9B%9E%E6%8F%92%E5%85%A5%E7%82%B9%20i%0A%20%20%20%20return%20i%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20%23%20%E5%8C%85%E5%90%AB%E9%87%8D%E5%A4%8D%E5%85%83%E7%B4%A0%E7%9A%84%E6%95%B0%E7%BB%84%0A%20%20%20%20nums%20%3D%20%5B1,%203,%206,%206,%206,%206,%206,%2010,%2012,%2015%5D%0A%20%20%20%20%23%20%E4%BA%8C%E5%88%86%E6%9F%A5%E6%89%BE%E6%8F%92%E5%85%A5%E7%82%B9%0A%20%20%20%20target%20%3D%206%0A%20%20%20%20index%20%3D%20binary_search_insertion%28nums,%20target%29%0A%20%20%20%20print%28f%22%E5%85%83%E7%B4%A0%20%7Btarget%7D%20%E7%9A%84%E6%8F%92%E5%85%A5%E7%82%B9%E7%9A%84%E7%B4%A2%E5%BC%95%E4%B8%BA%20%7Bindex%7D%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=5&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
docs/chapter_searching/replace_linear_by_hashing.md
-41
View File
@@ -241,26 +241,6 @@ comments: true
end
```
=== "Zig"
```zig title="two_sum.zig"
// 方法一：暴力枚举
fn twoSumBruteForce(nums: []i32, target: i32) ?[2]i32 {
var size: usize = nums.len;
var i: usize = 0;
// 两层循环，时间复杂度为 O(n^2)
while (i < size - 1) : (i += 1) {
var j = i + 1;
while (j < size) : (j += 1) {
if (nums[i] + nums[j] == target) {
return [_]i32{@intCast(i), @intCast(j)};
}
}
}
return null;
}
```
??? pythontutor "可视化运行"
<div style="height: 441px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20two_sum_brute_force%28nums%3A%20list%5Bint%5D,%20target%3A%20int%29%20-%3E%20list%5Bint%5D%3A%0A%20%20%20%20%22%22%22%E6%96%B9%E6%B3%95%E4%B8%80%EF%BC%9A%E6%9A%B4%E5%8A%9B%E6%9E%9A%E4%B8%BE%22%22%22%0A%20%20%20%20%23%20%E4%B8%A4%E5%B1%82%E5%BE%AA%E7%8E%AF%EF%BC%8C%E6%97%B6%E9%97%B4%E5%A4%8D%E6%9D%82%E5%BA%A6%E4%B8%BA%20O%28n%5E2%29%0A%20%20%20%20for%20i%20in%20range%28len%28nums%29%20-%201%29%3A%0A%20%20%20%20%20%20%20%20for%20j%20in%20range%28i%20%2B%201,%20len%28nums%29%29%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20if%20nums%5Bi%5D%20%2B%20nums%5Bj%5D%20%3D%3D%20target%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20return%20%5Bi,%20j%5D%0A%20%20%20%20return%20%5B%5D%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20nums%20%3D%20%5B2,%207,%2011,%2015%5D%0A%20%20%20%20target%20%3D%2013%0A%20%20%20%20res%20%3D%20two_sum_brute_force%28nums,%20target%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=5&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
@@ -556,27 +536,6 @@ comments: true
end
```
=== "Zig"
```zig title="two_sum.zig"
// 方法二：辅助哈希表
fn twoSumHashTable(nums: []i32, target: i32) !?[2]i32 {
var size: usize = nums.len;
// 辅助哈希表，空间复杂度为 O(n)
var dic = std.AutoHashMap(i32, i32).init(std.heap.page_allocator);
defer dic.deinit();
var i: usize = 0;
// 单层循环，时间复杂度为 O(n)
while (i < size) : (i += 1) {
if (dic.contains(target - nums[i])) {
return [_]i32{dic.get(target - nums[i]).?, @intCast(i)};
}
try dic.put(nums[i], @intCast(i));
}
return null;
}
```
??? pythontutor "可视化运行"
<div style="height: 477px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20two_sum_hash_table%28nums%3A%20list%5Bint%5D,%20target%3A%20int%29%20-%3E%20list%5Bint%5D%3A%0A%20%20%20%20%22%22%22%E6%96%B9%E6%B3%95%E4%BA%8C%EF%BC%9A%E8%BE%85%E5%8A%A9%E5%93%88%E5%B8%8C%E8%A1%A8%22%22%22%0A%20%20%20%20%23%20%E8%BE%85%E5%8A%A9%E5%93%88%E5%B8%8C%E8%A1%A8%EF%BC%8C%E7%A9%BA%E9%97%B4%E5%A4%8D%E6%9D%82%E5%BA%A6%E4%B8%BA%20O%28n%29%0A%20%20%20%20dic%20%3D%20%7B%7D%0A%20%20%20%20%23%20%E5%8D%95%E5%B1%82%E5%BE%AA%E7%8E%AF%EF%BC%8C%E6%97%B6%E9%97%B4%E5%A4%8D%E6%9D%82%E5%BA%A6%E4%B8%BA%20O%28n%29%0A%20%20%20%20for%20i%20in%20range%28len%28nums%29%29%3A%0A%20%20%20%20%20%20%20%20if%20target%20-%20nums%5Bi%5D%20in%20dic%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20return%20%5Bdic%5Btarget%20-%20nums%5Bi%5D%5D,%20i%5D%0A%20%20%20%20%20%20%20%20dic%5Bnums%5Bi%5D%5D%20%3D%20i%0A%20%20%20%20return%20%5B%5D%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20nums%20%3D%20%5B2,%207,%2011,%2015%5D%0A%20%20%20%20target%20%3D%2013%0A%20%20%20%20res%20%3D%20two_sum_hash_table%28nums,%20target%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=5&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
docs/chapter_searching/summary.md
+2
View File
@@ -4,6 +4,8 @@ comments: true
# 10.6 &nbsp; 小结
### 1. &nbsp; 重点回顾
- 二分查找依赖数据的有序性，通过循环逐步缩减一半搜索区间来进行查找。它要求输入数据有序，且仅适用于数组或基于数组实现的数据结构。
- 暴力搜索通过遍历数据结构来定位数据。线性搜索适用于数组和链表，广度优先搜索和深度优先搜索适用于图和树。此类算法通用性好，无须对数据进行预处理，但时间复杂度 $O(n)$ 较高。
- 哈希查找、树查找和二分查找属于高效搜索方法，可在特定数据结构中快速定位目标元素。此类算法效率高，时间复杂度可达 $O(\log n)$ 甚至 $O(1)$ ，但通常需要借助额外数据结构。
docs/chapter_sorting/bubble_sort.md
-47
View File
@@ -290,28 +290,6 @@ comments: true
end
```
=== "Zig"
```zig title="bubble_sort.zig"
// 冒泡排序
fn bubbleSort(nums: []i32) void {
// 外循环：未排序区间为 [0, i]
var i: usize = nums.len - 1;
while (i > 0) : (i -= 1) {
var j: usize = 0;
// 内循环：将未排序区间 [0, i] 中的最大元素交换至该区间的最右端
while (j < i) : (j += 1) {
if (nums[j] > nums[j + 1]) {
// 交换 nums[j] 与 nums[j + 1]
var tmp = nums[j];
nums[j] = nums[j + 1];
nums[j + 1] = tmp;
}
}
}
}
```
??? pythontutor "可视化运行"
<div style="height: 477px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20bubble_sort%28nums%3A%20list%5Bint%5D%29%3A%0A%20%20%20%20%22%22%22%E5%86%92%E6%B3%A1%E6%8E%92%E5%BA%8F%22%22%22%0A%20%20%20%20n%20%3D%20len%28nums%29%0A%20%20%20%20%23%20%E5%A4%96%E5%BE%AA%E7%8E%AF%EF%BC%9A%E6%9C%AA%E6%8E%92%E5%BA%8F%E5%8C%BA%E9%97%B4%E4%B8%BA%20%5B0,%20i%5D%0A%20%20%20%20for%20i%20in%20range%28n%20-%201,%200,%20-1%29%3A%0A%20%20%20%20%20%20%20%20%23%20%E5%86%85%E5%BE%AA%E7%8E%AF%EF%BC%9A%E5%B0%86%E6%9C%AA%E6%8E%92%E5%BA%8F%E5%8C%BA%E9%97%B4%20%5B0,%20i%5D%20%E4%B8%AD%E7%9A%84%E6%9C%80%E5%A4%A7%E5%85%83%E7%B4%A0%E4%BA%A4%E6%8D%A2%E8%87%B3%E8%AF%A5%E5%8C%BA%E9%97%B4%E7%9A%84%E6%9C%80%E5%8F%B3%E7%AB%AF%0A%20%20%20%20%20%20%20%20for%20j%20in%20range%28i%29%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20if%20nums%5Bj%5D%20%3E%20nums%5Bj%20%2B%201%5D%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%23%20%E4%BA%A4%E6%8D%A2%20nums%5Bj%5D%20%E4%B8%8E%20nums%5Bj%20%2B%201%5D%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20nums%5Bj%5D,%20nums%5Bj%20%2B%201%5D%20%3D%20nums%5Bj%20%2B%201%5D,%20nums%5Bj%5D%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20nums%20%3D%20%5B4,%201,%203,%201,%205,%202%5D%0A%20%20%20%20bubble_sort%28nums%29%0A%20%20%20%20print%28%22%E5%86%92%E6%B3%A1%E6%8E%92%E5%BA%8F%E5%AE%8C%E6%88%90%E5%90%8E%20nums%20%3D%22,%20nums%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=4&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
@@ -617,31 +595,6 @@ comments: true
end
```
=== "Zig"
```zig title="bubble_sort.zig"
// 冒泡排序（标志优化）
fn bubbleSortWithFlag(nums: []i32) void {
// 外循环：未排序区间为 [0, i]
var i: usize = nums.len - 1;
while (i > 0) : (i -= 1) {
var flag = false; // 初始化标志位
var j: usize = 0;
// 内循环：将未排序区间 [0, i] 中的最大元素交换至该区间的最右端
while (j < i) : (j += 1) {
if (nums[j] > nums[j + 1]) {
// 交换 nums[j] 与 nums[j + 1]
var tmp = nums[j];
nums[j] = nums[j + 1];
nums[j + 1] = tmp;
flag = true;
}
}
if (!flag) break; // 此轮“冒泡”未交换任何元素，直接跳出
}
}
```
??? pythontutor "可视化运行"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20bubble_sort_with_flag%28nums%3A%20list%5Bint%5D%29%3A%0A%20%20%20%20%22%22%22%E5%86%92%E6%B3%A1%E6%8E%92%E5%BA%8F%EF%BC%88%E6%A0%87%E5%BF%97%E4%BC%98%E5%8C%96%EF%BC%89%22%22%22%0A%20%20%20%20n%20%3D%20len%28nums%29%0A%20%20%20%20%23%20%E5%A4%96%E5%BE%AA%E7%8E%AF%EF%BC%9A%E6%9C%AA%E6%8E%92%E5%BA%8F%E5%8C%BA%E9%97%B4%E4%B8%BA%20%5B0,%20i%5D%0A%20%20%20%20for%20i%20in%20range%28n%20-%201,%200,%20-1%29%3A%0A%20%20%20%20%20%20%20%20flag%20%3D%20False%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%E6%A0%87%E5%BF%97%E4%BD%8D%0A%20%20%20%20%20%20%20%20%23%20%E5%86%85%E5%BE%AA%E7%8E%AF%EF%BC%9A%E5%B0%86%E6%9C%AA%E6%8E%92%E5%BA%8F%E5%8C%BA%E9%97%B4%20%5B0,%20i%5D%20%E4%B8%AD%E7%9A%84%E6%9C%80%E5%A4%A7%E5%85%83%E7%B4%A0%E4%BA%A4%E6%8D%A2%E8%87%B3%E8%AF%A5%E5%8C%BA%E9%97%B4%E7%9A%84%E6%9C%80%E5%8F%B3%E7%AB%AF%0A%20%20%20%20%20%20%20%20for%20j%20in%20range%28i%29%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20if%20nums%5Bj%5D%20%3E%20nums%5Bj%20%2B%201%5D%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%23%20%E4%BA%A4%E6%8D%A2%20nums%5Bj%5D%20%E4%B8%8E%20nums%5Bj%20%2B%201%5D%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20nums%5Bj%5D,%20nums%5Bj%20%2B%201%5D%20%3D%20nums%5Bj%20%2B%201%5D,%20nums%5Bj%5D%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20flag%20%3D%20True%20%20%23%20%E8%AE%B0%E5%BD%95%E4%BA%A4%E6%8D%A2%E5%85%83%E7%B4%A0%0A%20%20%20%20%20%20%20%20if%20not%20flag%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20break%20%20%23%20%E6%AD%A4%E8%BD%AE%E2%80%9C%E5%86%92%E6%B3%A1%E2%80%9D%E6%9C%AA%E4%BA%A4%E6%8D%A2%E4%BB%BB%E4%BD%95%E5%85%83%E7%B4%A0%EF%BC%8C%E7%9B%B4%E6%8E%A5%E8%B7%B3%E5%87%BA%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20nums%20%3D%20%5B4,%201,%203,%201,%205,%202%5D%0A%20%20%20%20bubble_sort_with_flag%28nums%29%0A%20%20%20%20print%28%22%E5%86%92%E6%B3%A1%E6%8E%92%E5%BA%8F%E5%AE%8C%E6%88%90%E5%90%8E%20nums%20%3D%22,%20nums%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=4&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
docs/chapter_sorting/bucket_sort.md
-6
View File
@@ -437,12 +437,6 @@ comments: true
end
```
=== "Zig"
```zig title="bucket_sort.zig"
[class]{}-[func]{bucketSort}
```
??? pythontutor "可视化运行"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20bucket_sort%28nums%3A%20list%5Bfloat%5D%29%3A%0A%20%20%20%20%22%22%22%E6%A1%B6%E6%8E%92%E5%BA%8F%22%22%22%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%20k%20%3D%20n/2%20%E4%B8%AA%E6%A1%B6%EF%BC%8C%E9%A2%84%E6%9C%9F%E5%90%91%E6%AF%8F%E4%B8%AA%E6%A1%B6%E5%88%86%E9%85%8D%202%20%E4%B8%AA%E5%85%83%E7%B4%A0%0A%20%20%20%20k%20%3D%20len%28nums%29%20//%202%0A%20%20%20%20buckets%20%3D%20%5B%5B%5D%20for%20_%20in%20range%28k%29%5D%0A%20%20%20%20%23%201.%20%E5%B0%86%E6%95%B0%E7%BB%84%E5%85%83%E7%B4%A0%E5%88%86%E9%85%8D%E5%88%B0%E5%90%84%E4%B8%AA%E6%A1%B6%E4%B8%AD%0A%20%20%20%20for%20num%20in%20nums%3A%0A%20%20%20%20%20%20%20%20%23%20%E8%BE%93%E5%85%A5%E6%95%B0%E6%8D%AE%E8%8C%83%E5%9B%B4%E4%B8%BA%20%5B0,%201%29%EF%BC%8C%E4%BD%BF%E7%94%A8%20num%20*%20k%20%E6%98%A0%E5%B0%84%E5%88%B0%E7%B4%A2%E5%BC%95%E8%8C%83%E5%9B%B4%20%5B0,%20k-1%5D%0A%20%20%20%20%20%20%20%20i%20%3D%20int%28num%20*%20k%29%0A%20%20%20%20%20%20%20%20%23%20%E5%B0%86%20num%20%E6%B7%BB%E5%8A%A0%E8%BF%9B%E6%A1%B6%20i%0A%20%20%20%20%20%20%20%20buckets%5Bi%5D.append%28num%29%0A%20%20%20%20%23%202.%20%E5%AF%B9%E5%90%84%E4%B8%AA%E6%A1%B6%E6%89%A7%E8%A1%8C%E6%8E%92%E5%BA%8F%0A%20%20%20%20for%20bucket%20in%20buckets%3A%0A%20%20%20%20%20%20%20%20%23%20%E4%BD%BF%E7%94%A8%E5%86%85%E7%BD%AE%E6%8E%92%E5%BA%8F%E5%87%BD%E6%95%B0%EF%BC%8C%E4%B9%9F%E5%8F%AF%E4%BB%A5%E6%9B%BF%E6%8D%A2%E6%88%90%E5%85%B6%E4%BB%96%E6%8E%92%E5%BA%8F%E7%AE%97%E6%B3%95%0A%20%20%20%20%20%20%20%20bucket.sort%28%29%0A%20%20%20%20%23%203.%20%E9%81%8D%E5%8E%86%E6%A1%B6%E5%90%88%E5%B9%B6%E7%BB%93%E6%9E%9C%0A%20%20%20%20i%20%3D%200%0A%20%20%20%20for%20bucket%20in%20buckets%3A%0A%20%20%20%20%20%20%20%20for%20num%20in%20bucket%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20nums%5Bi%5D%20%3D%20num%0A%20%20%20%20%20%20%20%20%20%20%20%20i%20%2B%3D%201%0A%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20%23%20%E8%AE%BE%E8%BE%93%E5%85%A5%E6%95%B0%E6%8D%AE%E4%B8%BA%E6%B5%AE%E7%82%B9%E6%95%B0%EF%BC%8C%E8%8C%83%E5%9B%B4%E4%B8%BA%20%5B0,%201%29%0A%20%20%20%20nums%20%3D%20%5B0.49,%200.96,%200.82,%200.09,%200.57,%200.43,%200.91,%200.75,%200.15,%200.37%5D%0A%20%20%20%20bucket_sort%28nums%29%0A%20%20%20%20print%28%22%E6%A1%B6%E6%8E%92%E5%BA%8F%E5%AE%8C%E6%88%90%E5%90%8E%20nums%20%3D%22,%20nums%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
docs/chapter_sorting/counting_sort.md
-12
View File
@@ -361,12 +361,6 @@ comments: true
end
```
=== "Zig"
```zig title="counting_sort.zig"
[class]{}-[func]{countingSortNaive}
```
??? pythontutor "可视化运行"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20counting_sort_naive%28nums%3A%20list%5Bint%5D%29%3A%0A%20%20%20%20%22%22%22%E8%AE%A1%E6%95%B0%E6%8E%92%E5%BA%8F%22%22%22%0A%20%20%20%20%23%20%E7%AE%80%E5%8D%95%E5%AE%9E%E7%8E%B0%EF%BC%8C%E6%97%A0%E6%B3%95%E7%94%A8%E4%BA%8E%E6%8E%92%E5%BA%8F%E5%AF%B9%E8%B1%A1%0A%20%20%20%20%23%201.%20%E7%BB%9F%E8%AE%A1%E6%95%B0%E7%BB%84%E6%9C%80%E5%A4%A7%E5%85%83%E7%B4%A0%20m%0A%20%20%20%20m%20%3D%200%0A%20%20%20%20for%20num%20in%20nums%3A%0A%20%20%20%20%20%20%20%20m%20%3D%20max%28m,%20num%29%0A%20%20%20%20%23%202.%20%E7%BB%9F%E8%AE%A1%E5%90%84%E6%95%B0%E5%AD%97%E7%9A%84%E5%87%BA%E7%8E%B0%E6%AC%A1%E6%95%B0%0A%20%20%20%20%23%20counter%5Bnum%5D%20%E4%BB%A3%E8%A1%A8%20num%20%E7%9A%84%E5%87%BA%E7%8E%B0%E6%AC%A1%E6%95%B0%0A%20%20%20%20counter%20%3D%20%5B0%5D%20*%20%28m%20%2B%201%29%0A%20%20%20%20for%20num%20in%20nums%3A%0A%20%20%20%20%20%20%20%20counter%5Bnum%5D%20%2B%3D%201%0A%20%20%20%20%23%203.%20%E9%81%8D%E5%8E%86%20counter%20%EF%BC%8C%E5%B0%86%E5%90%84%E5%85%83%E7%B4%A0%E5%A1%AB%E5%85%A5%E5%8E%9F%E6%95%B0%E7%BB%84%20nums%0A%20%20%20%20i%20%3D%200%0A%20%20%20%20for%20num%20in%20range%28m%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20for%20_%20in%20range%28counter%5Bnum%5D%29%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20nums%5Bi%5D%20%3D%20num%0A%20%20%20%20%20%20%20%20%20%20%20%20i%20%2B%3D%201%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20nums%20%3D%20%5B1,%200,%201,%202,%200,%204,%200,%202,%202,%204%5D%0A%20%20%20%20counting_sort_naive%28nums%29%0A%20%20%20%20print%28f%22%E8%AE%A1%E6%95%B0%E6%8E%92%E5%BA%8F%EF%BC%88%E6%97%A0%E6%B3%95%E6%8E%92%E5%BA%8F%E5%AF%B9%E8%B1%A1%EF%BC%89%E5%AE%8C%E6%88%90%E5%90%8E%20nums%20%3D%20%7Bnums%7D%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=4&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
@@ -883,12 +877,6 @@ $$
end
```
=== "Zig"
```zig title="counting_sort.zig"
[class]{}-[func]{countingSort}
```
??? pythontutor "可视化运行"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20counting_sort%28nums%3A%20list%5Bint%5D%29%3A%0A%20%20%20%20%22%22%22%E8%AE%A1%E6%95%B0%E6%8E%92%E5%BA%8F%22%22%22%0A%20%20%20%20%23%20%E5%AE%8C%E6%95%B4%E5%AE%9E%E7%8E%B0%EF%BC%8C%E5%8F%AF%E6%8E%92%E5%BA%8F%E5%AF%B9%E8%B1%A1%EF%BC%8C%E5%B9%B6%E4%B8%94%E6%98%AF%E7%A8%B3%E5%AE%9A%E6%8E%92%E5%BA%8F%0A%20%20%20%20%23%201.%20%E7%BB%9F%E8%AE%A1%E6%95%B0%E7%BB%84%E6%9C%80%E5%A4%A7%E5%85%83%E7%B4%A0%20m%0A%20%20%20%20m%20%3D%20max%28nums%29%0A%20%20%20%20%23%202.%20%E7%BB%9F%E8%AE%A1%E5%90%84%E6%95%B0%E5%AD%97%E7%9A%84%E5%87%BA%E7%8E%B0%E6%AC%A1%E6%95%B0%0A%20%20%20%20%23%20counter%5Bnum%5D%20%E4%BB%A3%E8%A1%A8%20num%20%E7%9A%84%E5%87%BA%E7%8E%B0%E6%AC%A1%E6%95%B0%0A%20%20%20%20counter%20%3D%20%5B0%5D%20*%20%28m%20%2B%201%29%0A%20%20%20%20for%20num%20in%20nums%3A%0A%20%20%20%20%20%20%20%20counter%5Bnum%5D%20%2B%3D%201%0A%20%20%20%20%23%203.%20%E6%B1%82%20counter%20%E7%9A%84%E5%89%8D%E7%BC%80%E5%92%8C%EF%BC%8C%E5%B0%86%E2%80%9C%E5%87%BA%E7%8E%B0%E6%AC%A1%E6%95%B0%E2%80%9D%E8%BD%AC%E6%8D%A2%E4%B8%BA%E2%80%9C%E5%B0%BE%E7%B4%A2%E5%BC%95%E2%80%9D%0A%20%20%20%20%23%20%E5%8D%B3%20counter%5Bnum%5D-1%20%E6%98%AF%20num%20%E5%9C%A8%20res%20%E4%B8%AD%E6%9C%80%E5%90%8E%E4%B8%80%E6%AC%A1%E5%87%BA%E7%8E%B0%E7%9A%84%E7%B4%A2%E5%BC%95%0A%20%20%20%20for%20i%20in%20range%28m%29%3A%0A%20%20%20%20%20%20%20%20counter%5Bi%20%2B%201%5D%20%2B%3D%20counter%5Bi%5D%0A%20%20%20%20%23%204.%20%E5%80%92%E5%BA%8F%E9%81%8D%E5%8E%86%20nums%20%EF%BC%8C%E5%B0%86%E5%90%84%E5%85%83%E7%B4%A0%E5%A1%AB%E5%85%A5%E7%BB%93%E6%9E%9C%E6%95%B0%E7%BB%84%20res%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%E6%95%B0%E7%BB%84%20res%20%E7%94%A8%E4%BA%8E%E8%AE%B0%E5%BD%95%E7%BB%93%E6%9E%9C%0A%20%20%20%20n%20%3D%20len%28nums%29%0A%20%20%20%20res%20%3D%20%5B0%5D%20*%20n%0A%20%20%20%20for%20i%20in%20range%28n%20-%201,%20-1,%20-1%29%3A%0A%20%20%20%20%20%20%20%20num%20%3D%20nums%5Bi%5D%0A%20%20%20%20%20%20%20%20res%5Bcounter%5Bnum%5D%20-%201%5D%20%3D%20num%20%20%23%20%E5%B0%86%20num%20%E6%94%BE%E7%BD%AE%E5%88%B0%E5%AF%B9%E5%BA%94%E7%B4%A2%E5%BC%95%E5%A4%84%0A%20%20%20%20%20%20%20%20counter%5Bnum%5D%20-%3D%201%20%20%23%20%E4%BB%A4%E5%89%8D%E7%BC%80%E5%92%8C%E8%87%AA%E5%87%8F%201%20%EF%BC%8C%E5%BE%97%E5%88%B0%E4%B8%8B%E6%AC%A1%E6%94%BE%E7%BD%AE%20num%20%E7%9A%84%E7%B4%A2%E5%BC%95%0A%20%20%20%20%23%20%E4%BD%BF%E7%94%A8%E7%BB%93%E6%9E%9C%E6%95%B0%E7%BB%84%20res%20%E8%A6%86%E7%9B%96%E5%8E%9F%E6%95%B0%E7%BB%84%20nums%0A%20%20%20%20for%20i%20in%20range%28n%29%3A%0A%20%20%20%20%20%20%20%20nums%5Bi%5D%20%3D%20res%5Bi%5D%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20nums%20%3D%20%5B1,%200,%201,%202,%200,%204,%200,%202,%202,%204%5D%0A%20%20%20%20counting_sort%28nums%29%0A%20%20%20%20print%28f%22%E8%AE%A1%E6%95%B0%E6%8E%92%E5%BA%8F%E5%AE%8C%E6%88%90%E5%90%8E%20nums%20%3D%20%7Bnums%7D%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=4&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
docs/chapter_sorting/heap_sort.md
-8
View File
@@ -612,14 +612,6 @@ comments: true
end
```
=== "Zig"
```zig title="heap_sort.zig"
[class]{}-[func]{siftDown}
[class]{}-[func]{heapSort}
```
??? pythontutor "可视化运行"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20sift_down%28nums%3A%20list%5Bint%5D,%20n%3A%20int,%20i%3A%20int%29%3A%0A%20%20%20%20%22%22%22%E5%A0%86%E7%9A%84%E9%95%BF%E5%BA%A6%E4%B8%BA%20n%20%EF%BC%8C%E4%BB%8E%E8%8A%82%E7%82%B9%20i%20%E5%BC%80%E5%A7%8B%EF%BC%8C%E4%BB%8E%E9%A1%B6%E8%87%B3%E5%BA%95%E5%A0%86%E5%8C%96%22%22%22%0A%20%20%20%20while%20True%3A%0A%20%20%20%20%20%20%20%20%23%20%E5%88%A4%E6%96%AD%E8%8A%82%E7%82%B9%20i,%20l,%20r%20%E4%B8%AD%E5%80%BC%E6%9C%80%E5%A4%A7%E7%9A%84%E8%8A%82%E7%82%B9%EF%BC%8C%E8%AE%B0%E4%B8%BA%20ma%0A%20%20%20%20%20%20%20%20l%20%3D%202%20*%20i%20%2B%201%0A%20%20%20%20%20%20%20%20r%20%3D%202%20*%20i%20%2B%202%0A%20%20%20%20%20%20%20%20ma%20%3D%20i%0A%20%20%20%20%20%20%20%20if%20l%20%3C%20n%20and%20nums%5Bl%5D%20%3E%20nums%5Bma%5D%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20ma%20%3D%20l%0A%20%20%20%20%20%20%20%20if%20r%20%3C%20n%20and%20nums%5Br%5D%20%3E%20nums%5Bma%5D%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20ma%20%3D%20r%0A%20%20%20%20%20%20%20%20%23%20%E8%8B%A5%E8%8A%82%E7%82%B9%20i%20%E6%9C%80%E5%A4%A7%E6%88%96%E7%B4%A2%E5%BC%95%20l,%20r%20%E8%B6%8A%E7%95%8C%EF%BC%8C%E5%88%99%E6%97%A0%E9%A1%BB%E7%BB%A7%E7%BB%AD%E5%A0%86%E5%8C%96%EF%BC%8C%E8%B7%B3%E5%87%BA%0A%20%20%20%20%20%20%20%20if%20ma%20%3D%3D%20i%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20break%0A%20%20%20%20%20%20%20%20%23%20%E4%BA%A4%E6%8D%A2%E4%B8%A4%E8%8A%82%E7%82%B9%0A%20%20%20%20%20%20%20%20nums%5Bi%5D,%20nums%5Bma%5D%20%3D%20nums%5Bma%5D,%20nums%5Bi%5D%0A%20%20%20%20%20%20%20%20%23%20%E5%BE%AA%E7%8E%AF%E5%90%91%E4%B8%8B%E5%A0%86%E5%8C%96%0A%20%20%20%20%20%20%20%20i%20%3D%20ma%0A%0Adef%20heap_sort%28nums%3A%20list%5Bint%5D%29%3A%0A%20%20%20%20%22%22%22%E5%A0%86%E6%8E%92%E5%BA%8F%22%22%22%0A%20%20%20%20%23%20%E5%BB%BA%E5%A0%86%E6%93%8D%E4%BD%9C%EF%BC%9A%E5%A0%86%E5%8C%96%E9%99%A4%E5%8F%B6%E8%8A%82%E7%82%B9%E4%BB%A5%E5%A4%96%E7%9A%84%E5%85%B6%E4%BB%96%E6%89%80%E6%9C%89%E8%8A%82%E7%82%B9%0A%20%20%20%20for%20i%20in%20range%28len%28nums%29%20//%202%20-%201,%20-1,%20-1%29%3A%0A%20%20%20%20%20%20%20%20sift_down%28nums,%20len%28nums%29,%20i%29%0A%20%20%20%20%23%20%E4%BB%8E%E5%A0%86%E4%B8%AD%E6%8F%90%E5%8F%96%E6%9C%80%E5%A4%A7%E5%85%83%E7%B4%A0%EF%BC%8C%E5%BE%AA%E7%8E%AF%20n-1%20%E8%BD%AE%0A%20%20%20%20for%20i%20in%20range%28len%28nums%29%20-%201,%200,%20-1%29%3A%0A%20%20%20%20%20%20%20%20%23%20%E4%BA%A4%E6%8D%A2%E6%A0%B9%E8%8A%82%E7%82%B9%E4%B8%8E%E6%9C%80%E5%8F%B3%E5%8F%B6%E8%8A%82%E7%82%B9%EF%BC%88%E4%BA%A4%E6%8D%A2%E9%A6%96%E5%85%83%E7%B4%A0%E4%B8%8E%E5%B0%BE%E5%85%83%E7%B4%A0%EF%BC%89%0A%20%20%20%20%20%20%20%20nums%5B0%5D,%20nums%5Bi%5D%20%3D%20nums%5Bi%5D,%20nums%5B0%5D%0A%20%20%20%20%20%20%20%20%23%20%E4%BB%A5%E6%A0%B9%E8%8A%82%E7%82%B9%E4%B8%BA%E8%B5%B7%E7%82%B9%EF%BC%8C%E4%BB%8E%E9%A1%B6%E8%87%B3%E5%BA%95%E8%BF%9B%E8%A1%8C%E5%A0%86%E5%8C%96%0A%20%20%20%20%20%20%20%20sift_down%28nums,%20i,%200%29%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20nums%20%3D%20%5B4,%201,%203,%201,%205,%202%5D%0A%20%20%20%20heap_sort%28nums%29%0A%20%20%20%20print%28%22%E5%A0%86%E6%8E%92%E5%BA%8F%E5%AE%8C%E6%88%90%E5%90%8E%20nums%20%3D%22,%20nums%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=5&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
docs/chapter_sorting/insertion_sort.md
-19
View File
@@ -270,25 +270,6 @@ comments: true
end
```
=== "Zig"
```zig title="insertion_sort.zig"
// 插入排序
fn insertionSort(nums: []i32) void {
// 外循环：已排序区间为 [0, i-1]
var i: usize = 1;
while (i < nums.len) : (i += 1) {
var base = nums[i];
var j: usize = i;
// 内循环：将 base 插入到已排序区间 [0, i-1] 中的正确位置
while (j >= 1 and nums[j - 1] > base) : (j -= 1) {
nums[j] = nums[j - 1]; // 将 nums[j] 向右移动一位
}
nums[j] = base; // 将 base 赋值到正确位置
}
}
```
??? pythontutor "可视化运行"
<div style="height: 513px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20insertion_sort%28nums%3A%20list%5Bint%5D%29%3A%0A%20%20%20%20%22%22%22%E6%8F%92%E5%85%A5%E6%8E%92%E5%BA%8F%22%22%22%0A%20%20%20%20%23%20%E5%A4%96%E5%BE%AA%E7%8E%AF%EF%BC%9A%E5%B7%B2%E6%8E%92%E5%BA%8F%E5%8C%BA%E9%97%B4%E4%B8%BA%20%5B0,%20i-1%5D%0A%20%20%20%20for%20i%20in%20range%281,%20len%28nums%29%29%3A%0A%20%20%20%20%20%20%20%20base%20%3D%20nums%5Bi%5D%0A%20%20%20%20%20%20%20%20j%20%3D%20i%20-%201%0A%20%20%20%20%20%20%20%20%23%20%E5%86%85%E5%BE%AA%E7%8E%AF%EF%BC%9A%E5%B0%86%20base%20%E6%8F%92%E5%85%A5%E5%88%B0%E5%B7%B2%E6%8E%92%E5%BA%8F%E5%8C%BA%E9%97%B4%20%5B0,%20i-1%5D%20%E4%B8%AD%E7%9A%84%E6%AD%A3%E7%A1%AE%E4%BD%8D%E7%BD%AE%0A%20%20%20%20%20%20%20%20while%20j%20%3E%3D%200%20and%20nums%5Bj%5D%20%3E%20base%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20nums%5Bj%20%2B%201%5D%20%3D%20nums%5Bj%5D%20%20%23%20%E5%B0%86%20nums%5Bj%5D%20%E5%90%91%E5%8F%B3%E7%A7%BB%E5%8A%A8%E4%B8%80%E4%BD%8D%0A%20%20%20%20%20%20%20%20%20%20%20%20j%20-%3D%201%0A%20%20%20%20%20%20%20%20nums%5Bj%20%2B%201%5D%20%3D%20base%20%20%23%20%E5%B0%86%20base%20%E8%B5%8B%E5%80%BC%E5%88%B0%E6%AD%A3%E7%A1%AE%E4%BD%8D%E7%BD%AE%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20nums%20%3D%20%5B4,%201,%203,%201,%205,%202%5D%0A%20%20%20%20insertion_sort%28nums%29%0A%20%20%20%20print%28%22%E6%8F%92%E5%85%A5%E6%8E%92%E5%BA%8F%E5%AE%8C%E6%88%90%E5%90%8E%20nums%20%3D%22,%20nums%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=4&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
docs/chapter_sorting/merge_sort.md
-54
View File
@@ -679,60 +679,6 @@ comments: true
end
```
=== "Zig"
```zig title="merge_sort.zig"
// 合并左子数组和右子数组
// 左子数组区间 [left, mid]
// 右子数组区间 [mid + 1, right]
fn merge(nums: []i32, left: usize, mid: usize, right: usize) !void {
// 初始化辅助数组
var mem_arena = std.heap.ArenaAllocator.init(std.heap.page_allocator);
defer mem_arena.deinit();
const mem_allocator = mem_arena.allocator();
var tmp = try mem_allocator.alloc(i32, right + 1 - left);
std.mem.copy(i32, tmp, nums[left..right+1]);
// 左子数组的起始索引和结束索引
var leftStart = left - left;
var leftEnd = mid - left;
// 右子数组的起始索引和结束索引
var rightStart = mid + 1 - left;
var rightEnd = right - left;
// i, j 分别指向左子数组、右子数组的首元素
var i = leftStart;
var j = rightStart;
// 通过覆盖原数组 nums 来合并左子数组和右子数组
var k = left;
while (k <= right) : (k += 1) {
// 若“左子数组已全部合并完”，则选取右子数组元素，并且 j++
if (i > leftEnd) {
nums[k] = tmp[j];
j += 1;
// 否则，若“右子数组已全部合并完”或“左子数组元素 <= 右子数组元素”，则选取左子数组元素，并且 i++
} else if (j > rightEnd or tmp[i] <= tmp[j]) {
nums[k] = tmp[i];
i += 1;
// 否则，若“左右子数组都未全部合并完”且“左子数组元素 > 右子数组元素”，则选取右子数组元素，并且 j++
} else {
nums[k] = tmp[j];
j += 1;
}
}
}
// 归并排序
fn mergeSort(nums: []i32, left: usize, right: usize) !void {
// 终止条件
if (left >= right) return; // 当子数组长度为 1 时终止递归
// 划分阶段
var mid = left + (right - left) / 2; // 计算中点
try mergeSort(nums, left, mid); // 递归左子数组
try mergeSort(nums, mid + 1, right); // 递归右子数组
// 合并阶段
try merge(nums, left, mid, right);
}
```
??? pythontutor "可视化运行"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20merge%28nums%3A%20list%5Bint%5D,%20left%3A%20int,%20mid%3A%20int,%20right%3A%20int%29%3A%0A%20%20%20%20%22%22%22%E5%90%88%E5%B9%B6%E5%B7%A6%E5%AD%90%E6%95%B0%E7%BB%84%E5%92%8C%E5%8F%B3%E5%AD%90%E6%95%B0%E7%BB%84%22%22%22%0A%20%20%20%20%23%20%E5%B7%A6%E5%AD%90%E6%95%B0%E7%BB%84%E5%8C%BA%E9%97%B4%E4%B8%BA%20%5Bleft,%20mid%5D,%20%E5%8F%B3%E5%AD%90%E6%95%B0%E7%BB%84%E5%8C%BA%E9%97%B4%E4%B8%BA%20%5Bmid%2B1,%20right%5D%0A%20%20%20%20%23%20%E5%88%9B%E5%BB%BA%E4%B8%80%E4%B8%AA%E4%B8%B4%E6%97%B6%E6%95%B0%E7%BB%84%20tmp%20%EF%BC%8C%E7%94%A8%E4%BA%8E%E5%AD%98%E6%94%BE%E5%90%88%E5%B9%B6%E5%90%8E%E7%9A%84%E7%BB%93%E6%9E%9C%0A%20%20%20%20tmp%20%3D%20%5B0%5D%20*%20%28right%20-%20left%20%2B%201%29%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%E5%B7%A6%E5%AD%90%E6%95%B0%E7%BB%84%E5%92%8C%E5%8F%B3%E5%AD%90%E6%95%B0%E7%BB%84%E7%9A%84%E8%B5%B7%E5%A7%8B%E7%B4%A2%E5%BC%95%0A%20%20%20%20i,%20j,%20k%20%3D%20left,%20mid%20%2B%201,%200%0A%20%20%20%20%23%20%E5%BD%93%E5%B7%A6%E5%8F%B3%E5%AD%90%E6%95%B0%E7%BB%84%E9%83%BD%E8%BF%98%E6%9C%89%E5%85%83%E7%B4%A0%E6%97%B6%EF%BC%8C%E8%BF%9B%E8%A1%8C%E6%AF%94%E8%BE%83%E5%B9%B6%E5%B0%86%E8%BE%83%E5%B0%8F%E7%9A%84%E5%85%83%E7%B4%A0%E5%A4%8D%E5%88%B6%E5%88%B0%E4%B8%B4%E6%97%B6%E6%95%B0%E7%BB%84%E4%B8%AD%0A%20%20%20%20while%20i%20%3C%3D%20mid%20and%20j%20%3C%3D%20right%3A%0A%20%20%20%20%20%20%20%20if%20nums%5Bi%5D%20%3C%3D%20nums%5Bj%5D%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20tmp%5Bk%5D%20%3D%20nums%5Bi%5D%0A%20%20%20%20%20%20%20%20%20%20%20%20i%20%2B%3D%201%0A%20%20%20%20%20%20%20%20else%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20tmp%5Bk%5D%20%3D%20nums%5Bj%5D%0A%20%20%20%20%20%20%20%20%20%20%20%20j%20%2B%3D%201%0A%20%20%20%20%20%20%20%20k%20%2B%3D%201%0A%20%20%20%20%23%20%E5%B0%86%E5%B7%A6%E5%AD%90%E6%95%B0%E7%BB%84%E5%92%8C%E5%8F%B3%E5%AD%90%E6%95%B0%E7%BB%84%E7%9A%84%E5%89%A9%E4%BD%99%E5%85%83%E7%B4%A0%E5%A4%8D%E5%88%B6%E5%88%B0%E4%B8%B4%E6%97%B6%E6%95%B0%E7%BB%84%E4%B8%AD%0A%20%20%20%20while%20i%20%3C%3D%20mid%3A%0A%20%20%20%20%20%20%20%20tmp%5Bk%5D%20%3D%20nums%5Bi%5D%0A%20%20%20%20%20%20%20%20i%20%2B%3D%201%0A%20%20%20%20%20%20%20%20k%20%2B%3D%201%0A%20%20%20%20while%20j%20%3C%3D%20right%3A%0A%20%20%20%20%20%20%20%20tmp%5Bk%5D%20%3D%20nums%5Bj%5D%0A%20%20%20%20%20%20%20%20j%20%2B%3D%201%0A%20%20%20%20%20%20%20%20k%20%2B%3D%201%0A%20%20%20%20%23%20%E5%B0%86%E4%B8%B4%E6%97%B6%E6%95%B0%E7%BB%84%20tmp%20%E4%B8%AD%E7%9A%84%E5%85%83%E7%B4%A0%E5%A4%8D%E5%88%B6%E5%9B%9E%E5%8E%9F%E6%95%B0%E7%BB%84%20nums%20%E7%9A%84%E5%AF%B9%E5%BA%94%E5%8C%BA%E9%97%B4%0A%20%20%20%20for%20k%20in%20range%280,%20len%28tmp%29%29%3A%0A%20%20%20%20%20%20%20%20nums%5Bleft%20%2B%20k%5D%20%3D%20tmp%5Bk%5D%0A%0A%0Adef%20merge_sort%28nums%3A%20list%5Bint%5D,%20left%3A%20int,%20right%3A%20int%29%3A%0A%20%20%20%20%22%22%22%E5%BD%92%E5%B9%B6%E6%8E%92%E5%BA%8F%22%22%22%0A%20%20%20%20%23%20%E7%BB%88%E6%AD%A2%E6%9D%A1%E4%BB%B6%0A%20%20%20%20if%20left%20%3E%3D%20right%3A%0A%20%20%20%20%20%20%20%20return%20%20%23%20%E5%BD%93%E5%AD%90%E6%95%B0%E7%BB%84%E9%95%BF%E5%BA%A6%E4%B8%BA%201%20%E6%97%B6%E7%BB%88%E6%AD%A2%E9%80%92%E5%BD%92%0A%20%20%20%20%23%20%E5%88%92%E5%88%86%E9%98%B6%E6%AE%B5%0A%20%20%20%20mid%20%3D%20%28left%20%2B%20right%29%20//%202%20%20%23%20%E8%AE%A1%E7%AE%97%E4%B8%AD%E7%82%B9%0A%20%20%20%20merge_sort%28nums,%20left,%20mid%29%20%20%23%20%E9%80%92%E5%BD%92%E5%B7%A6%E5%AD%90%E6%95%B0%E7%BB%84%0A%20%20%20%20merge_sort%28nums,%20mid%20%2B%201,%20right%29%20%20%23%20%E9%80%92%E5%BD%92%E5%8F%B3%E5%AD%90%E6%95%B0%E7%BB%84%0A%20%20%20%20%23%20%E5%90%88%E5%B9%B6%E9%98%B6%E6%AE%B5%0A%20%20%20%20merge%28nums,%20left,%20mid,%20right%29%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20nums%20%3D%20%5B7,%203,%202,%206,%200,%201,%205,%204%5D%0A%20%20%20%20merge_sort%28nums,%200,%20len%28nums%29%20-%201%29%0A%20%20%20%20print%28%22%E5%BD%92%E5%B9%B6%E6%8E%92%E5%BA%8F%E5%AE%8C%E6%88%90%E5%90%8E%20nums%20%3D%22,%20nums%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=5&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
docs/chapter_sorting/quick_sort.md
-97
View File
@@ -366,31 +366,6 @@ comments: true
end
```
=== "Zig"
```zig title="quick_sort.zig"
// 元素交换
fn swap(nums: []i32, i: usize, j: usize) void {
var tmp = nums[i];
nums[i] = nums[j];
nums[j] = tmp;
}
// 哨兵划分
fn partition(nums: []i32, left: usize, right: usize) usize {
// 以 nums[left] 为基准数
var i = left;
var j = right;
while (i < j) {
while (i < j and nums[j] >= nums[left]) j -= 1; // 从右向左找首个小于基准数的元素
while (i < j and nums[i] <= nums[left]) i += 1; // 从左向右找首个大于基准数的元素
swap(nums, i, j); // 交换这两个元素
}
swap(nums, i, left); // 将基准数交换至两子数组的分界线
return i; // 返回基准数的索引
}
```
??? pythontutor "可视化运行"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20partition%28nums%3A%20list%5Bint%5D,%20left%3A%20int,%20right%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E5%93%A8%E5%85%B5%E5%88%92%E5%88%86%22%22%22%0A%20%20%20%20%23%20%E4%BB%A5%20nums%5Bleft%5D%20%E4%B8%BA%E5%9F%BA%E5%87%86%E6%95%B0%0A%20%20%20%20i,%20j%20%3D%20left,%20right%0A%20%20%20%20while%20i%20%3C%20j%3A%0A%20%20%20%20%20%20%20%20while%20i%20%3C%20j%20and%20nums%5Bj%5D%20%3E%3D%20nums%5Bleft%5D%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20j%20-%3D%201%20%20%23%20%E4%BB%8E%E5%8F%B3%E5%90%91%E5%B7%A6%E6%89%BE%E9%A6%96%E4%B8%AA%E5%B0%8F%E4%BA%8E%E5%9F%BA%E5%87%86%E6%95%B0%E7%9A%84%E5%85%83%E7%B4%A0%0A%20%20%20%20%20%20%20%20while%20i%20%3C%20j%20and%20nums%5Bi%5D%20%3C%3D%20nums%5Bleft%5D%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20i%20%2B%3D%201%20%20%23%20%E4%BB%8E%E5%B7%A6%E5%90%91%E5%8F%B3%E6%89%BE%E9%A6%96%E4%B8%AA%E5%A4%A7%E4%BA%8E%E5%9F%BA%E5%87%86%E6%95%B0%E7%9A%84%E5%85%83%E7%B4%A0%0A%20%20%20%20%20%20%20%20%23%20%E5%85%83%E7%B4%A0%E4%BA%A4%E6%8D%A2%0A%20%20%20%20%20%20%20%20nums%5Bi%5D,%20nums%5Bj%5D%20%3D%20nums%5Bj%5D,%20nums%5Bi%5D%0A%20%20%20%20%23%20%E5%B0%86%E5%9F%BA%E5%87%86%E6%95%B0%E4%BA%A4%E6%8D%A2%E8%87%B3%E4%B8%A4%E5%AD%90%E6%95%B0%E7%BB%84%E7%9A%84%E5%88%86%E7%95%8C%E7%BA%BF%0A%20%20%20%20nums%5Bi%5D,%20nums%5Bleft%5D%20%3D%20nums%5Bleft%5D,%20nums%5Bi%5D%0A%20%20%20%20return%20i%20%20%23%20%E8%BF%94%E5%9B%9E%E5%9F%BA%E5%87%86%E6%95%B0%E7%9A%84%E7%B4%A2%E5%BC%95%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20nums%20%3D%20%5B2,%204,%201,%200,%203,%205%5D%0A%20%20%20%20partition%28nums,%200,%20len%28nums%29%20-%201%29%0A%20%20%20%20print%28%22%E5%93%A8%E5%85%B5%E5%88%92%E5%88%86%E5%AE%8C%E6%88%90%E5%90%8E%20nums%20%3D%22,%20nums%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=4&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
@@ -618,21 +593,6 @@ comments: true
end
```
=== "Zig"
```zig title="quick_sort.zig"
// 快速排序
fn quickSort(nums: []i32, left: usize, right: usize) void {
// 子数组长度为 1 时终止递归
if (left >= right) return;
// 哨兵划分
var pivot = partition(nums, left, right);
// 递归左子数组、右子数组
quickSort(nums, left, pivot - 1);
quickSort(nums, pivot + 1, right);
}
```
??? pythontutor "可视化运行"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20partition%28nums%3A%20list%5Bint%5D,%20left%3A%20int,%20right%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E5%93%A8%E5%85%B5%E5%88%92%E5%88%86%22%22%22%0A%20%20%20%20%23%20%E4%BB%A5%20nums%5Bleft%5D%20%E4%B8%BA%E5%9F%BA%E5%87%86%E6%95%B0%0A%20%20%20%20i,%20j%20%3D%20left,%20right%0A%20%20%20%20while%20i%20%3C%20j%3A%0A%20%20%20%20%20%20%20%20while%20i%20%3C%20j%20and%20nums%5Bj%5D%20%3E%3D%20nums%5Bleft%5D%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20j%20-%3D%201%20%20%23%20%E4%BB%8E%E5%8F%B3%E5%90%91%E5%B7%A6%E6%89%BE%E9%A6%96%E4%B8%AA%E5%B0%8F%E4%BA%8E%E5%9F%BA%E5%87%86%E6%95%B0%E7%9A%84%E5%85%83%E7%B4%A0%0A%20%20%20%20%20%20%20%20while%20i%20%3C%20j%20and%20nums%5Bi%5D%20%3C%3D%20nums%5Bleft%5D%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20i%20%2B%3D%201%20%20%23%20%E4%BB%8E%E5%B7%A6%E5%90%91%E5%8F%B3%E6%89%BE%E9%A6%96%E4%B8%AA%E5%A4%A7%E4%BA%8E%E5%9F%BA%E5%87%86%E6%95%B0%E7%9A%84%E5%85%83%E7%B4%A0%0A%20%20%20%20%20%20%20%20%23%20%E5%85%83%E7%B4%A0%E4%BA%A4%E6%8D%A2%0A%20%20%20%20%20%20%20%20nums%5Bi%5D,%20nums%5Bj%5D%20%3D%20nums%5Bj%5D,%20nums%5Bi%5D%0A%20%20%20%20%23%20%E5%B0%86%E5%9F%BA%E5%87%86%E6%95%B0%E4%BA%A4%E6%8D%A2%E8%87%B3%E4%B8%A4%E5%AD%90%E6%95%B0%E7%BB%84%E7%9A%84%E5%88%86%E7%95%8C%E7%BA%BF%0A%20%20%20%20nums%5Bi%5D,%20nums%5Bleft%5D%20%3D%20nums%5Bleft%5D,%20nums%5Bi%5D%0A%20%20%20%20return%20i%20%20%23%20%E8%BF%94%E5%9B%9E%E5%9F%BA%E5%87%86%E6%95%B0%E7%9A%84%E7%B4%A2%E5%BC%95%0A%0Adef%20quick_sort%28nums%3A%20list%5Bint%5D,%20left%3A%20int,%20right%3A%20int%29%3A%0A%20%20%20%20%22%22%22%E5%BF%AB%E9%80%9F%E6%8E%92%E5%BA%8F%22%22%22%0A%20%20%20%20%23%20%E5%AD%90%E6%95%B0%E7%BB%84%E9%95%BF%E5%BA%A6%E4%B8%BA%201%20%E6%97%B6%E7%BB%88%E6%AD%A2%E9%80%92%E5%BD%92%0A%20%20%20%20if%20left%20%3E%3D%20right%3A%0A%20%20%20%20%20%20%20%20return%0A%20%20%20%20%23%20%E5%93%A8%E5%85%B5%E5%88%92%E5%88%86%0A%20%20%20%20pivot%20%3D%20partition%28nums,%20left,%20right%29%0A%20%20%20%20%23%20%E9%80%92%E5%BD%92%E5%B7%A6%E5%AD%90%E6%95%B0%E7%BB%84%E3%80%81%E5%8F%B3%E5%AD%90%E6%95%B0%E7%BB%84%0A%20%20%20%20quick_sort%28nums,%20left,%20pivot%20-%201%29%0A%20%20%20%20quick_sort%28nums,%20pivot%20%2B%201,%20right%29%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20%23%20%E5%BF%AB%E9%80%9F%E6%8E%92%E5%BA%8F%0A%20%20%20%20nums%20%3D%20%5B2,%204,%201,%200,%203,%205%5D%0A%20%20%20%20quick_sort%28nums,%200,%20len%28nums%29%20-%201%29%0A%20%20%20%20print%28%22%E5%BF%AB%E9%80%9F%E6%8E%92%E5%BA%8F%E5%AE%8C%E6%88%90%E5%90%8E%20nums%20%3D%22,%20nums%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=5&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
@@ -1122,40 +1082,6 @@ comments: true
end
```
=== "Zig"
```zig title="quick_sort.zig"
// 选取三个候选元素的中位数
fn medianThree(nums: []i32, left: usize, mid: usize, right: usize) usize {
var l = nums[left];
var m = nums[mid];
var r = nums[right];
if ((l <= m && m <= r) || (r <= m && m <= l))
return mid; // m 在 l 和 r 之间
if ((m <= l && l <= r) || (r <= l && l <= m))
return left; // l 在 m 和 r 之间
return right;
}
// 哨兵划分（三数取中值）
fn partition(nums: []i32, left: usize, right: usize) usize {
// 选取三个候选元素的中位数
var med = medianThree(nums, left, (left + right) / 2, right);
// 将中位数交换至数组最左端
swap(nums, left, med);
// 以 nums[left] 为基准数
var i = left;
var j = right;
while (i < j) {
while (i < j and nums[j] >= nums[left]) j -= 1; // 从右向左找首个小于基准数的元素
while (i < j and nums[i] <= nums[left]) i += 1; // 从左向右找首个大于基准数的元素
swap(nums, i, j); // 交换这两个元素
}
swap(nums, i, left); // 将基准数交换至两子数组的分界线
return i; // 返回基准数的索引
}
```
??? pythontutor "可视化运行"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20median_three%28nums%3A%20list%5Bint%5D,%20left%3A%20int,%20mid%3A%20int,%20right%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E9%80%89%E5%8F%96%E4%B8%89%E4%B8%AA%E5%80%99%E9%80%89%E5%85%83%E7%B4%A0%E7%9A%84%E4%B8%AD%E4%BD%8D%E6%95%B0%22%22%22%0A%20%20%20%20l,%20m,%20r%20%3D%20nums%5Bleft%5D,%20nums%5Bmid%5D,%20nums%5Bright%5D%0A%20%20%20%20if%20%28l%20%3C%3D%20m%20%3C%3D%20r%29%20or%20%28r%20%3C%3D%20m%20%3C%3D%20l%29%3A%0A%20%20%20%20%20%20%20%20return%20mid%20%20%23%20m%20%E5%9C%A8%20l%20%E5%92%8C%20r%20%E4%B9%8B%E9%97%B4%0A%20%20%20%20if%20%28m%20%3C%3D%20l%20%3C%3D%20r%29%20or%20%28r%20%3C%3D%20l%20%3C%3D%20m%29%3A%0A%20%20%20%20%20%20%20%20return%20left%20%20%23%20l%20%E5%9C%A8%20m%20%E5%92%8C%20r%20%E4%B9%8B%E9%97%B4%0A%20%20%20%20return%20right%0A%0Adef%20partition%28nums%3A%20list%5Bint%5D,%20left%3A%20int,%20right%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E5%93%A8%E5%85%B5%E5%88%92%E5%88%86%EF%BC%88%E4%B8%89%E6%95%B0%E5%8F%96%E4%B8%AD%E5%80%BC%EF%BC%89%22%22%22%0A%20%20%20%20%23%20%E4%BB%A5%20nums%5Bleft%5D%20%E4%B8%BA%E5%9F%BA%E5%87%86%E6%95%B0%0A%20%20%20%20med%20%3D%20median_three%28nums,%20left,%20%28left%20%2B%20right%29%20//%202,%20right%29%0A%20%20%20%20%23%20%E5%B0%86%E4%B8%AD%E4%BD%8D%E6%95%B0%E4%BA%A4%E6%8D%A2%E8%87%B3%E6%95%B0%E7%BB%84%E6%9C%80%E5%B7%A6%E7%AB%AF%0A%20%20%20%20nums%5Bleft%5D,%20nums%5Bmed%5D%20%3D%20nums%5Bmed%5D,%20nums%5Bleft%5D%0A%20%20%20%20%23%20%E4%BB%A5%20nums%5Bleft%5D%20%E4%B8%BA%E5%9F%BA%E5%87%86%E6%95%B0%0A%20%20%20%20i,%20j%20%3D%20left,%20right%0A%20%20%20%20while%20i%20%3C%20j%3A%0A%20%20%20%20%20%20%20%20while%20i%20%3C%20j%20and%20nums%5Bj%5D%20%3E%3D%20nums%5Bleft%5D%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20j%20-%3D%201%20%20%23%20%E4%BB%8E%E5%8F%B3%E5%90%91%E5%B7%A6%E6%89%BE%E9%A6%96%E4%B8%AA%E5%B0%8F%E4%BA%8E%E5%9F%BA%E5%87%86%E6%95%B0%E7%9A%84%E5%85%83%E7%B4%A0%0A%20%20%20%20%20%20%20%20while%20i%20%3C%20j%20and%20nums%5Bi%5D%20%3C%3D%20nums%5Bleft%5D%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20i%20%2B%3D%201%20%20%23%20%E4%BB%8E%E5%B7%A6%E5%90%91%E5%8F%B3%E6%89%BE%E9%A6%96%E4%B8%AA%E5%A4%A7%E4%BA%8E%E5%9F%BA%E5%87%86%E6%95%B0%E7%9A%84%E5%85%83%E7%B4%A0%0A%20%20%20%20%20%20%20%20%23%20%E5%85%83%E7%B4%A0%E4%BA%A4%E6%8D%A2%0A%20%20%20%20%20%20%20%20nums%5Bi%5D,%20nums%5Bj%5D%20%3D%20nums%5Bj%5D,%20nums%5Bi%5D%0A%20%20%20%20%23%20%E5%B0%86%E5%9F%BA%E5%87%86%E6%95%B0%E4%BA%A4%E6%8D%A2%E8%87%B3%E4%B8%A4%E5%AD%90%E6%95%B0%E7%BB%84%E7%9A%84%E5%88%86%E7%95%8C%E7%BA%BF%0A%20%20%20%20nums%5Bi%5D,%20nums%5Bleft%5D%20%3D%20nums%5Bleft%5D,%20nums%5Bi%5D%0A%20%20%20%20return%20i%20%20%23%20%E8%BF%94%E5%9B%9E%E5%9F%BA%E5%87%86%E6%95%B0%E7%9A%84%E7%B4%A2%E5%BC%95%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20%23%20%E4%B8%AD%E4%BD%8D%E5%9F%BA%E5%87%86%E6%95%B0%E4%BC%98%E5%8C%96%0A%20%20%20%20nums%20%3D%20%5B2,%204,%201,%200,%203,%205%5D%0A%20%20%20%20partition%28nums,%200,%20len%28nums%29%20-%201%29%0A%20%20%20%20print%28%22%E5%93%A8%E5%85%B5%E5%88%92%E5%88%86%EF%BC%88%E4%B8%AD%E4%BD%8D%E5%9F%BA%E5%87%86%E6%95%B0%E4%BC%98%E5%8C%96%EF%BC%89%E5%AE%8C%E6%88%90%E5%90%8E%20nums%20%3D%22,%20nums%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=5&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
@@ -1445,29 +1371,6 @@ comments: true
end
```
=== "Zig"
```zig title="quick_sort.zig"
// 快速排序（递归深度优化）
fn quickSort(nums: []i32, left_: usize, right_: usize) void {
var left = left_;
var right = right_;
// 子数组长度为 1 时终止递归
while (left < right) {
// 哨兵划分操作
var pivot = partition(nums, left, right);
// 对两个子数组中较短的那个执行快速排序
if (pivot - left < right - pivot) {
quickSort(nums, left, pivot - 1); // 递归排序左子数组
left = pivot + 1; // 剩余未排序区间为 [pivot + 1, right]
} else {
quickSort(nums, pivot + 1, right); // 递归排序右子数组
right = pivot - 1; // 剩余未排序区间为 [left, pivot - 1]
}
}
}
```
??? pythontutor "可视化运行"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20partition%28nums%3A%20list%5Bint%5D,%20left%3A%20int,%20right%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E5%93%A8%E5%85%B5%E5%88%92%E5%88%86%22%22%22%0A%20%20%20%20%23%20%E4%BB%A5%20nums%5Bleft%5D%20%E4%B8%BA%E5%9F%BA%E5%87%86%E6%95%B0%0A%20%20%20%20i,%20j%20%3D%20left,%20right%0A%20%20%20%20while%20i%20%3C%20j%3A%0A%20%20%20%20%20%20%20%20while%20i%20%3C%20j%20and%20nums%5Bj%5D%20%3E%3D%20nums%5Bleft%5D%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20j%20-%3D%201%20%20%23%20%E4%BB%8E%E5%8F%B3%E5%90%91%E5%B7%A6%E6%89%BE%E9%A6%96%E4%B8%AA%E5%B0%8F%E4%BA%8E%E5%9F%BA%E5%87%86%E6%95%B0%E7%9A%84%E5%85%83%E7%B4%A0%0A%20%20%20%20%20%20%20%20while%20i%20%3C%20j%20and%20nums%5Bi%5D%20%3C%3D%20nums%5Bleft%5D%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20i%20%2B%3D%201%20%20%23%20%E4%BB%8E%E5%B7%A6%E5%90%91%E5%8F%B3%E6%89%BE%E9%A6%96%E4%B8%AA%E5%A4%A7%E4%BA%8E%E5%9F%BA%E5%87%86%E6%95%B0%E7%9A%84%E5%85%83%E7%B4%A0%0A%20%20%20%20%20%20%20%20%23%20%E5%85%83%E7%B4%A0%E4%BA%A4%E6%8D%A2%0A%20%20%20%20%20%20%20%20nums%5Bi%5D,%20nums%5Bj%5D%20%3D%20nums%5Bj%5D,%20nums%5Bi%5D%0A%20%20%20%20%23%20%E5%B0%86%E5%9F%BA%E5%87%86%E6%95%B0%E4%BA%A4%E6%8D%A2%E8%87%B3%E4%B8%A4%E5%AD%90%E6%95%B0%E7%BB%84%E7%9A%84%E5%88%86%E7%95%8C%E7%BA%BF%0A%20%20%20%20nums%5Bi%5D,%20nums%5Bleft%5D%20%3D%20nums%5Bleft%5D,%20nums%5Bi%5D%0A%20%20%20%20return%20i%20%20%23%20%E8%BF%94%E5%9B%9E%E5%9F%BA%E5%87%86%E6%95%B0%E7%9A%84%E7%B4%A2%E5%BC%95%0A%0Adef%20quick_sort%28nums%3A%20list%5Bint%5D,%20left%3A%20int,%20right%3A%20int%29%3A%0A%20%20%20%20%22%22%22%E5%BF%AB%E9%80%9F%E6%8E%92%E5%BA%8F%EF%BC%88%E5%B0%BE%E9%80%92%E5%BD%92%E4%BC%98%E5%8C%96%EF%BC%89%22%22%22%0A%20%20%20%20%23%20%E5%AD%90%E6%95%B0%E7%BB%84%E9%95%BF%E5%BA%A6%E4%B8%BA%201%20%E6%97%B6%E7%BB%88%E6%AD%A2%0A%20%20%20%20while%20left%20%3C%20right%3A%0A%20%20%20%20%20%20%20%20%23%20%E5%93%A8%E5%85%B5%E5%88%92%E5%88%86%E6%93%8D%E4%BD%9C%0A%20%20%20%20%20%20%20%20pivot%20%3D%20partition%28nums,%20left,%20right%29%0A%20%20%20%20%20%20%20%20%23%20%E5%AF%B9%E4%B8%A4%E4%B8%AA%E5%AD%90%E6%95%B0%E7%BB%84%E4%B8%AD%E8%BE%83%E7%9F%AD%E7%9A%84%E9%82%A3%E4%B8%AA%E6%89%A7%E8%A1%8C%E5%BF%AB%E9%80%9F%E6%8E%92%E5%BA%8F%0A%20%20%20%20%20%20%20%20if%20pivot%20-%20left%20%3C%20right%20-%20pivot%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20quick_sort%28nums,%20left,%20pivot%20-%201%29%20%20%23%20%E9%80%92%E5%BD%92%E6%8E%92%E5%BA%8F%E5%B7%A6%E5%AD%90%E6%95%B0%E7%BB%84%0A%20%20%20%20%20%20%20%20%20%20%20%20left%20%3D%20pivot%20%2B%201%20%20%23%20%E5%89%A9%E4%BD%99%E6%9C%AA%E6%8E%92%E5%BA%8F%E5%8C%BA%E9%97%B4%E4%B8%BA%20%5Bpivot%20%2B%201,%20right%5D%0A%20%20%20%20%20%20%20%20else%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20quick_sort%28nums,%20pivot%20%2B%201,%20right%29%20%20%23%20%E9%80%92%E5%BD%92%E6%8E%92%E5%BA%8F%E5%8F%B3%E5%AD%90%E6%95%B0%E7%BB%84%0A%20%20%20%20%20%20%20%20%20%20%20%20right%20%3D%20pivot%20-%201%20%20%23%20%E5%89%A9%E4%BD%99%E6%9C%AA%E6%8E%92%E5%BA%8F%E5%8C%BA%E9%97%B4%E4%B8%BA%20%5Bleft,%20pivot%20-%201%5D%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20%23%20%E5%BF%AB%E9%80%9F%E6%8E%92%E5%BA%8F%EF%BC%88%E5%B0%BE%E9%80%92%E5%BD%92%E4%BC%98%E5%8C%96%EF%BC%89%0A%20%20%20%20nums%20%3D%20%5B2,%204,%201,%200,%203,%205%5D%0A%20%20%20%20quick_sort%28nums,%200,%20len%28nums%29%20-%201%29%0A%20%20%20%20print%28%22%E5%BF%AB%E9%80%9F%E6%8E%92%E5%BA%8F%EF%BC%88%E5%B0%BE%E9%80%92%E5%BD%92%E4%BC%98%E5%8C%96%EF%BC%89%E5%AE%8C%E6%88%90%E5%90%8E%20nums%20%3D%22,%20nums%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=5&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
docs/chapter_sorting/radix_sort.md
-64
View File
@@ -716,70 +716,6 @@ $$
end
```
=== "Zig"
```zig title="radix_sort.zig"
// 获取元素 num 的第 k 位，其中 exp = 10^(k-1)
fn digit(num: i32, exp: i32) i32 {
// 传入 exp 而非 k 可以避免在此重复执行昂贵的次方计算
return @mod(@divFloor(num, exp), 10);
}
// 计数排序（根据 nums 第 k 位排序）
fn countingSortDigit(nums: []i32, exp: i32) !void {
// 十进制的位范围为 0~9 ，因此需要长度为 10 的桶数组
var mem_arena = std.heap.ArenaAllocator.init(std.heap.page_allocator);
// defer mem_arena.deinit();
const mem_allocator = mem_arena.allocator();
var counter = try mem_allocator.alloc(usize, 10);
@memset(counter, 0);
var n = nums.len;
// 统计 0~9 各数字的出现次数
for (nums) |num| {
var d: u32 = @bitCast(digit(num, exp)); // 获取 nums[i] 第 k 位，记为 d
counter[d] += 1; // 统计数字 d 的出现次数
}
// 求前缀和，将“出现个数”转换为“数组索引”
var i: usize = 1;
while (i < 10) : (i += 1) {
counter[i] += counter[i - 1];
}
// 倒序遍历，根据桶内统计结果，将各元素填入 res
var res = try mem_allocator.alloc(i32, n);
i = n - 1;
while (i >= 0) : (i -= 1) {
var d: u32 = @bitCast(digit(nums[i], exp));
var j = counter[d] - 1; // 获取 d 在数组中的索引 j
res[j] = nums[i]; // 将当前元素填入索引 j
counter[d] -= 1; // 将 d 的数量减 1
if (i == 0) break;
}
// 使用结果覆盖原数组 nums
i = 0;
while (i < n) : (i += 1) {
nums[i] = res[i];
}
}
// 基数排序
fn radixSort(nums: []i32) !void {
// 获取数组的最大元素，用于判断最大位数
var m: i32 = std.math.minInt(i32);
for (nums) |num| {
if (num > m) m = num;
}
// 按照从低位到高位的顺序遍历
var exp: i32 = 1;
while (exp <= m) : (exp *= 10) {
// 对数组元素的第 k 位执行计数排序
// k = 1 -> exp = 1
// k = 2 -> exp = 10
// 即 exp = 10^(k-1)
try countingSortDigit(nums, exp);
}
}
```
??? pythontutor "可视化运行"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20digit%28num%3A%20int,%20exp%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E8%8E%B7%E5%8F%96%E5%85%83%E7%B4%A0%20num%20%E7%9A%84%E7%AC%AC%20k%20%E4%BD%8D%EF%BC%8C%E5%85%B6%E4%B8%AD%20exp%20%3D%2010%5E%28k-1%29%22%22%22%0A%20%20%20%20%23%20%E4%BC%A0%E5%85%A5%20exp%20%E8%80%8C%E9%9D%9E%20k%20%E5%8F%AF%E4%BB%A5%E9%81%BF%E5%85%8D%E5%9C%A8%E6%AD%A4%E9%87%8D%E5%A4%8D%E6%89%A7%E8%A1%8C%E6%98%82%E8%B4%B5%E7%9A%84%E6%AC%A1%E6%96%B9%E8%AE%A1%E7%AE%97%0A%20%20%20%20return%20%28num%20//%20exp%29%20%25%2010%0A%0Adef%20counting_sort_digit%28nums%3A%20list%5Bint%5D,%20exp%3A%20int%29%3A%0A%20%20%20%20%22%22%22%E8%AE%A1%E6%95%B0%E6%8E%92%E5%BA%8F%EF%BC%88%E6%A0%B9%E6%8D%AE%20nums%20%E7%AC%AC%20k%20%E4%BD%8D%E6%8E%92%E5%BA%8F%EF%BC%89%22%22%22%0A%20%20%20%20%23%20%E5%8D%81%E8%BF%9B%E5%88%B6%E7%9A%84%E4%BD%8D%E8%8C%83%E5%9B%B4%E4%B8%BA%200~9%20%EF%BC%8C%E5%9B%A0%E6%AD%A4%E9%9C%80%E8%A6%81%E9%95%BF%E5%BA%A6%E4%B8%BA%2010%20%E7%9A%84%E6%A1%B6%E6%95%B0%E7%BB%84%0A%20%20%20%20counter%20%3D%20%5B0%5D%20*%2010%0A%20%20%20%20n%20%3D%20len%28nums%29%0A%20%20%20%20%23%20%E7%BB%9F%E8%AE%A1%200~9%20%E5%90%84%E6%95%B0%E5%AD%97%E7%9A%84%E5%87%BA%E7%8E%B0%E6%AC%A1%E6%95%B0%0A%20%20%20%20for%20i%20in%20range%28n%29%3A%0A%20%20%20%20%20%20%20%20d%20%3D%20digit%28nums%5Bi%5D,%20exp%29%20%20%23%20%E8%8E%B7%E5%8F%96%20nums%5Bi%5D%20%E7%AC%AC%20k%20%E4%BD%8D%EF%BC%8C%E8%AE%B0%E4%B8%BA%20d%0A%20%20%20%20%20%20%20%20counter%5Bd%5D%20%2B%3D%201%20%20%23%20%E7%BB%9F%E8%AE%A1%E6%95%B0%E5%AD%97%20d%20%E7%9A%84%E5%87%BA%E7%8E%B0%E6%AC%A1%E6%95%B0%0A%20%20%20%20%23%20%E6%B1%82%E5%89%8D%E7%BC%80%E5%92%8C%EF%BC%8C%E5%B0%86%E2%80%9C%E5%87%BA%E7%8E%B0%E4%B8%AA%E6%95%B0%E2%80%9D%E8%BD%AC%E6%8D%A2%E4%B8%BA%E2%80%9C%E6%95%B0%E7%BB%84%E7%B4%A2%E5%BC%95%E2%80%9D%0A%20%20%20%20for%20i%20in%20range%281,%2010%29%3A%0A%20%20%20%20%20%20%20%20counter%5Bi%5D%20%2B%3D%20counter%5Bi%20-%201%5D%0A%20%20%20%20%23%20%E5%80%92%E5%BA%8F%E9%81%8D%E5%8E%86%EF%BC%8C%E6%A0%B9%E6%8D%AE%E6%A1%B6%E5%86%85%E7%BB%9F%E8%AE%A1%E7%BB%93%E6%9E%9C%EF%BC%8C%E5%B0%86%E5%90%84%E5%85%83%E7%B4%A0%E5%A1%AB%E5%85%A5%20res%0A%20%20%20%20res%20%3D%20%5B0%5D%20*%20n%0A%20%20%20%20for%20i%20in%20range%28n%20-%201,%20-1,%20-1%29%3A%0A%20%20%20%20%20%20%20%20d%20%3D%20digit%28nums%5Bi%5D,%20exp%29%0A%20%20%20%20%20%20%20%20j%20%3D%20counter%5Bd%5D%20-%201%20%20%23%20%E8%8E%B7%E5%8F%96%20d%20%E5%9C%A8%E6%95%B0%E7%BB%84%E4%B8%AD%E7%9A%84%E7%B4%A2%E5%BC%95%20j%0A%20%20%20%20%20%20%20%20res%5Bj%5D%20%3D%20nums%5Bi%5D%20%20%23%20%E5%B0%86%E5%BD%93%E5%89%8D%E5%85%83%E7%B4%A0%E5%A1%AB%E5%85%A5%E7%B4%A2%E5%BC%95%20j%0A%20%20%20%20%20%20%20%20counter%5Bd%5D%20-%3D%201%20%20%23%20%E5%B0%86%20d%20%E7%9A%84%E6%95%B0%E9%87%8F%E5%87%8F%201%0A%20%20%20%20%23%20%E4%BD%BF%E7%94%A8%E7%BB%93%E6%9E%9C%E8%A6%86%E7%9B%96%E5%8E%9F%E6%95%B0%E7%BB%84%20nums%0A%20%20%20%20for%20i%20in%20range%28n%29%3A%0A%20%20%20%20%20%20%20%20nums%5Bi%5D%20%3D%20res%5Bi%5D%0A%0Adef%20radix_sort%28nums%3A%20list%5Bint%5D%29%3A%0A%20%20%20%20%22%22%22%E5%9F%BA%E6%95%B0%E6%8E%92%E5%BA%8F%22%22%22%0A%20%20%20%20%23%20%E8%8E%B7%E5%8F%96%E6%95%B0%E7%BB%84%E7%9A%84%E6%9C%80%E5%A4%A7%E5%85%83%E7%B4%A0%EF%BC%8C%E7%94%A8%E4%BA%8E%E5%88%A4%E6%96%AD%E6%9C%80%E5%A4%A7%E4%BD%8D%E6%95%B0%0A%20%20%20%20m%20%3D%20max%28nums%29%0A%20%20%20%20%23%20%E6%8C%89%E7%85%A7%E4%BB%8E%E4%BD%8E%E4%BD%8D%E5%88%B0%E9%AB%98%E4%BD%8D%E7%9A%84%E9%A1%BA%E5%BA%8F%E9%81%8D%E5%8E%86%0A%20%20%20%20exp%20%3D%201%0A%20%20%20%20while%20exp%20%3C%3D%20m%3A%0A%20%20%20%20%20%20%20%20%23%20%E5%AF%B9%E6%95%B0%E7%BB%84%E5%85%83%E7%B4%A0%E7%9A%84%E7%AC%AC%20k%20%E4%BD%8D%E6%89%A7%E8%A1%8C%E8%AE%A1%E6%95%B0%E6%8E%92%E5%BA%8F%0A%20%20%20%20%20%20%20%20%23%20k%20%3D%201%20-%3E%20exp%20%3D%201%0A%20%20%20%20%20%20%20%20%23%20k%20%3D%202%20-%3E%20exp%20%3D%2010%0A%20%20%20%20%20%20%20%20%23%20%E5%8D%B3%20exp%20%3D%2010%5E%28k-1%29%0A%20%20%20%20%20%20%20%20counting_sort_digit%28nums,%20exp%29%0A%20%20%20%20%20%20%20%20exp%20*%3D%2010%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20%23%20%E5%9F%BA%E6%95%B0%E6%8E%92%E5%BA%8F%0A%20%20%20%20nums%20%3D%20%5B%0A%20%20%20%20%20%20%20%20105,%0A%20%20%20%20%20%20%20%20356,%0A%20%20%20%20%20%20%20%20428,%0A%20%20%20%20%20%20%20%20348,%0A%20%20%20%20%20%20%20%20818,%0A%20%20%20%20%5D%0A%20%20%20%20radix_sort%28nums%29%0A%20%20%20%20print%28%22%E5%9F%BA%E6%95%B0%E6%8E%92%E5%BA%8F%E5%AE%8C%E6%88%90%E5%90%8E%20nums%20%3D%22,%20nums%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=6&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
docs/chapter_sorting/selection_sort.md
-6
View File
@@ -324,12 +324,6 @@ comments: true
end
```
=== "Zig"
```zig title="selection_sort.zig"
[class]{}-[func]{selectionSort}
```
??? pythontutor "可视化运行"
<div style="height: 531px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20selection_sort%28nums%3A%20list%5Bint%5D%29%3A%0A%20%20%20%20%22%22%22%E9%80%89%E6%8B%A9%E6%8E%92%E5%BA%8F%22%22%22%0A%20%20%20%20n%20%3D%20len%28nums%29%0A%20%20%20%20%23%20%E5%A4%96%E5%BE%AA%E7%8E%AF%EF%BC%9A%E6%9C%AA%E6%8E%92%E5%BA%8F%E5%8C%BA%E9%97%B4%E4%B8%BA%20%5Bi,%20n-1%5D%0A%20%20%20%20for%20i%20in%20range%28n%20-%201%29%3A%0A%20%20%20%20%20%20%20%20%23%20%E5%86%85%E5%BE%AA%E7%8E%AF%EF%BC%9A%E6%89%BE%E5%88%B0%E6%9C%AA%E6%8E%92%E5%BA%8F%E5%8C%BA%E9%97%B4%E5%86%85%E7%9A%84%E6%9C%80%E5%B0%8F%E5%85%83%E7%B4%A0%0A%20%20%20%20%20%20%20%20k%20%3D%20i%0A%20%20%20%20%20%20%20%20for%20j%20in%20range%28i%20%2B%201,%20n%29%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20if%20nums%5Bj%5D%20%3C%20nums%5Bk%5D%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20k%20%3D%20j%20%20%23%20%E8%AE%B0%E5%BD%95%E6%9C%80%E5%B0%8F%E5%85%83%E7%B4%A0%E7%9A%84%E7%B4%A2%E5%BC%95%0A%20%20%20%20%20%20%20%20%23%20%E5%B0%86%E8%AF%A5%E6%9C%80%E5%B0%8F%E5%85%83%E7%B4%A0%E4%B8%8E%E6%9C%AA%E6%8E%92%E5%BA%8F%E5%8C%BA%E9%97%B4%E7%9A%84%E9%A6%96%E4%B8%AA%E5%85%83%E7%B4%A0%E4%BA%A4%E6%8D%A2%0A%20%20%20%20%20%20%20%20nums%5Bi%5D,%20nums%5Bk%5D%20%3D%20nums%5Bk%5D,%20nums%5Bi%5D%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20nums%20%3D%20%5B4,%201,%203,%201,%205,%202%5D%0A%20%20%20%20selection_sort%28nums%29%0A%20%20%20%20print%28%22%E9%80%89%E6%8B%A9%E6%8E%92%E5%BA%8F%E5%AE%8C%E6%88%90%E5%90%8E%20nums%20%3D%22,%20nums%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=4&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
docs/chapter_stack_and_queue/deque.md
-172
View File
@@ -393,12 +393,6 @@ comments: true
is_empty = size.zero?
```
=== "Zig"
```zig title="deque.zig"
```
??? pythontutor "可视化运行"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=from%20collections%20import%20deque%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%E5%8F%8C%E5%90%91%E9%98%9F%E5%88%97%0A%20%20%20%20deq%20%3D%20deque%28%29%0A%0A%20%20%20%20%23%20%E5%85%83%E7%B4%A0%E5%85%A5%E9%98%9F%0A%20%20%20%20deq.append%282%29%20%20%23%20%E6%B7%BB%E5%8A%A0%E8%87%B3%E9%98%9F%E5%B0%BE%0A%20%20%20%20deq.append%285%29%0A%20%20%20%20deq.append%284%29%0A%20%20%20%20deq.appendleft%283%29%20%20%23%20%E6%B7%BB%E5%8A%A0%E8%87%B3%E9%98%9F%E9%A6%96%0A%20%20%20%20deq.appendleft%281%29%0A%20%20%20%20print%28%22%E5%8F%8C%E5%90%91%E9%98%9F%E5%88%97%20deque%20%3D%22,%20deq%29%0A%0A%20%20%20%20%23%20%E8%AE%BF%E9%97%AE%E5%85%83%E7%B4%A0%0A%20%20%20%20front%20%3D%20deq%5B0%5D%20%20%23%20%E9%98%9F%E9%A6%96%E5%85%83%E7%B4%A0%0A%20%20%20%20print%28%22%E9%98%9F%E9%A6%96%E5%85%83%E7%B4%A0%20front%20%3D%22,%20front%29%0A%20%20%20%20rear%20%3D%20deq%5B-1%5D%20%20%23%20%E9%98%9F%E5%B0%BE%E5%85%83%E7%B4%A0%0A%20%20%20%20print%28%22%E9%98%9F%E5%B0%BE%E5%85%83%E7%B4%A0%20rear%20%3D%22,%20rear%29%0A%0A%20%20%20%20%23%20%E5%85%83%E7%B4%A0%E5%87%BA%E9%98%9F%0A%20%20%20%20pop_front%20%3D%20deq.popleft%28%29%20%20%23%20%E9%98%9F%E9%A6%96%E5%85%83%E7%B4%A0%E5%87%BA%E9%98%9F%0A%20%20%20%20print%28%22%E9%98%9F%E9%A6%96%E5%87%BA%E9%98%9F%E5%85%83%E7%B4%A0%20%20pop_front%20%3D%22,%20pop_front%29%0A%20%20%20%20print%28%22%E9%98%9F%E9%A6%96%E5%87%BA%E9%98%9F%E5%90%8E%20deque%20%3D%22,%20deq%29%0A%20%20%20%20pop_rear%20%3D%20deq.pop%28%29%20%20%23%20%E9%98%9F%E5%B0%BE%E5%85%83%E7%B4%A0%E5%87%BA%E9%98%9F%0A%20%20%20%20print%28%22%E9%98%9F%E5%B0%BE%E5%87%BA%E9%98%9F%E5%85%83%E7%B4%A0%20%20pop_rear%20%3D%22,%20pop_rear%29%0A%20%20%20%20print%28%22%E9%98%9F%E5%B0%BE%E5%87%BA%E9%98%9F%E5%90%8E%20deque%20%3D%22,%20deq%29%0A%0A%20%20%20%20%23%20%E8%8E%B7%E5%8F%96%E5%8F%8C%E5%90%91%E9%98%9F%E5%88%97%E7%9A%84%E9%95%BF%E5%BA%A6%0A%20%20%20%20size%20%3D%20len%28deq%29%0A%20%20%20%20print%28%22%E5%8F%8C%E5%90%91%E9%98%9F%E5%88%97%E9%95%BF%E5%BA%A6%20size%20%3D%22,%20size%29%0A%0A%20%20%20%20%23%20%E5%88%A4%E6%96%AD%E5%8F%8C%E5%90%91%E9%98%9F%E5%88%97%E6%98%AF%E5%90%A6%E4%B8%BA%E7%A9%BA%0A%20%20%20%20is_empty%20%3D%20len%28deq%29%20%3D%3D%200%0A%20%20%20%20print%28%22%E5%8F%8C%E5%90%91%E9%98%9F%E5%88%97%E6%98%AF%E5%90%A6%E4%B8%BA%E7%A9%BA%20%3D%22,%20is_empty%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
@@ -2160,166 +2154,6 @@ comments: true
end
```
=== "Zig"
```zig title="linkedlist_deque.zig"
// 双向链表节点
fn ListNode(comptime T: type) type {
return struct {
const Self = @This();
val: T = undefined, // 节点值
next: ?*Self = null, // 后继节点指针
prev: ?*Self = null, // 前驱节点指针
// Initialize a list node with specific value
pub fn init(self: *Self, x: i32) void {
self.val = x;
self.next = null;
self.prev = null;
}
};
}
// 基于双向链表实现的双向队列
fn LinkedListDeque(comptime T: type) type {
return struct {
const Self = @This();
front: ?*ListNode(T) = null, // 头节点 front
rear: ?*ListNode(T) = null, // 尾节点 rear
que_size: usize = 0, // 双向队列的长度
mem_arena: ?std.heap.ArenaAllocator = null,
mem_allocator: std.mem.Allocator = undefined, // 内存分配器
// 构造函数（分配内存+初始化队列）
pub fn init(self: *Self, allocator: std.mem.Allocator) !void {
if (self.mem_arena == null) {
self.mem_arena = std.heap.ArenaAllocator.init(allocator);
self.mem_allocator = self.mem_arena.?.allocator();
}
self.front = null;
self.rear = null;
self.que_size = 0;
}
// 析构函数（释放内存）
pub fn deinit(self: *Self) void {
if (self.mem_arena == null) return;
self.mem_arena.?.deinit();
}
// 获取双向队列的长度
pub fn size(self: *Self) usize {
return self.que_size;
}
// 判断双向队列是否为空
pub fn isEmpty(self: *Self) bool {
return self.size() == 0;
}
// 入队操作
pub fn push(self: *Self, num: T, is_front: bool) !void {
var node = try self.mem_allocator.create(ListNode(T));
node.init(num);
// 若链表为空，则令 front 和 rear 都指向 node
if (self.isEmpty()) {
self.front = node;
self.rear = node;
// 队首入队操作
} else if (is_front) {
// 将 node 添加至链表头部
self.front.?.prev = node;
node.next = self.front;
self.front = node; // 更新头节点
// 队尾入队操作
} else {
// 将 node 添加至链表尾部
self.rear.?.next = node;
node.prev = self.rear;
self.rear = node; // 更新尾节点
}
self.que_size += 1; // 更新队列长度
}
// 队首入队
pub fn pushFirst(self: *Self, num: T) !void {
try self.push(num, true);
}
// 队尾入队
pub fn pushLast(self: *Self, num: T) !void {
try self.push(num, false);
}
// 出队操作
pub fn pop(self: *Self, is_front: bool) T {
if (self.isEmpty()) @panic("双向队列为空");
var val: T = undefined;
// 队首出队操作
if (is_front) {
val = self.front.?.val; // 暂存头节点值
// 删除头节点
var fNext = self.front.?.next;
if (fNext != null) {
fNext.?.prev = null;
self.front.?.next = null;
}
self.front = fNext; // 更新头节点
// 队尾出队操作
} else {
val = self.rear.?.val; // 暂存尾节点值
// 删除尾节点
var rPrev = self.rear.?.prev;
if (rPrev != null) {
rPrev.?.next = null;
self.rear.?.prev = null;
}
self.rear = rPrev; // 更新尾节点
}
self.que_size -= 1; // 更新队列长度
return val;
}
// 队首出队
pub fn popFirst(self: *Self) T {
return self.pop(true);
}
// 队尾出队
pub fn popLast(self: *Self) T {
return self.pop(false);
}
// 访问队首元素
pub fn peekFirst(self: *Self) T {
if (self.isEmpty()) @panic("双向队列为空");
return self.front.?.val;
}
// 访问队尾元素
pub fn peekLast(self: *Self) T {
if (self.isEmpty()) @panic("双向队列为空");
return self.rear.?.val;
}
// 返回数组用于打印
pub fn toArray(self: *Self) ![]T {
var node = self.front;
var res = try self.mem_allocator.alloc(T, self.size());
@memset(res, @as(T, 0));
var i: usize = 0;
while (i < res.len) : (i += 1) {
res[i] = node.?.val;
node = node.?.next;
}
return res;
}
};
}
```
### 2. &nbsp; 基于数组的实现
如图 5-9 所示，与基于数组实现队列类似，我们也可以使用环形数组来实现双向队列。
@@ -3768,12 +3602,6 @@ comments: true
end
```
=== "Zig"
```zig title="array_deque.zig"
[class]{ArrayDeque}-[func]{}
```
## 5.3.3 &nbsp; 双向队列应用
双向队列兼具栈与队列的逻辑，**因此它可以实现这两者的所有应用场景，同时提供更高的自由度**。
docs/chapter_stack_and_queue/queue.md
-187
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docs/chapter_stack_and_queue/stack.md
-142
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@@ -359,12 +359,6 @@ comments: true
is_empty = stack.empty?
```
=== "Zig"
```zig title="stack.zig"
```
??? pythontutor "可视化运行"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%E6%A0%88%0A%20%20%20%20%23%20Python%20%E6%B2%A1%E6%9C%89%E5%86%85%E7%BD%AE%E7%9A%84%E6%A0%88%E7%B1%BB%EF%BC%8C%E5%8F%AF%E4%BB%A5%E6%8A%8A%20list%20%E5%BD%93%E4%BD%9C%E6%A0%88%E6%9D%A5%E4%BD%BF%E7%94%A8%0A%20%20%20%20stack%20%3D%20%5B%5D%0A%0A%20%20%20%20%23%20%E5%85%83%E7%B4%A0%E5%85%A5%E6%A0%88%0A%20%20%20%20stack.append%281%29%0A%20%20%20%20stack.append%283%29%0A%20%20%20%20stack.append%282%29%0A%20%20%20%20stack.append%285%29%0A%20%20%20%20stack.append%284%29%0A%20%20%20%20print%28%22%E6%A0%88%20stack%20%3D%22,%20stack%29%0A%0A%20%20%20%20%23%20%E8%AE%BF%E9%97%AE%E6%A0%88%E9%A1%B6%E5%85%83%E7%B4%A0%0A%20%20%20%20peek%20%3D%20stack%5B-1%5D%0A%20%20%20%20print%28%22%E6%A0%88%E9%A1%B6%E5%85%83%E7%B4%A0%20peek%20%3D%22,%20peek%29%0A%0A%20%20%20%20%23%20%E5%85%83%E7%B4%A0%E5%87%BA%E6%A0%88%0A%20%20%20%20pop%20%3D%20stack.pop%28%29%0A%20%20%20%20print%28%22%E5%87%BA%E6%A0%88%E5%85%83%E7%B4%A0%20pop%20%3D%22,%20pop%29%0A%20%20%20%20print%28%22%E5%87%BA%E6%A0%88%E5%90%8E%20stack%20%3D%22,%20stack%29%0A%0A%20%20%20%20%23%20%E8%8E%B7%E5%8F%96%E6%A0%88%E7%9A%84%E9%95%BF%E5%BA%A6%0A%20%20%20%20size%20%3D%20len%28stack%29%0A%20%20%20%20print%28%22%E6%A0%88%E7%9A%84%E9%95%BF%E5%BA%A6%20size%20%3D%22,%20size%29%0A%0A%20%20%20%20%23%20%E5%88%A4%E6%96%AD%E6%98%AF%E5%90%A6%E4%B8%BA%E7%A9%BA%0A%20%20%20%20is_empty%20%3D%20len%28stack%29%20%3D%3D%200%0A%20%20%20%20print%28%22%E6%A0%88%E6%98%AF%E5%90%A6%E4%B8%BA%E7%A9%BA%20%3D%22,%20is_empty%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=2&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
@@ -1166,84 +1160,6 @@ comments: true
end
```
=== "Zig"
```zig title="linkedlist_stack.zig"
// 基于链表实现的栈
fn LinkedListStack(comptime T: type) type {
return struct {
const Self = @This();
stack_top: ?*inc.ListNode(T) = null, // 将头节点作为栈顶
stk_size: usize = 0, // 栈的长度
mem_arena: ?std.heap.ArenaAllocator = null,
mem_allocator: std.mem.Allocator = undefined, // 内存分配器
// 构造函数（分配内存+初始化栈）
pub fn init(self: *Self, allocator: std.mem.Allocator) !void {
if (self.mem_arena == null) {
self.mem_arena = std.heap.ArenaAllocator.init(allocator);
self.mem_allocator = self.mem_arena.?.allocator();
}
self.stack_top = null;
self.stk_size = 0;
}
// 析构函数（释放内存）
pub fn deinit(self: *Self) void {
if (self.mem_arena == null) return;
self.mem_arena.?.deinit();
}
// 获取栈的长度
pub fn size(self: *Self) usize {
return self.stk_size;
}
// 判断栈是否为空
pub fn isEmpty(self: *Self) bool {
return self.size() == 0;
}
// 访问栈顶元素
pub fn peek(self: *Self) T {
if (self.size() == 0) @panic("栈为空");
return self.stack_top.?.val;
}
// 入栈
pub fn push(self: *Self, num: T) !void {
var node = try self.mem_allocator.create(inc.ListNode(T));
node.init(num);
node.next = self.stack_top;
self.stack_top = node;
self.stk_size += 1;
}
// 出栈
pub fn pop(self: *Self) T {
var num = self.peek();
self.stack_top = self.stack_top.?.next;
self.stk_size -= 1;
return num;
}
// 将栈转换为数组
pub fn toArray(self: *Self) ![]T {
var node = self.stack_top;
var res = try self.mem_allocator.alloc(T, self.size());
@memset(res, @as(T, 0));
var i: usize = 0;
while (i < res.len) : (i += 1) {
res[res.len - i - 1] = node.?.val;
node = node.?.next;
}
return res;
}
};
}
```
??? pythontutor "可视化运行"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=class%20ListNode%3A%0A%20%20%20%20%22%22%22%E9%93%BE%E8%A1%A8%E8%8A%82%E7%82%B9%E7%B1%BB%22%22%22%0A%20%20%20%20def%20__init__%28self,%20val%3A%20int%29%3A%0A%20%20%20%20%20%20%20%20self.val%3A%20int%20%3D%20val%20%20%23%20%E8%8A%82%E7%82%B9%E5%80%BC%0A%20%20%20%20%20%20%20%20self.next%3A%20ListNode%20%7C%20None%20%3D%20None%20%20%23%20%E5%90%8E%E7%BB%A7%E8%8A%82%E7%82%B9%E5%BC%95%E7%94%A8%0A%0A%0Aclass%20LinkedListStack%3A%0A%20%20%20%20%22%22%22%E5%9F%BA%E4%BA%8E%E9%93%BE%E8%A1%A8%E5%AE%9E%E7%8E%B0%E7%9A%84%E6%A0%88%22%22%22%0A%0A%20%20%20%20def%20__init__%28self%29%3A%0A%20%20%20%20%20%20%20%20%22%22%22%E6%9E%84%E9%80%A0%E6%96%B9%E6%B3%95%22%22%22%0A%20%20%20%20%20%20%20%20self._peek%3A%20ListNode%20%7C%20None%20%3D%20None%0A%20%20%20%20%20%20%20%20self._size%3A%20int%20%3D%200%0A%0A%20%20%20%20def%20size%28self%29%20-%3E%20int%3A%0A%20%20%20%20%20%20%20%20%22%22%22%E8%8E%B7%E5%8F%96%E6%A0%88%E7%9A%84%E9%95%BF%E5%BA%A6%22%22%22%0A%20%20%20%20%20%20%20%20return%20self._size%0A%0A%20%20%20%20def%20is_empty%28self%29%20-%3E%20bool%3A%0A%20%20%20%20%20%20%20%20%22%22%22%E5%88%A4%E6%96%AD%E6%A0%88%E6%98%AF%E5%90%A6%E4%B8%BA%E7%A9%BA%22%22%22%0A%20%20%20%20%20%20%20%20return%20not%20self._peek%0A%0A%20%20%20%20def%20push%28self,%20val%3A%20int%29%3A%0A%20%20%20%20%20%20%20%20%22%22%22%E5%85%A5%E6%A0%88%22%22%22%0A%20%20%20%20%20%20%20%20node%20%3D%20ListNode%28val%29%0A%20%20%20%20%20%20%20%20node.next%20%3D%20self._peek%0A%20%20%20%20%20%20%20%20self._peek%20%3D%20node%0A%20%20%20%20%20%20%20%20self._size%20%2B%3D%201%0A%0A%20%20%20%20def%20pop%28self%29%20-%3E%20int%3A%0A%20%20%20%20%20%20%20%20%22%22%22%E5%87%BA%E6%A0%88%22%22%22%0A%20%20%20%20%20%20%20%20num%20%3D%20self.peek%28%29%0A%20%20%20%20%20%20%20%20self._peek%20%3D%20self._peek.next%0A%20%20%20%20%20%20%20%20self._size%20-%3D%201%0A%20%20%20%20%20%20%20%20return%20num%0A%0A%20%20%20%20def%20peek%28self%29%20-%3E%20int%3A%0A%20%20%20%20%20%20%20%20%22%22%22%E8%AE%BF%E9%97%AE%E6%A0%88%E9%A1%B6%E5%85%83%E7%B4%A0%22%22%22%0A%20%20%20%20%20%20%20%20if%20self.is_empty%28%29%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20raise%20IndexError%28%22%E6%A0%88%E4%B8%BA%E7%A9%BA%22%29%0A%20%20%20%20%20%20%20%20return%20self._peek.val%0A%0A%20%20%20%20def%20to_list%28self%29%20-%3E%20list%5Bint%5D%3A%0A%20%20%20%20%20%20%20%20%22%22%22%E8%BD%AC%E5%8C%96%E4%B8%BA%E5%88%97%E8%A1%A8%E7%94%A8%E4%BA%8E%E6%89%93%E5%8D%B0%22%22%22%0A%20%20%20%20%20%20%20%20arr%20%3D%20%5B%5D%0A%20%20%20%20%20%20%20%20node%20%3D%20self._peek%0A%20%20%20%20%20%20%20%20while%20node%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20arr.append%28node.val%29%0A%20%20%20%20%20%20%20%20%20%20%20%20node%20%3D%20node.next%0A%20%20%20%20%20%20%20%20arr.reverse%28%29%0A%20%20%20%20%20%20%20%20return%20arr%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%E6%A0%88%0A%20%20%20%20stack%20%3D%20LinkedListStack%28%29%0A%0A%20%20%20%20%23%20%E5%85%83%E7%B4%A0%E5%85%A5%E6%A0%88%0A%20%20%20%20stack.push%281%29%0A%20%20%20%20stack.push%283%29%0A%20%20%20%20stack.push%282%29%0A%20%20%20%20stack.push%285%29%0A%20%20%20%20stack.push%284%29%0A%20%20%20%20print%28%22%E6%A0%88%20stack%20%3D%22,%20stack.to_list%28%29%29%0A%0A%20%20%20%20%23%20%E8%AE%BF%E9%97%AE%E6%A0%88%E9%A1%B6%E5%85%83%E7%B4%A0%0A%20%20%20%20peek%20%3D%20stack.peek%28%29%0A%20%20%20%20print%28%22%E6%A0%88%E9%A1%B6%E5%85%83%E7%B4%A0%20peek%20%3D%22,%20peek%29%0A%0A%20%20%20%20%23%20%E5%85%83%E7%B4%A0%E5%87%BA%E6%A0%88%0A%20%20%20%20pop%20%3D%20stack.pop%28%29%0A%20%20%20%20print%28%22%E5%87%BA%E6%A0%88%E5%85%83%E7%B4%A0%20pop%20%3D%22,%20pop%29%0A%20%20%20%20print%28%22%E5%87%BA%E6%A0%88%E5%90%8E%20stack%20%3D%22,%20stack.to_list%28%29%29%0A%0A%20%20%20%20%23%20%E8%8E%B7%E5%8F%96%E6%A0%88%E7%9A%84%E9%95%BF%E5%BA%A6%0A%20%20%20%20size%20%3D%20stack.size%28%29%0A%20%20%20%20print%28%22%E6%A0%88%E7%9A%84%E9%95%BF%E5%BA%A6%20size%20%3D%22,%20size%29%0A%0A%20%20%20%20%23%20%E5%88%A4%E6%96%AD%E6%98%AF%E5%90%A6%E4%B8%BA%E7%A9%BA%0A%20%20%20%20is_empty%20%3D%20stack.is_empty%28%29%0A%20%20%20%20print%28%22%E6%A0%88%E6%98%AF%E5%90%A6%E4%B8%BA%E7%A9%BA%20%3D%22,%20is_empty%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=4&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
@@ -1886,64 +1802,6 @@ comments: true
end
```
=== "Zig"
```zig title="array_stack.zig"
// 基于数组实现的栈
fn ArrayStack(comptime T: type) type {
return struct {
const Self = @This();
stack: ?std.ArrayList(T) = null,
// 构造方法（分配内存+初始化栈）
pub fn init(self: *Self, allocator: std.mem.Allocator) void {
if (self.stack == null) {
self.stack = std.ArrayList(T).init(allocator);
}
}
// 析构方法（释放内存）
pub fn deinit(self: *Self) void {
if (self.stack == null) return;
self.stack.?.deinit();
}
// 获取栈的长度
pub fn size(self: *Self) usize {
return self.stack.?.items.len;
}
// 判断栈是否为空
pub fn isEmpty(self: *Self) bool {
return self.size() == 0;
}
// 访问栈顶元素
pub fn peek(self: *Self) T {
if (self.isEmpty()) @panic("栈为空");
return self.stack.?.items[self.size() - 1];
}
// 入栈
pub fn push(self: *Self, num: T) !void {
try self.stack.?.append(num);
}
// 出栈
pub fn pop(self: *Self) T {
var num = self.stack.?.pop();
return num;
}
// 返回 ArrayList
pub fn toList(self: *Self) std.ArrayList(T) {
return self.stack.?;
}
};
}
```
??? pythontutor "可视化运行"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=class%20ArrayStack%3A%0A%20%20%20%20%22%22%22%E5%9F%BA%E4%BA%8E%E6%95%B0%E7%BB%84%E5%AE%9E%E7%8E%B0%E7%9A%84%E6%A0%88%22%22%22%0A%0A%20%20%20%20def%20__init__%28self%29%3A%0A%20%20%20%20%20%20%20%20%22%22%22%E6%9E%84%E9%80%A0%E6%96%B9%E6%B3%95%22%22%22%0A%20%20%20%20%20%20%20%20self._stack%3A%20list%5Bint%5D%20%3D%20%5B%5D%0A%0A%20%20%20%20def%20size%28self%29%20-%3E%20int%3A%0A%20%20%20%20%20%20%20%20%22%22%22%E8%8E%B7%E5%8F%96%E6%A0%88%E7%9A%84%E9%95%BF%E5%BA%A6%22%22%22%0A%20%20%20%20%20%20%20%20return%20len%28self._stack%29%0A%0A%20%20%20%20def%20is_empty%28self%29%20-%3E%20bool%3A%0A%20%20%20%20%20%20%20%20%22%22%22%E5%88%A4%E6%96%AD%E6%A0%88%E6%98%AF%E5%90%A6%E4%B8%BA%E7%A9%BA%22%22%22%0A%20%20%20%20%20%20%20%20return%20self._stack%20%3D%3D%20%5B%5D%0A%0A%20%20%20%20def%20push%28self,%20item%3A%20int%29%3A%0A%20%20%20%20%20%20%20%20%22%22%22%E5%85%A5%E6%A0%88%22%22%22%0A%20%20%20%20%20%20%20%20self._stack.append%28item%29%0A%0A%20%20%20%20def%20pop%28self%29%20-%3E%20int%3A%0A%20%20%20%20%20%20%20%20%22%22%22%E5%87%BA%E6%A0%88%22%22%22%0A%20%20%20%20%20%20%20%20if%20self.is_empty%28%29%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20raise%20IndexError%28%22%E6%A0%88%E4%B8%BA%E7%A9%BA%22%29%0A%20%20%20%20%20%20%20%20return%20self._stack.pop%28%29%0A%0A%20%20%20%20def%20peek%28self%29%20-%3E%20int%3A%0A%20%20%20%20%20%20%20%20%22%22%22%E8%AE%BF%E9%97%AE%E6%A0%88%E9%A1%B6%E5%85%83%E7%B4%A0%22%22%22%0A%20%20%20%20%20%20%20%20if%20self.is_empty%28%29%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20raise%20IndexError%28%22%E6%A0%88%E4%B8%BA%E7%A9%BA%22%29%0A%20%20%20%20%20%20%20%20return%20self._stack%5B-1%5D%0A%0A%20%20%20%20def%20to_list%28self%29%20-%3E%20list%5Bint%5D%3A%0A%20%20%20%20%20%20%20%20%22%22%22%E8%BF%94%E5%9B%9E%E5%88%97%E8%A1%A8%E7%94%A8%E4%BA%8E%E6%89%93%E5%8D%B0%22%22%22%0A%20%20%20%20%20%20%20%20return%20self._stack%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%E6%A0%88%0A%20%20%20%20stack%20%3D%20ArrayStack%28%29%0A%0A%20%20%20%20%23%20%E5%85%83%E7%B4%A0%E5%85%A5%E6%A0%88%0A%20%20%20%20stack.push%281%29%0A%20%20%20%20stack.push%283%29%0A%20%20%20%20stack.push%282%29%0A%20%20%20%20stack.push%285%29%0A%20%20%20%20stack.push%284%29%0A%20%20%20%20print%28%22%E6%A0%88%20stack%20%3D%22,%20stack.to_list%28%29%29%0A%0A%20%20%20%20%23%20%E8%AE%BF%E9%97%AE%E6%A0%88%E9%A1%B6%E5%85%83%E7%B4%A0%0A%20%20%20%20peek%20%3D%20stack.peek%28%29%0A%20%20%20%20print%28%22%E6%A0%88%E9%A1%B6%E5%85%83%E7%B4%A0%20peek%20%3D%22,%20peek%29%0A%0A%20%20%20%20%23%20%E5%85%83%E7%B4%A0%E5%87%BA%E6%A0%88%0A%20%20%20%20pop%20%3D%20stack.pop%28%29%0A%20%20%20%20print%28%22%E5%87%BA%E6%A0%88%E5%85%83%E7%B4%A0%20pop%20%3D%22,%20pop%29%0A%20%20%20%20print%28%22%E5%87%BA%E6%A0%88%E5%90%8E%20stack%20%3D%22,%20stack.to_list%28%29%29%0A%0A%20%20%20%20%23%20%E8%8E%B7%E5%8F%96%E6%A0%88%E7%9A%84%E9%95%BF%E5%BA%A6%0A%20%20%20%20size%20%3D%20stack.size%28%29%0A%20%20%20%20print%28%22%E6%A0%88%E7%9A%84%E9%95%BF%E5%BA%A6%20size%20%3D%22,%20size%29%0A%0A%20%20%20%20%23%20%E5%88%A4%E6%96%AD%E6%98%AF%E5%90%A6%E4%B8%BA%E7%A9%BA%0A%20%20%20%20is_empty%20%3D%20stack.is_empty%28%29%0A%20%20%20%20print%28%22%E6%A0%88%E6%98%AF%E5%90%A6%E4%B8%BA%E7%A9%BA%20%3D%22,%20is_empty%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
docs/chapter_tree/array_representation_of_tree.md
-12
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docs/chapter_tree/avl_tree.md
-182
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@@ -236,12 +236,6 @@ AVL 树既是二叉搜索树，也是平衡二叉树，同时满足这两类二
end
```
=== "Zig"
```zig title=""
```
“节点高度”是指从该节点到它的最远叶节点的距离，即所经过的“边”的数量。需要特别注意的是，叶节点的高度为 $0$ ，而空节点的高度为 $-1$ 。我们将创建两个工具函数，分别用于获取和更新节点的高度：
=== "Python"
@@ -481,23 +475,6 @@ AVL 树既是二叉搜索树，也是平衡二叉树，同时满足这两类二
end
```
=== "Zig"
```zig title="avl_tree.zig"
// 获取节点高度
fn height(self: *Self, node: ?*inc.TreeNode(T)) i32 {
_ = self;
// 空节点高度为 -1 ，叶节点高度为 0
return if (node == null) -1 else node.?.height;
}
// 更新节点高度
fn updateHeight(self: *Self, node: ?*inc.TreeNode(T)) void {
// 节点高度等于最高子树高度 + 1
node.?.height = @max(self.height(node.?.left), self.height(node.?.right)) + 1;
}
```
### 2. &nbsp; 节点平衡因子
节点的<u>平衡因子（balance factor）</u>定义为节点左子树的高度减去右子树的高度，同时规定空节点的平衡因子为 $0$ 。我们同样将获取节点平衡因子的功能封装成函数，方便后续使用：
@@ -669,18 +646,6 @@ AVL 树既是二叉搜索树，也是平衡二叉树，同时满足这两类二
end
```
=== "Zig"
```zig title="avl_tree.zig"
// 获取平衡因子
fn balanceFactor(self: *Self, node: ?*inc.TreeNode(T)) i32 {
// 空节点平衡因子为 0
if (node == null) return 0;
// 节点平衡因子 = 左子树高度 - 右子树高度
return self.height(node.?.left) - self.height(node.?.right);
}
```
!!! tip
设平衡因子为 $f$ ，则一棵 AVL 树的任意节点的平衡因子皆满足 $-1 \le f \le 1$ 。
@@ -956,24 +921,6 @@ AVL 树的特点在于“旋转”操作，它能够在不影响二叉树的中
end
```
=== "Zig"
```zig title="avl_tree.zig"
// 右旋操作
fn rightRotate(self: *Self, node: ?*inc.TreeNode(T)) ?*inc.TreeNode(T) {
var child = node.?.left;
var grandChild = child.?.right;
// 以 child 为原点，将 node 向右旋转
child.?.right = node;
node.?.left = grandChild;
// 更新节点高度
self.updateHeight(node);
self.updateHeight(child);
// 返回旋转后子树的根节点
return child;
}
```
### 2. &nbsp; 左旋
相应地，如果考虑上述失衡二叉树的“镜像”，则需要执行图 7-28 所示的“左旋”操作。
@@ -1229,24 +1176,6 @@ AVL 树的特点在于“旋转”操作，它能够在不影响二叉树的中
end
```
=== "Zig"
```zig title="avl_tree.zig"
// 左旋操作
fn leftRotate(self: *Self, node: ?*inc.TreeNode(T)) ?*inc.TreeNode(T) {
var child = node.?.right;
var grandChild = child.?.left;
// 以 child 为原点，将 node 向左旋转
child.?.left = node;
node.?.right = grandChild;
// 更新节点高度
self.updateHeight(node);
self.updateHeight(child);
// 返回旋转后子树的根节点
return child;
}
```
### 3. &nbsp; 先左旋后右旋
对于图 7-30 中的失衡节点 3 ，仅使用左旋或右旋都无法使子树恢复平衡。此时需要先对 `child` 执行“左旋”，再对 `node` 执行“右旋”。
@@ -1730,40 +1659,6 @@ AVL 树的特点在于“旋转”操作，它能够在不影响二叉树的中
end
```
=== "Zig"
```zig title="avl_tree.zig"
// 执行旋转操作，使该子树重新恢复平衡
fn rotate(self: *Self, node: ?*inc.TreeNode(T)) ?*inc.TreeNode(T) {
// 获取节点 node 的平衡因子
var balance_factor = self.balanceFactor(node);
// 左偏树
if (balance_factor > 1) {
if (self.balanceFactor(node.?.left) >= 0) {
// 右旋
return self.rightRotate(node);
} else {
// 先左旋后右旋
node.?.left = self.leftRotate(node.?.left);
return self.rightRotate(node);
}
}
// 右偏树
if (balance_factor < -1) {
if (self.balanceFactor(node.?.right) <= 0) {
// 左旋
return self.leftRotate(node);
} else {
// 先右旋后左旋
node.?.right = self.rightRotate(node.?.right);
return self.leftRotate(node);
}
}
// 平衡树，无须旋转，直接返回
return node;
}
```
## 7.5.3 &nbsp; AVL 树常用操作
### 1. &nbsp; 插入节点
@@ -2142,38 +2037,6 @@ AVL 树的节点插入操作与二叉搜索树在主体上类似。唯一的区
end
```
=== "Zig"
```zig title="avl_tree.zig"
// 插入节点
fn insert(self: *Self, val: T) !void {
self.root = (try self.insertHelper(self.root, val)).?;
}
// 递归插入节点（辅助方法）
fn insertHelper(self: *Self, node_: ?*inc.TreeNode(T), val: T) !?*inc.TreeNode(T) {
var node = node_;
if (node == null) {
var tmp_node = try self.mem_allocator.create(inc.TreeNode(T));
tmp_node.init(val);
return tmp_node;
}
// 1. 查找插入位置并插入节点
if (val < node.?.val) {
node.?.left = try self.insertHelper(node.?.left, val);
} else if (val > node.?.val) {
node.?.right = try self.insertHelper(node.?.right, val);
} else {
return node; // 重复节点不插入，直接返回
}
self.updateHeight(node); // 更新节点高度
// 2. 执行旋转操作，使该子树重新恢复平衡
node = self.rotate(node);
// 返回子树的根节点
return node;
}
```
### 2. &nbsp; 删除节点
类似地，在二叉搜索树的删除节点方法的基础上，需要从底至顶执行旋转操作，使所有失衡节点恢复平衡。代码如下所示：
@@ -2775,51 +2638,6 @@ AVL 树的节点插入操作与二叉搜索树在主体上类似。唯一的区
end
```
=== "Zig"
```zig title="avl_tree.zig"
// 删除节点
fn remove(self: *Self, val: T) void {
self.root = self.removeHelper(self.root, val).?;
}
// 递归删除节点（辅助方法）
fn removeHelper(self: *Self, node_: ?*inc.TreeNode(T), val: T) ?*inc.TreeNode(T) {
var node = node_;
if (node == null) return null;
// 1. 查找节点并删除
if (val < node.?.val) {
node.?.left = self.removeHelper(node.?.left, val);
} else if (val > node.?.val) {
node.?.right = self.removeHelper(node.?.right, val);
} else {
if (node.?.left == null or node.?.right == null) {
var child = if (node.?.left != null) node.?.left else node.?.right;
// 子节点数量 = 0 ，直接删除 node 并返回
if (child == null) {
return null;
// 子节点数量 = 1 ，直接删除 node
} else {
node = child;
}
} else {
// 子节点数量 = 2 ，则将中序遍历的下个节点删除，并用该节点替换当前节点
var temp = node.?.right;
while (temp.?.left != null) {
temp = temp.?.left;
}
node.?.right = self.removeHelper(node.?.right, temp.?.val);
node.?.val = temp.?.val;
}
}
self.updateHeight(node); // 更新节点高度
// 2. 执行旋转操作，使该子树重新恢复平衡
node = self.rotate(node);
// 返回子树的根节点
return node;
}
```
### 3. &nbsp; 查找节点
AVL 树的节点查找操作与二叉搜索树一致，在此不再赘述。
docs/chapter_tree/binary_search_tree.md
-110
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docs/chapter_tree/binary_tree.md
-18
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@@ -205,12 +205,6 @@ comments: true
end
```
=== "Zig"
```zig title=""
```
每个节点都有两个引用（指针），分别指向<u>左子节点（left-child node）</u>和<u>右子节点（right-child node）</u>，该节点被称为这两个子节点的<u>父节点（parent node）</u>。当给定一个二叉树的节点时，我们将该节点的左子节点及其以下节点形成的树称为该节点的<u>左子树（left subtree）</u>，同理可得<u>右子树（right subtree）</u>。
**在二叉树中，除叶节点外，其他所有节点都包含子节点和非空子树**。如图 7-1 所示，如果将“节点 2”视为父节点，则其左子节点和右子节点分别是“节点 4”和“节点 5”，左子树是“节点 4 及其以下节点形成的树”，右子树是“节点 5 及其以下节点形成的树”。
@@ -463,12 +457,6 @@ comments: true
n2.right = n5
```
=== "Zig"
```zig title="binary_tree.zig"
```
??? pythontutor "可视化运行"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=class%20TreeNode%3A%0A%20%20%20%20%22%22%22%E4%BA%8C%E5%8F%89%E6%A0%91%E8%8A%82%E7%82%B9%E7%B1%BB%22%22%22%0A%20%20%20%20def%20__init__%28self,%20val%3A%20int%29%3A%0A%20%20%20%20%20%20%20%20self.val%3A%20int%20%3D%20val%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%23%20%E8%8A%82%E7%82%B9%E5%80%BC%0A%20%20%20%20%20%20%20%20self.left%3A%20TreeNode%20%7C%20None%20%3D%20None%20%20%23%20%E5%B7%A6%E5%AD%90%E8%8A%82%E7%82%B9%E5%BC%95%E7%94%A8%0A%20%20%20%20%20%20%20%20self.right%3A%20TreeNode%20%7C%20None%20%3D%20None%20%23%20%E5%8F%B3%E5%AD%90%E8%8A%82%E7%82%B9%E5%BC%95%E7%94%A8%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%E4%BA%8C%E5%8F%89%E6%A0%91%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%E8%8A%82%E7%82%B9%0A%20%20%20%20n1%20%3D%20TreeNode%28val%3D1%29%0A%20%20%20%20n2%20%3D%20TreeNode%28val%3D2%29%0A%20%20%20%20n3%20%3D%20TreeNode%28val%3D3%29%0A%20%20%20%20n4%20%3D%20TreeNode%28val%3D4%29%0A%20%20%20%20n5%20%3D%20TreeNode%28val%3D5%29%0A%20%20%20%20%23%20%E6%9E%84%E5%BB%BA%E8%8A%82%E7%82%B9%E4%B9%8B%E9%97%B4%E7%9A%84%E5%BC%95%E7%94%A8%EF%BC%88%E6%8C%87%E9%92%88%EF%BC%89%0A%20%20%20%20n1.left%20%3D%20n2%0A%20%20%20%20n1.right%20%3D%20n3%0A%20%20%20%20n2.left%20%3D%20n4%0A%20%20%20%20n2.right%20%3D%20n5&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
@@ -638,12 +626,6 @@ comments: true
n1.left = n2
```
=== "Zig"
```zig title="binary_tree.zig"
```
??? pythontutor "可视化运行"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=class%20TreeNode%3A%0A%20%20%20%20%22%22%22%E4%BA%8C%E5%8F%89%E6%A0%91%E8%8A%82%E7%82%B9%E7%B1%BB%22%22%22%0A%20%20%20%20def%20__init__%28self,%20val%3A%20int%29%3A%0A%20%20%20%20%20%20%20%20self.val%3A%20int%20%3D%20val%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%23%20%E8%8A%82%E7%82%B9%E5%80%BC%0A%20%20%20%20%20%20%20%20self.left%3A%20TreeNode%20%7C%20None%20%3D%20None%20%20%23%20%E5%B7%A6%E5%AD%90%E8%8A%82%E7%82%B9%E5%BC%95%E7%94%A8%0A%20%20%20%20%20%20%20%20self.right%3A%20TreeNode%20%7C%20None%20%3D%20None%20%23%20%E5%8F%B3%E5%AD%90%E8%8A%82%E7%82%B9%E5%BC%95%E7%94%A8%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%E4%BA%8C%E5%8F%89%E6%A0%91%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%E8%8A%82%E7%82%B9%0A%20%20%20%20n1%20%3D%20TreeNode%28val%3D1%29%0A%20%20%20%20n2%20%3D%20TreeNode%28val%3D2%29%0A%20%20%20%20n3%20%3D%20TreeNode%28val%3D3%29%0A%20%20%20%20n4%20%3D%20TreeNode%28val%3D4%29%0A%20%20%20%20n5%20%3D%20TreeNode%28val%3D5%29%0A%20%20%20%20%23%20%E6%9E%84%E5%BB%BA%E8%8A%82%E7%82%B9%E4%B9%8B%E9%97%B4%E7%9A%84%E5%BC%95%E7%94%A8%EF%BC%88%E6%8C%87%E9%92%88%EF%BC%89%0A%20%20%20%20n1.left%20%3D%20n2%0A%20%20%20%20n1.right%20%3D%20n3%0A%20%20%20%20n2.left%20%3D%20n4%0A%20%20%20%20n2.right%20%3D%20n5%0A%0A%20%20%20%20%23%20%E6%8F%92%E5%85%A5%E4%B8%8E%E5%88%A0%E9%99%A4%E8%8A%82%E7%82%B9%0A%20%20%20%20p%20%3D%20TreeNode%280%29%0A%20%20%20%20%23%20%E5%9C%A8%20n1%20-%3E%20n2%20%E4%B8%AD%E9%97%B4%E6%8F%92%E5%85%A5%E8%8A%82%E7%82%B9%20P%0A%20%20%20%20n1.left%20%3D%20p%0A%20%20%20%20p.left%20%3D%20n2%0A%20%20%20%20%23%20%E5%88%A0%E9%99%A4%E8%8A%82%E7%82%B9%20P%0A%20%20%20%20n1.left%20%3D%20n2&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=37&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
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---
# 16.2 &nbsp; Contributing
# 16.2 &nbsp; Contributing Together
Due to the limited abilities of the author, some omissions and errors are inevitable in this book. Please understand. If you discover any typos, broken links, missing content, textual ambiguities, unclear explanations, or unreasonable text structures, please assist us in making corrections to provide readers with better quality learning resources.
Due to limited capacity, there may be inevitable omissions and errors in this book. We appreciate your understanding and are grateful for your help in correcting them. If you discover typos, broken links, missing content, ambiguous wording, unclear explanations, or structural issues, please help us make corrections to provide readers with higher-quality learning resources.
The GitHub IDs of all [contributors](https://github.com/krahets/hello-algo/graphs/contributors) will be displayed on the repository, web, and PDF versions of the homepage of this book to thank them for their selfless contributions to the open-source community.
The GitHub IDs of all [contributors](https://github.com/krahets/hello-algo/graphs/contributors) will be displayed on the homepage of the book repository, the web version, and the PDF version to acknowledge their selfless contributions to the open source community.
!!! success "The charm of open source"
!!! success "The Charm of Open Source"
The interval between two printings of a paper book is often long, making content updates very inconvenient.
In this open-source book, however, the content update cycle is shortened to just a few days or even hours.
The interval between two printings of a physical book is often quite long, making content updates very inconvenient.
### 1. &nbsp; Content fine-tuning
In this open source book, the time for content updates has been shortened to just days or even hours.
As shown in Figure 16-3, there is an "edit icon" in the upper right corner of each page. You can follow these steps to modify text or code.
### 1. &nbsp; Minor Content Adjustments
1. Click the "edit icon". If prompted to "fork this repository", please agree to do so.
2. Modify the Markdown source file content, check the accuracy of the content, and try to keep the formatting consistent.
3. Fill in the modification description at the bottom of the page, then click the "Propose file change" button. After the page redirects, click the "Create pull request" button to initiate the pull request.
As shown in Figure 16-3, there is an "edit icon" in the top-right corner of each page. You can modify text or code by following these steps.
![Edit page button](contribution.assets/edit_markdown.png){ class="animation-figure" }
1. Click the "edit icon". If you encounter a prompt asking you to "Fork this repository", please approve the operation.
2. Modify the content of the Markdown source file, verify the correctness of the content, and maintain consistent formatting as much as possible.
3. Fill in a description of your changes at the bottom of the page, then click the "Propose file change" button. After the page transitions, click the "Create pull request" button to submit your pull request.
<p align="center"> Figure 16-3 &nbsp; Edit page button </p>
![Page edit button](contribution.assets/edit_markdown.png){ class="animation-figure" }
Figures cannot be directly modified and require the creation of a new [Issue](https://github.com/krahets/hello-algo/issues) or a comment to describe the problem. We will redraw and replace the figures as soon as possible.
<p align="center"> Figure 16-3 &nbsp; Page edit button </p>
### 2. &nbsp; Content creation
Images cannot be directly modified. Please describe the issue by creating a new [Issue](https://github.com/krahets/hello-algo/issues) or leaving a comment. We will promptly redraw and replace the images.
If you are interested in participating in this open-source project, including translating code into other programming languages or expanding article content, then the following Pull Request workflow needs to be implemented.
### 2. &nbsp; Content Creation
1. Log in to GitHub and Fork the [code repository](https://github.com/krahets/hello-algo) of this book to your personal account.
2. Go to your Forked repository web page and use the `git clone` command to clone the repository to your local machine.
3. Create content locally and perform complete tests to verify the correctness of the code.
4. Commit the changes made locally, then push them to the remote repository.
5. Refresh the repository webpage and click the "Create pull request" button to initiate the pull request.
If you are interested in contributing to this open source project, including translating code into other programming languages or expanding article content, you will need to follow the Pull Request workflow below.
### 3. &nbsp; Docker deployment
1. Log in to GitHub and Fork the book's [code repository](https://github.com/krahets/hello-algo) to your personal account.
2. Enter your forked repository webpage and use the `git clone` command to clone the repository to your local machine.
3. Create content locally and conduct comprehensive tests to verify code correctness.
4. Commit your local changes and push them to the remote repository.
5. Refresh the repository webpage and click the "Create pull request" button to submit your pull request.
In the `hello-algo` root directory, execute the following Docker script to access the project at `http://localhost:8000`:
### 3. &nbsp; Docker Deployment
From the root directory of `hello-algo`, run the following Docker script to access the project at `http://localhost:8000`:
```shell
docker-compose up -d
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## Chapter contents
- [16.1 &nbsp; Installation](installation.md)
- [16.2 &nbsp; Contributing](contribution.md)
- [16.3 &nbsp; Terminology](terminology.md)
- [16.1 &nbsp; Programming Environment Installation](installation.md)
- [16.2 &nbsp; Contributing Together](contribution.md)
- [16.3 &nbsp; Terminology Table](terminology.md)
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---
# 16.1 &nbsp; Installation
# 16.1 &nbsp; Programming Environment Installation
## 16.1.1 &nbsp; Install IDE
## 16.1.1 &nbsp; Installing Ide
We recommend using the open-source, lightweight VS Code as your local Integrated Development Environment (IDE). Visit the [VS Code official website](https://code.visualstudio.com/) and choose the version of VS Code appropriate for your operating system to download and install.
We recommend using the open-source and lightweight VS Code as the local integrated development environment (IDE). Visit the [VS Code official website](https://code.visualstudio.com/), and download and install the appropriate version of VS Code according to your operating system.
![Download VS Code from the official website](installation.assets/vscode_installation.png){ class="animation-figure" }
![Download VS Code from the Official Website](installation.assets/vscode_installation.png){ class="animation-figure" }
<p align="center"> Figure 16-1 &nbsp; Download VS Code from the official website </p>
<p align="center"> Figure 16-1 &nbsp; Download VS Code from the Official Website </p>
VS Code has a powerful extension ecosystem, supporting the execution and debugging of most programming languages. For example, after installing the "Python Extension Pack," you can debug Python code. The installation steps are shown in Figure 16-2.
VS Code has a powerful ecosystem of extensions that supports running and debugging most programming languages. For example, after installing the "Python Extension Pack" extension, you can debug Python code. The installation steps are shown in the following figure.
![Install VS Code Extension Pack](installation.assets/vscode_extension_installation.png){ class="animation-figure" }
![Install VS Code Extensions](installation.assets/vscode_extension_installation.png){ class="animation-figure" }
<p align="center"> Figure 16-2 &nbsp; Install VS Code Extension Pack </p>
<p align="center"> Figure 16-2 &nbsp; Install VS Code Extensions </p>
## 16.1.2 &nbsp; Install language environments
## 16.1.2 &nbsp; Installing Language Environments
### 1. &nbsp; Python environment
### 1. &nbsp; Python Environment
1. Download and install [Miniconda3](https://docs.conda.io/en/latest/miniconda.html), requiring Python 3.10 or newer.
2. In the VS Code extension marketplace, search for `python` and install the Python Extension Pack.
3. (Optional) Enter `pip install black` in the command line to install the code formatting tool.
1. Download and install [Miniconda3](https://docs.conda.io/en/latest/miniconda.html), which requires Python 3.10 or newer.
2. Search for `python` in the VS Code extension marketplace and install the Python Extension Pack.
3. (Optional) Enter `pip install black` on the command line to install the code formatter.
### 2. &nbsp; C/C++ environment
### 2. &nbsp; C/c++ Environment
1. Windows systems need to install [MinGW](https://sourceforge.net/projects/mingw-w64/files/) ([Configuration tutorial](https://blog.csdn.net/qq_33698226/article/details/129031241)); MacOS comes with Clang, so no installation is necessary.
2. In the VS Code extension marketplace, search for `c++` and install the C/C++ Extension Pack.
1. Windows systems need to install [MinGW](https://sourceforge.net/projects/mingw-w64/files/) ([configuration tutorial](https://blog.csdn.net/qq_33698226/article/details/129031241)); macOS comes with Clang built-in and does not require installation.
2. Search for `c++` in the VS Code extension marketplace and install the C/C++ Extension Pack.
3. (Optional) Open the Settings page, search for the `Clang_format_fallback Style` code formatting option, and set it to `{ BasedOnStyle: Microsoft, BreakBeforeBraces: Attach }`.
### 3. &nbsp; Java environment
### 3. &nbsp; Java Environment
1. Download and install [OpenJDK](https://jdk.java.net/18/) (version must be > JDK 9).
2. In the VS Code extension marketplace, search for `java` and install the Extension Pack for Java.
2. Search for `java` in the VS Code extension marketplace and install the Extension Pack for Java.
### 4. &nbsp; C# environment
### 4. &nbsp; C# Environment
1. Download and install [.Net 8.0](https://dotnet.microsoft.com/en-us/download).
2. In the VS Code extension marketplace, search for `C# Dev Kit` and install the C# Dev Kit ([Configuration tutorial](https://code.visualstudio.com/docs/csharp/get-started)).
3. You can also use Visual Studio ([Installation tutorial](https://learn.microsoft.com/zh-cn/visualstudio/install/install-visual-studio?view=vs-2022)).
2. Search for `C# Dev Kit` in the VS Code extension marketplace and install C# Dev Kit ([configuration tutorial](https://code.visualstudio.com/docs/csharp/get-started)).
3. You can also use Visual Studio ([installation tutorial](https://learn.microsoft.com/zh-cn/visualstudio/install/install-visual-studio?view=vs-2022)).
### 5. &nbsp; Go environment
### 5. &nbsp; Go Environment
1. Download and install [go](https://go.dev/dl/).
2. In the VS Code extension marketplace, search for `go` and install Go.
3. Press `Ctrl + Shift + P` to call up the command bar, enter go, choose `Go: Install/Update Tools`, select all and install.
1. Download and install [Go](https://go.dev/dl/).
2. Search for `go` in the VS Code extension marketplace and install Go.
3. Press `Ctrl + Shift + P` to open the command palette, type `go`, select `Go: Install/Update Tools`, check all options and install.
### 6. &nbsp; Swift environment
### 6. &nbsp; Swift Environment
1. Download and install [Swift](https://www.swift.org/download/).
2. In the VS Code extension marketplace, search for `swift` and install [Swift for Visual Studio Code](https://marketplace.visualstudio.com/items?itemName=sswg.swift-lang).
2. Search for `swift` in the VS Code extension marketplace and install [Swift for Visual Studio Code](https://marketplace.visualstudio.com/items?itemName=sswg.swift-lang).
### 7. &nbsp; JavaScript environment
### 7. &nbsp; Javascript Environment
1. Download and install [Node.js](https://nodejs.org/en/).
2. (Optional) In the VS Code extension marketplace, search for `Prettier` and install the code formatting tool.
2. (Optional) Search for `Prettier` in the VS Code extension marketplace and install the code formatter.
### 8. &nbsp; TypeScript environment
### 8. &nbsp; Typescript Environment
1. Follow the same installation steps as the JavaScript environment.
2. Install [TypeScript Execute (tsx)](https://github.com/privatenumber/tsx?tab=readme-ov-file#global-installation).
3. In the VS Code extension marketplace, search for `typescript` and install [Pretty TypeScript Errors](https://marketplace.visualstudio.com/items?itemName=yoavbls.pretty-ts-errors).
3. Search for `typescript` in the VS Code extension marketplace and install [Pretty TypeScript Errors](https://marketplace.visualstudio.com/items?itemName=yoavbls.pretty-ts-errors).
### 9. &nbsp; Dart environment
### 9. &nbsp; Dart Environment
1. Download and install [Dart](https://dart.dev/get-dart).
2. In the VS Code extension marketplace, search for `dart` and install [Dart](https://marketplace.visualstudio.com/items?itemName=Dart-Code.dart-code).
2. Search for `dart` in the VS Code extension marketplace and install [Dart](https://marketplace.visualstudio.com/items?itemName=Dart-Code.dart-code).
### 10. &nbsp; Rust environment
### 10. &nbsp; Rust Environment
1. Download and install [Rust](https://www.rust-lang.org/tools/install).
2. In the VS Code extension marketplace, search for `rust` and install [rust-analyzer](https://marketplace.visualstudio.com/items?itemName=rust-lang.rust-analyzer).
2. Search for `rust` in the VS Code extension marketplace and install [rust-analyzer](https://marketplace.visualstudio.com/items?itemName=rust-lang.rust-analyzer).
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---
# 16.3 &nbsp; Glossary
# 16.3 &nbsp; Terminology Table
Table 16-1 lists the important terms that appear in the book, and it is worth noting the following points.
The following table lists important terms that appear in this book. It is worth noting the following points:
- It is recommended to remember the English names of the terms to facilitate reading English literature.
- Some terms have different names in Simplified and Traditional Chinese.
- We recommend remembering the English names of terms to help with reading English literature.
- Some terms have different names in Simplified Chinese and Traditional Chinese.
<p align="center"> Table 16-1 &nbsp; Important Terms in Data Structures and Algorithms </p>
<div class="center-table" markdown>
| English | 简体中文 | 繁体中文 |
| ------------------------------ | -------------- | -------------- |
| English | Simplified Chinese | Traditional Chinese |
| ------------------------------ | ------------------ | ------------------- |
| algorithm | 算法 | 演算法 |
| data structure | 数据结构 | 資料結構 |
| code | 代码 | 程式碼 |
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---
# Chapter 4. &nbsp; Arrays and linked lists
# Chapter 4. &nbsp; Array and Linked List
![Arrays and linked lists](../assets/covers/chapter_array_and_linkedlist.jpg){ class="cover-image" }
![Array and Linked List](../assets/covers/chapter_array_and_linkedlist.jpg){ class="cover-image" }
!!! abstract
The world of data structures resembles a sturdy brick wall.
The world of data structures is like a solid brick wall.
In arrays, envision bricks snugly aligned, each resting seamlessly beside the next, creating a unified formation. Meanwhile, in linked lists, these bricks disperse freely, embraced by vines gracefully knitting connections between them.
Array bricks are neatly arranged, tightly packed one by one. Linked list bricks are scattered everywhere, with connecting vines freely weaving through the gaps between bricks.
## Chapter contents
- [4.1 &nbsp; Array](array.md)
- [4.2 &nbsp; Linked list](linked_list.md)
- [4.2 &nbsp; Linked List](linked_list.md)
- [4.3 &nbsp; List](list.md)
- [4.4 &nbsp; Memory and cache *](ram_and_cache.md)
- [4.4 &nbsp; Memory and Cache *](ram_and_cache.md)
- [4.5 &nbsp; Summary](summary.md)
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# 4.4 &nbsp; Memory and cache *
# 4.4 &nbsp; Random-Access Memory and Cache *
In the first two sections of this chapter, we explored arrays and linked lists, two fundamental data structures that represent "continuous storage" and "dispersed storage," respectively.
In the first two sections of this chapter, we explored arrays and linked lists, two fundamental and important data structures that represent "contiguous storage" and "distributed storage" as two physical structures, respectively.
In fact, **the physical structure largely determines how efficiently a program utilizes memory and cache**, which in turn affects the overall performance of the algorithm.
In fact, **physical structure largely determines the efficiency with which programs utilize memory and cache**, which in turn affects the overall performance of algorithmic programs.
## 4.4.1 &nbsp; Computer storage devices
## 4.4.1 &nbsp; Computer Storage Devices
There are three types of storage devices in computers: <u>hard disk</u>, <u>random-access memory (RAM)</u>, and <u>cache memory</u>. The following table shows their respective roles and performance characteristics in computer systems.
Computers include three types of storage devices: <u>hard disk</u>, <u>random-access memory (RAM)</u>, and <u>cache memory</u>. The following table shows their different roles and performance characteristics in a computer system.
<p align="center"> Table 4-2 &nbsp; Computer storage devices </p>
<p align="center"> Table 4-2 &nbsp; Computer Storage Devices </p>
<div class="center-table" markdown>
| | Hard Disk | Memory | Cache |
| ----------- | -------------------------------------------------------------- | ------------------------------------------------------------------------ | ----------------------------------------------------------------------------------------------- |
| Usage | Long-term storage of data, including OS, programs, files, etc. | Temporary storage of currently running programs and data being processed | Stores frequently accessed data and instructions, reducing the number of CPU accesses to memory |
| Volatility | Data is not lost after power off | Data is lost after power off | Data is lost after power off |
| Capacity | Larger, TB level | Smaller, GB level | Very small, MB level |
| Speed | Slower, several hundred to thousands MB/s | Faster, several tens of GB/s | Very fast, several tens to hundreds of GB/s |
| Price (USD) | Cheaper, a few cents / GB | More expensive, a few dollars / GB | Very expensive, priced with CPU |
| | Hard Disk | RAM | Cache |
| -------------- | ------------------------------------------------------------- | ------------------------------------------------ | -------------------------------------------------------------- |
| Purpose | Long-term storage of data, including operating systems, programs, and files | Temporary storage of currently running programs and data being processed | Storage of frequently accessed data and instructions to reduce CPU's accesses to memory |
| Volatility | Data is not lost after power-off | Data is lost after power-off | Data is lost after power-off |
| Capacity | Large, on the order of terabytes (TB) | Small, on the order of gigabytes (GB) | Very small, on the order of megabytes (MB) |
| Speed | Slow, hundreds to thousands of MB/s | Fast, tens of GB/s | Very fast, tens to hundreds of GB/s |
| Cost (USD/GB) | Inexpensive, fractions of a dollar to a few dollars per GB | Expensive, tens to hundreds of dollars per GB | Very expensive, priced as part of the CPU package |
</div>
The computer storage system can be visualized as a pyramid, as shown in Figure 4-9. The storage devices at the top of the pyramid are faster, have smaller capacities, and are more expensive. This multi-level design is not accidental, but a deliberate outcome of careful consideration by computer scientists and engineers.
We can imagine the computer storage system as a pyramid structure as shown in the diagram below. Storage devices closer to the top of the pyramid are faster, have smaller capacity, and are more expensive. This multi-layered design is not by accident, but rather the result of careful consideration by computer scientists and engineers.
- **Replacing hard disks with memory is challenging**. Firstly, data in memory is lost after power off, making it unsuitable for long-term data storage; secondly, memory is significantly more expensive than hard disks, limiting its feasibility for widespread use in the consumer market.
- **Caches face a trade-off between large capacity and high speed**. As the capacity of L1, L2, and L3 caches increases, their physical size grows, increasing the distance from the CPU core. This results in longer data transfer times and higher access latency. With current technology, a multi-level cache structure provides the optimal balance between capacity, speed, and cost.
- **Hard disk cannot be easily replaced by RAM**. First, data in memory is lost after power-off, making it unsuitable for long-term data storage. Second, memory is tens of times more expensive than hard disk, which makes it difficult to popularize in the consumer market.
- **Cache cannot simultaneously achieve large capacity and high speed**. As the capacity of L1, L2, and L3 caches increases, their physical size becomes larger, and the physical distance between them and the CPU core increases, resulting in longer data transmission time and higher element access latency. With current technology, the multi-layered cache structure represents the best balance point between capacity, speed, and cost.
![Computer storage system](ram_and_cache.assets/storage_pyramid.png){ class="animation-figure" }
![Computer Storage System](ram_and_cache.assets/storage_pyramid.png){ class="animation-figure" }
<p align="center"> Figure 4-9 &nbsp; Computer storage system </p>
<p align="center"> Figure 4-9 &nbsp; Computer Storage System </p>
!!! tip
The storage hierarchy in computers reflects a careful balance between speed, capacity, and cost. This type of trade-off is common across various industries, where finding the optimal balance between benefits and limitations is essential.
The storage hierarchy of computers embodies a delicate balance among speed, capacity, and cost. In fact, such trade-offs are common across all industrial fields, requiring us to find the optimal balance point between different advantages and constraints.
Overall, **hard disks provide long-term storage for large volumes of data, memory serves as temporary storage for data being processed during program execution, and cache stores frequently accessed data and instructions to enhance execution efficiency**. Together, they ensure the efficient operation of computer systems.
In summary, **hard disk is used for long-term storage of large amounts of data, RAM is used for temporary storage of data being processed during program execution, and cache is used for storage of frequently accessed data and instructions**, to improve program execution efficiency. The three work together to ensure efficient operation of the computer system.
As shown in Figure 4-10, during program execution, data is read from the hard disk into memory for CPU computation. The cache, acting as an extension of the CPU, **intelligently preloads data from memory**, enabling faster data access for the CPU. This greatly improves program execution efficiency while reducing reliance on slower memory.
As shown in the diagram below, during program execution, data is read from the hard disk into RAM for CPU computation. Cache can be viewed as part of the CPU, **it intelligently loads data from RAM**, providing the CPU with high-speed data reading, thereby significantly improving program execution efficiency and reducing reliance on slower RAM.
![Data flow between hard disk, memory, and cache](ram_and_cache.assets/computer_storage_devices.png){ class="animation-figure" }
![Data Flow Among Hard Disk, RAM, and Cache](ram_and_cache.assets/computer_storage_devices.png){ class="animation-figure" }
<p align="center"> Figure 4-10 &nbsp; Data flow between hard disk, memory, and cache </p>
<p align="center"> Figure 4-10 &nbsp; Data Flow Among Hard Disk, RAM, and Cache </p>
## 4.4.2 &nbsp; Memory efficiency of data structures
## 4.4.2 &nbsp; Memory Efficiency of Data Structures
In terms of memory space utilization, arrays and linked lists have their advantages and limitations.
In terms of memory space utilization, arrays and linked lists each have advantages and limitations.
On one hand, **memory is limited and cannot be shared by multiple programs**, so optimizing space usage in data structures is crucial. Arrays are space-efficient because their elements are tightly packed, without requiring extra memory for references (pointers) as in linked lists. However, arrays require pre-allocating a contiguous block of memory, which can lead to waste if the allocated space exceeds the actual need. Expanding an array also incurs additional time and space overhead. In contrast, linked lists allocate and free memory dynamically for each node, offering greater flexibility at the cost of additional memory for pointers.
On one hand, **memory is limited, and the same memory cannot be shared by multiple programs**, so we hope data structures can utilize space as efficiently as possible. Array elements are tightly packed and do not require additional space to store references (pointers) between linked list nodes, thus having higher space efficiency. However, arrays need to allocate sufficient contiguous memory space at once, which may lead to memory waste, and array expansion requires additional time and space costs. In comparison, linked lists perform dynamic memory allocation and deallocation on a "node" basis, providing greater flexibility.
On the other hand, during program execution, **repeated memory allocation and deallocation increase memory fragmentation**, reducing memory utilization efficiency. Arrays, due to their continuous storage method, are relatively less likely to cause memory fragmentation. In contrast, linked lists store elements in non-contiguous locations, and frequent insertions and deletions can exacerbate memory fragmentation.
On the other hand, during program execution, **as memory is repeatedly allocated and freed, the degree of fragmentation of free memory becomes increasingly severe**, leading to reduced memory utilization efficiency. Arrays, due to their contiguous storage approach, are relatively less prone to memory fragmentation. Conversely, linked list elements are distributed in storage, and frequent insertion and deletion operations are more likely to cause memory fragmentation.
## 4.4.3 &nbsp; Cache efficiency of data structures
## 4.4.3 &nbsp; Cache Efficiency of Data Structures
Although caches are much smaller in space capacity than memory, they are much faster and play a crucial role in program execution speed. Due to their limited capacity, caches can only store a subset of frequently accessed data. When the CPU attempts to access data not present in the cache, a <u>cache miss</u> occurs, requiring the CPU to retrieve the needed data from slower memory, which can impact performance.
Although cache has much smaller space capacity than memory, it is much faster than memory and plays a crucial role in program execution speed. Since cache capacity is limited and can only store a small portion of frequently accessed data, when the CPU attempts to access data that is not in the cache, a <u>cache miss</u> occurs, and the CPU must load the required data from the slower memory.
Clearly, **the fewer the cache misses, the higher the CPU's data read-write efficiency**, and the better the program performance. The proportion of successful data retrieval from the cache by the CPU is called the <u>cache hit rate</u>, a metric often used to measure cache efficiency.
Clearly, **the fewer "cache misses," the higher the efficiency of CPU data reads and writes**, and the better the program performance. We call the proportion of data that the CPU successfully obtains from the cache the <u>cache hit rate</u>, a metric typically used to measure cache efficiency.
To achieve higher efficiency, caches adopt the following data loading mechanisms.
To achieve the highest efficiency possible, cache employs the following data loading mechanisms.
- **Cache lines**: Caches operate by storing and loading data in units called cache lines, rather than individual bytes. This approach improves efficiency by transferring larger blocks of data at once.
- **Prefetch mechanism**: Processors predict data access patterns (e.g., sequential or fixed-stride access) and preload data into the cache based on these patterns to increase the cache hit rate.
- **Spatial locality**: When a specific piece of data is accessed, nearby data is likely to be accessed soon. To leverage this, caches load adjacent data along with the requested data, improving hit rates.
- **Temporal locality**: If data is accessed, it's likely to be accessed again in the near future. Caches use this principle to retain recently accessed data to improve the hit rate.
- **Cache lines**: The cache does not store and load data on a byte-by-byte basis, but rather as cache lines. Compared to byte-by-byte transmission, cache line transmission is more efficient.
- **Prefetching mechanism**: The processor attempts to predict data access patterns (e.g., sequential access, fixed-stride jumping access, etc.) and loads data into the cache according to specific patterns, thereby improving hit rate.
- **Spatial locality**: If a piece of data is accessed, nearby data may also be accessed in the near future. Therefore, when the cache loads a particular piece of data, it also loads nearby data to improve hit rate.
- **Temporal locality**: If a piece of data is accessed, it is likely to be accessed again in the near future. Cache leverages this principle by retaining recently accessed data to improve hit rate.
In fact, **arrays and linked lists have different cache utilization efficiencies**, which is mainly reflected in the following aspects.
In fact, **arrays and linked lists have different efficiencies in utilizing cache**, manifested in the following aspects.
- **Occupied space**: Linked list elements take up more space than array elements, resulting in less effective data being held in the cache.
- **Cache lines**: Linked list data is scattered throughout the memory, and cache is "loaded by row", so the proportion of invalid data loaded is higher.
- **Prefetch mechanism**: The data access pattern of arrays is more "predictable" than that of linked lists, that is, it is easier for the system to guess the data that is about to be loaded.
- **Spatial locality**: Arrays are stored in a continuous memory space, so data near the data being loaded is more likely to be accessed soon.
- **Space occupied**: Linked list elements occupy more space than array elements, resulting in fewer effective data in the cache.
- **Cache lines**: Linked list data are scattered throughout memory, while cache loads "by lines," so the proportion of invalid data loaded is higher.
- **Prefetching mechanism**: Arrays have more "predictable" data access patterns than linked lists, making it easier for the system to guess which data will be loaded next.
- **Spatial locality**: Arrays are stored in centralized memory space, so data near loaded data is more likely to be accessed soon.
Overall, **arrays have a higher cache hit rate and are generally more efficient in operation than linked lists**. This makes data structures based on arrays more popular in solving algorithmic problems.
Overall, **arrays have higher cache hit rates, thus they usually outperform linked lists in operation efficiency**. This makes data structures implemented based on arrays more popular when solving algorithmic problems.
It should be noted that **high cache efficiency does not mean that arrays are always better than linked lists**. The choice of data structure should depend on specific application requirements. For example, both arrays and linked lists can implement the "stack" data structure (which will be detailed in the next chapter), but they are suitable for different scenarios.
It is important to note that **high cache efficiency does not mean arrays are superior to linked lists in all cases**. In practical applications, which data structure to choose should be determined based on specific requirements. For example, both arrays and linked lists can implement the "stack" data structure (which will be discussed in detail in the next chapter), but they are suitable for different scenarios.
- In algorithm problems, we tend to choose stacks based on arrays because they provide higher operational efficiency and random access capabilities, with the only cost being the need to pre-allocate a certain amount of memory space for the array.
- If the data volume is very large, highly dynamic, and the expected size of the stack is difficult to estimate, then a stack based on a linked list is a better choice. Linked lists can distribute a large amount of data in different parts of the memory and avoid the additional overhead of array expansion.
- When solving algorithm problems, we tend to prefer stack implementations based on arrays, because they provide higher operation efficiency and the ability of random access, at the cost of needing to pre-allocate a certain amount of memory space for the array.
- If the data volume is very large, the dynamic nature is high, and the expected size of the stack is difficult to estimate, then a stack implementation based on linked lists is more suitable. Linked lists can distribute large amounts of data across different parts of memory and avoid the additional overhead produced by array expansion.
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# 4.5 &nbsp; Summary
### 1. &nbsp; Key review
### 1. &nbsp; Key Review
- Arrays and linked lists are two basic data structures, representing two storage methods in computer memory: contiguous space storage and non-contiguous space storage. Their characteristics complement each other.
- Arrays support random access and use less memory; however, they are inefficient in inserting and deleting elements and have a fixed length after initialization.
- Linked lists implement efficient node insertion and deletion through changing references (pointers) and can flexibly adjust their length; however, they have lower node access efficiency and consume more memory.
- Common types of linked lists include singly linked lists, circular linked lists, and doubly linked lists, each with its own application scenarios.
- Lists are ordered collections of elements that support addition, deletion, and modification, typically implemented based on dynamic arrays, retaining the advantages of arrays while allowing flexible length adjustment.
- The advent of lists significantly enhanced the practicality of arrays but may lead to some memory space wastage.
- During program execution, data is mainly stored in memory. Arrays provide higher memory space efficiency, while linked lists are more flexible in memory usage.
- Caches provide fast data access to CPUs through mechanisms like cache lines, prefetching, spatial locality, and temporal locality, significantly enhancing program execution efficiency.
- Due to higher cache hit rates, arrays are generally more efficient than linked lists. When choosing a data structure, the appropriate choice should be made based on specific needs and scenarios.
- Arrays and linked lists are two fundamental data structures, representing two different ways data can be stored in computer memory: contiguous memory storage and scattered memory storage. The characteristics of the two complement each other.
- Arrays support random access and use less memory; however, inserting and deleting elements is inefficient, and the length is immutable after initialization.
- Linked lists achieve efficient insertion and deletion of nodes by modifying references (pointers), and can flexibly adjust length; however, node access is inefficient and memory consumption is higher. Common linked list types include singly linked lists, circular linked lists, and doubly linked lists.
- A list is an ordered collection of elements that supports insertion, deletion, search, and modification, typically implemented based on dynamic arrays. It retains the advantages of arrays while allowing flexible adjustment of length.
- The emergence of lists has greatly improved the practicality of arrays, but may result in some wasted memory space.
- During program execution, data is primarily stored in memory. Arrays provide higher memory space efficiency, while linked lists offer greater flexibility in memory usage.
- Caches provide fast data access to the CPU through mechanisms such as cache lines, prefetching, and spatial and temporal locality, significantly improving program execution efficiency.
- Because arrays have higher cache hit rates, they are generally more efficient than linked lists. When choosing a data structure, appropriate selection should be made based on specific requirements and scenarios.
### 2. &nbsp; Q & A
**Q**: Does storing arrays on the stack versus the heap affect time and space efficiency?
**Q**: Does storing an array on the stack versus on the heap affect time efficiency and space efficiency?
Arrays stored on both the stack and heap are stored in contiguous memory spaces, and data operation efficiency is essentially the same. However, stacks and heaps have their own characteristics, leading to the following differences.
Arrays stored on the stack and on the heap are both stored in contiguous memory space, so data operation efficiency is basically the same. However, the stack and heap have their own characteristics, leading to the following differences.
1. Allocation and release efficiency: The stack is a smaller memory block, allocated automatically by the compiler; the heap memory is relatively larger and can be dynamically allocated in the code, more prone to fragmentation. Therefore, allocation and release operations on the heap are generally slower than on the stack.
2. Size limitation: Stack memory is relatively small, while the heap size is generally limited by available memory. Therefore, the heap is more suitable for storing large arrays.
3. Flexibility: The size of arrays on the stack needs to be determined at compile-time, while the size of arrays on the heap can be dynamically determined at runtime.
1. Allocation and deallocation efficiency: The stack is a relatively small piece of memory, with allocation automatically handled by the compiler; the heap is relatively larger and can be dynamically allocated in code, more prone to fragmentation. Therefore, allocation and deallocation operations on the heap are usually slower than on the stack.
2. Size limitations: Stack memory is relatively small, and the heap size is generally limited by available memory. Therefore, the heap is more suitable for storing large arrays.
3. Flexibility: The size of an array on the stack must be determined at compile time, while the size of an array on the heap can be determined dynamically at runtime.
**Q**: Why do arrays require elements of the same type, while linked lists do not emphasize same-type elements?
**Q**: Why do arrays require elements of the same type, while linked lists do not emphasize this requirement?
Linked lists consist of nodes connected by references (pointers), and each node can store data of different types, such as int, double, string, object, etc.
Linked lists are composed of nodes, with nodes connected through references (pointers), and each node can store different types of data, such as `int`, `double`, `string`, `object`, etc.
In contrast, array elements must be of the same type, allowing the calculation of offsets to access the corresponding element positions. For example, an array containing both int and long types, with single elements occupying 4 bytes and 8 bytes respectively, cannot use the following formula to calculate offsets, as the array contains elements of two different lengths.
In contrast, array elements must be of the same type, so that the corresponding element position can be obtained by calculating the offset. For example, if an array contains both `int` and `long` types, with individual elements occupying 4 bytes and 8 bytes respectively, then the following formula cannot be used to calculate the offset, because the array contains two different "element lengths".
```shell
# Element memory address = array memory address + element length * element index
# Element Memory Address = Array Memory Address (first Element Memory address) + Element Length * Element Index
```
**Q**: After deleting a node, is it necessary to set `P.next` to `None`?
**Q**: After deleting node `P`, do we need to set `P.next` to `None`?
Not modifying `P.next` is also acceptable. From the perspective of the linked list, traversing from the head node to the tail node will no longer encounter `P`. This means that node `P` has been effectively removed from the list, and where `P` points no longer affects the list.
It is not necessary to modify `P.next`. From the perspective of the linked list, traversing from the head node to the tail node will no longer encounter `P`. This means that node `P` has been removed from the linked list, and it doesn't matter where node `P` points to at this time—it won't affect the linked list.
From a garbage collection perspective, for languages with automatic garbage collection mechanisms like Java, Python, and Go, whether node `P` is collected depends on whether there are still references pointing to it, not on the value of `P.next`. In languages like C and C++, we need to manually free the node's memory.
From a data structures and algorithms perspective (problem-solving), not disconnecting the pointer doesn't matter as long as the program logic is correct. From the perspective of standard libraries, disconnecting is safer and the logic is clearer. If not disconnected, assuming the deleted node is not properly reclaimed, it may affect the memory reclamation of its successor nodes.
**Q**: In linked lists, the time complexity for insertion and deletion operations is `O(1)`. But searching for the element before insertion or deletion takes `O(n)` time, so why isn't the time complexity `O(n)`?
**Q**: In a linked list, the time complexity of insertion and deletion operations is $O(1)$. However, both insertion and deletion require $O(n)$ time to find the element; why isn't the time complexity $O(n)$?
If an element is searched first and then deleted, the time complexity is indeed `O(n)`. However, the `O(1)` advantage of linked lists in insertion and deletion can be realized in other applications. For example, in the implementation of double-ended queues using linked lists, we maintain pointers always pointing to the head and tail nodes, making each insertion and deletion operation `O(1)`.
If the element is first found and then deleted, the time complexity is indeed $O(n)$. However, the advantage of $O(1)$ insertion and deletion in linked lists can be demonstrated in other applications. For example, a deque is well-suited for linked list implementation, where we maintain pointer variables always pointing to the head and tail nodes, with each insertion and deletion operation being $O(1)$.
**Q**: In the figure "Linked List Definition and Storage Method", do the light blue storage nodes occupy a single memory address, or do they share half with the node value?
**Q**: In the diagram "Linked List Definition and Storage Methods", does the light blue pointer node occupy a single memory address, or does it share equally with the node value?
The figure is just a qualitative representation; quantitative analysis depends on specific situations.
This diagram is a qualitative representation; a quantitative representation requires analysis based on the specific situation.
- Different types of node values occupy different amounts of space, such as int, long, double, and object instances.
- The memory space occupied by pointer variables depends on the operating system and compilation environment used, usually 8 bytes or 4 bytes.
- Different types of node values occupy different amounts of space, such as `int`, `long`, `double`, and instance objects, etc.
- The amount of memory space occupied by pointer variables depends on the operating system and compilation environment used, usually 8 bytes or 4 bytes.
**Q**: Is adding elements to the end of a list always `O(1)`?
**Q**: Is appending an element at the end of a list always $O(1)$?
If adding an element exceeds the list length, the list needs to be expanded first. The system will request a new memory block and move all elements of the original list over, in which case the time complexity becomes `O(n)`.
If appending an element exceeds the list length, the list must first be expanded before adding. The system allocates a new block of memory and moves all elements from the original list to it, in which case the time complexity becomes $O(n)$.
**Q**: The statement "The emergence of lists greatly improves the practicality of arrays, but may lead to some memory space wastage" - does this refer to the memory occupied by additional variables like capacity, length, and expansion multiplier?
**Q**: "The emergence of lists has greatly improved the practicality of arrays, but may result in some wasted memory space"—does this space waste refer to the memory occupied by additional variables such as capacity, length, and expansion factor?
The space wastage here mainly refers to two aspects: on the one hand, lists are set with an initial length, which we may not always need; on the other hand, to prevent frequent expansion, expansion usually multiplies by a coefficient, such as $\times 1.5$. This results in many empty slots, which we typically cannot fully fill.
This space waste mainly has two aspects: on one hand, lists typically set an initial length, which we may not need to fully utilize; on the other hand, to prevent frequent expansion, expansion generally multiplies by a coefficient, such as $\times 1.5$. As a result, there will be many empty positions that we typically cannot completely fill.
**Q**: In Python, after initializing `n = [1, 2, 3]`, the addresses of these 3 elements are contiguous, but initializing `m = [2, 1, 3]` shows that each element's `id` is not consecutive but identical to those in `n`. If the addresses of these elements are not contiguous, is `m` still an array?
**Q**: In Python, after initializing `n = [1, 2, 3]`, the addresses of these 3 elements are contiguous, but initializing `m = [2, 1, 3]` reveals that each element's id is not continuous; rather, they are the same as those in `n`. Since the addresses of these elements are not contiguous, is `m` still an array?
If we replace list elements with linked list nodes `n = [n1, n2, n3, n4, n5]`, these 5 node objects are also typically dispersed throughout memory. However, given a list index, we can still access the node's memory address in `O(1)` time, thereby accessing the corresponding node. This is because the array stores references to the nodes, not the nodes themselves.
If we replace list elements with linked list nodes `n = [n1, n2, n3, n4, n5]`, usually these 5 node objects are also scattered throughout memory. However, given a list index, we can still obtain the node memory address in $O(1)$ time, thereby accessing the corresponding node. This is because the array stores references to nodes, not the nodes themselves.
Unlike many languages, in Python, numbers are also wrapped as objects, and lists store references to these numbers, not the numbers themselves. Therefore, we find that the same number in two arrays has the same `id`, and these numbers' memory addresses need not be contiguous.
Unlike many languages, numbers in Python are wrapped as objects, and lists store not the numbers themselves, but references to the numbers. Therefore, we find that the same numbers in two arrays have the same id, and the memory addresses of these numbers need not be contiguous.
**Q**: The `std::list` in C++ STL has already implemented a doubly linked list, but it seems that some algorithm books don't directly use it. Is there any limitation?
**Q**: C++ STL has `std::list` which has already implemented a doubly linked list, but it seems that some algorithm books don't use it directly. Is there a limitation?
On the one hand, we often prefer to use arrays to implement algorithms, only using linked lists when necessary, mainly for two reasons.
On one hand, we often prefer to use arrays for implementing algorithms and only use linked lists when necessary, mainly for two reasons.
- Space overhead: Since each element requires two additional pointers (one for the previous element and one for the next), `std::list` usually occupies more space than `std::vector`.
- Cache unfriendly: As the data is not stored continuously, `std::list` has a lower cache utilization rate. Generally, `std::vector` performs better.
- Space overhead: Since each element requires two additional pointers (one for the previous element and one for the next element), `std::list` typically consumes more space than `std::vector`.
- Cache unfriendliness: Since data is not stored contiguously, `std::list` has lower cache utilization. In general, `std::vector` has better performance.
On the other hand, linked lists are primarily necessary for binary trees and graphs. Stacks and queues are often implemented using the programming language's `stack` and `queue` classes, rather than linked lists.
On the other hand, cases where linked lists are necessary mainly involve binary trees and graphs. Stacks and queues usually use the `stack` and `queue` provided by the programming language, rather than linked lists.
**Q**: Does initializing a list `res = [0] * self.size()` result in each element of `res` referencing the same address?
**Q**: Does the operation `res = [[0]] * n` create a 2D list where each `[0]` is independent?
No. However, this issue arises with two-dimensional arrays, for example, initializing a two-dimensional list `res = [[0]] * self.size()` would reference the same list `[0]` multiple times.
No, they are not independent. In this 2D list, all the `[0]` are actually references to the same object. If we modify one element, we will find that all corresponding elements change accordingly.
**Q**: In deleting a node, is it necessary to break the reference to its successor node?
If we want each `[0]` in the 2D list to be independent, we can use `res = [[0] for _ in range(n)]` to achieve this. The principle of this approach is to initialize $n$ independent `[0]` list objects.
From the perspective of data structures and algorithms (problem-solving), it's okay not to break the link, as long as the program's logic is correct. From the perspective of standard libraries, breaking the link is safer and more logically clear. If the link is not broken, and the deleted node is not properly recycled, it could affect the recycling of the successor node's memory.
**Q**: Does the operation `res = [0] * n` create a list where each integer 0 is independent?
In this list, all integer 0s are references to the same object. This is because Python uses a caching mechanism for small integers (typically -5 to 256) to maximize object reuse and improve performance.
Although they point to the same object, we can still independently modify each element in the list. This is because Python integers are "immutable objects". When we modify an element, we are actually switching to a reference of another object, rather than changing the original object itself.
However, when list elements are "mutable objects" (such as lists, dictionaries, or class instances), modifying an element directly changes the object itself, and all elements referencing that object will have the same change.
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!!! abstract
Like explorers in a maze, we may encounter obstacles on our path forward.
We are like explorers in a maze, and may encounter difficulties on the path forward.
The power of backtracking lets us begin anew, keep trying, and eventually find the exit leading to the light.
The power of backtracking allows us to start over, keep trying, and eventually find the exit leading to light.
## Chapter contents
- [13.1 &nbsp; Backtracking algorithms](backtracking_algorithm.md)
- [13.2 &nbsp; Permutation problem](permutations_problem.md)
- [13.3 &nbsp; Subset sum problem](subset_sum_problem.md)
- [13.4 &nbsp; n queens problem](n_queens_problem.md)
- [13.1 &nbsp; Backtracking Algorithm](backtracking_algorithm.md)
- [13.2 &nbsp; Permutations Problem](permutations_problem.md)
- [13.3 &nbsp; Subset-Sum Problem](subset_sum_problem.md)
- [13.4 &nbsp; N-Queens Problem](n_queens_problem.md)
- [13.5 &nbsp; Summary](summary.md)
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---
# 13.4 &nbsp; n queens problem
# 13.4 &nbsp; N-Queens Problem
!!! question
According to the rules of chess, a queen can attack pieces in the same row, column, or diagonal line. Given $n$ queens and an $n \times n$ chessboard, find arrangements where no two queens can attack each other.
According to the rules of chess, a queen can attack pieces that share the same row, column, or diagonal line. Given $n$ queens and an $n \times n$ chessboard, find a placement scheme such that no two queens can attack each other.
As shown in Figure 13-15, there are two solutions when $n = 4$. From the perspective of the backtracking algorithm, an $n \times n$ chessboard has $n^2$ squares, presenting all possible choices `choices`. The state of the chessboard `state` changes continuously as each queen is placed.
As shown in Figure 13-15, when $n = 4$, there are two solutions that can be found. From the perspective of the backtracking algorithm, an $n \times n$ chessboard has $n^2$ squares, which provide all the choices `choices`. During the process of placing queens one by one, the chessboard state changes continuously, and the chessboard at each moment represents the state `state`.
![Solution to the 4 queens problem](n_queens_problem.assets/solution_4_queens.png){ class="animation-figure" }
![Solution to the 4-queens problem](n_queens_problem.assets/solution_4_queens.png){ class="animation-figure" }
<p align="center"> Figure 13-15 &nbsp; Solution to the 4 queens problem </p>
<p align="center"> Figure 13-15 &nbsp; Solution to the 4-queens problem </p>
Figure 13-16 shows the three constraints of this problem: **multiple queens cannot occupy the same row, column, or diagonal**. It is important to note that diagonals are divided into the main diagonal `\` and the secondary diagonal `/`.
Figure 13-16 illustrates the three constraints of this problem: **multiple queens cannot be in the same row, the same column, or on the same diagonal**. It is worth noting that diagonals are divided into two types: the main diagonal `\` and the anti-diagonal `/`.
![Constraints of the n queens problem](n_queens_problem.assets/n_queens_constraints.png){ class="animation-figure" }
![Constraints of the n-queens problem](n_queens_problem.assets/n_queens_constraints.png){ class="animation-figure" }
<p align="center"> Figure 13-16 &nbsp; Constraints of the n queens problem </p>
<p align="center"> Figure 13-16 &nbsp; Constraints of the n-queens problem </p>
### 1. &nbsp; Row-by-row placing strategy
### 1. &nbsp; Row-By-Row Placement Strategy
As the number of queens equals the number of rows on the chessboard, both being $n$, it is easy to conclude that **each row on the chessboard allows and only allows one queen to be placed**.
Since both the number of queens and the number of rows on the chessboard are $n$, we can easily derive a conclusion: **each row of the chessboard allows and only allows exactly one queen to be placed**.
This means that we can adopt a row-by-row placing strategy: starting from the first row, place one queen per row until the last row is reached.
This means we can adopt a row-by-row placement strategy: starting from the first row, place one queen in each row until the last row is completed.
Figure 13-17 shows the row-by-row placing process for the 4 queens problem. Due to space limitations, the figure only expands one search branch of the first row, and prunes any placements that do not meet the column and diagonal constraints.
Figure 13-17 shows the row-by-row placement process for the 4-queens problem. Due to space limitations, the figure only expands one search branch of the first row, and all schemes that do not satisfy the column constraint and diagonal constraints are pruned.
![Row-by-row placing strategy](n_queens_problem.assets/n_queens_placing.png){ class="animation-figure" }
![Row-by-row placement strategy](n_queens_problem.assets/n_queens_placing.png){ class="animation-figure" }
<p align="center"> Figure 13-17 &nbsp; Row-by-row placing strategy </p>
<p align="center"> Figure 13-17 &nbsp; Row-by-row placement strategy </p>
Essentially, **the row-by-row placing strategy serves as a pruning function**, eliminating all search branches that would place multiple queens in the same row.
Essentially, **the row-by-row placement strategy serves a pruning function**, as it avoids all search branches where multiple queens appear in the same row.
### 2. &nbsp; Column and diagonal pruning
### 2. &nbsp; Column and Diagonal Pruning
To satisfy column constraints, we can use a boolean array `cols` of length $n$ to track whether a queen occupies each column. Before each placement decision, `cols` is used to prune the columns that already have queens, and it is dynamically updated during backtracking.
To satisfy the column constraint, we can use a boolean array `cols` of length $n$ to record whether each column has a queen. Before each placement decision, we use `cols` to prune columns that already have queens, and dynamically update the state of `cols` during backtracking.
!!! tip
Note that the origin of the matrix is located in the upper left corner, where the row index increases from top to bottom, and the column index increases from left to right.
Please note that the origin of the matrix is located in the upper-left corner, where the row index increases from top to bottom, and the column index increases from left to right.
How about the diagonal constraints? Let the row and column indices of a certain cell on the chessboard be $(row, col)$. By selecting a specific main diagonal, we notice that the difference $row - col$ is the same for all cells on that diagonal, **meaning that $row - col$ is a constant value on the main diagonal**.
So how do we handle diagonal constraints? Consider a square on the chessboard with row and column indices $(row, col)$. If we select a specific main diagonal in the matrix, we find that all squares on that diagonal have the same difference between their row and column indices, **meaning that $row - col$ is a constant value for all squares on the main diagonal**.
In other words, if two cells satisfy $row_1 - col_1 = row_2 - col_2$, they are definitely on the same main diagonal. Using this pattern, we can utilize the array `diags1` shown in Figure 13-18 to track whether a queen is on any main diagonal.
In other words, if two squares satisfy $row_1 - col_1 = row_2 - col_2$, they must be on the same main diagonal. Using this pattern, we can use the array `diags1` shown in Figure 13-18 to record whether there is a queen on each main diagonal.
Similarly, **the sum of $row + col$ is a constant value for all cells on the secondary diagonal**. We can also use the array `diags2` to handle secondary diagonal constraints.
Similarly, **for all squares on an anti-diagonal, the sum $row + col$ is a constant value**. We can likewise use the array `diags2` to handle anti-diagonal constraints.
![Handling column and diagonal constraints](n_queens_problem.assets/n_queens_cols_diagonals.png){ class="animation-figure" }
<p align="center"> Figure 13-18 &nbsp; Handling column and diagonal constraints </p>
### 3. &nbsp; Code implementation
### 3. &nbsp; Code Implementation
Please note, in an $n$-dimensional square matrix, the range of $row - col$ is $[-n + 1, n - 1]$, and the range of $row + col$ is $[0, 2n - 2]$. Consequently, the number of both main and secondary diagonals is $2n - 1$, meaning the length of the arrays `diags1` and `diags2` is $2n - 1$.
Please note that in an $n$-dimensional square matrix, the range of $row - col$ is $[-n + 1, n - 1]$, and the range of $row + col$ is $[0, 2n - 2]$. Therefore, the number of both main diagonals and anti-diagonals is $2n - 1$, meaning the length of both arrays `diags1` and `diags2` is $2n - 1$.
=== "Python"
@@ -68,34 +68,34 @@ Please note, in an $n$-dimensional square matrix, the range of $row - col$ is $[
diags1: list[bool],
diags2: list[bool],
):
"""Backtracking algorithm: n queens"""
"""Backtracking algorithm: N queens"""
# When all rows are placed, record the solution
if row == n:
res.append([list(row) for row in state])
return
# Traverse all columns
for col in range(n):
# Calculate the main and minor diagonals corresponding to the cell
# Calculate the main diagonal and anti-diagonal corresponding to this cell
diag1 = row - col + n - 1
diag2 = row + col
# Pruning: do not allow queens on the column, main diagonal, or minor diagonal of the cell
# Pruning: do not allow queens to exist in the column, main diagonal, and anti-diagonal of this cell
if not cols[col] and not diags1[diag1] and not diags2[diag2]:
# Attempt: place the queen in the cell
# Attempt: place the queen in this cell
state[row][col] = "Q"
cols[col] = diags1[diag1] = diags2[diag2] = True
# Place the next row
backtrack(row + 1, n, state, res, cols, diags1, diags2)
# Retract: restore the cell to an empty spot
# Backtrack: restore this cell to an empty cell
state[row][col] = "#"
cols[col] = diags1[diag1] = diags2[diag2] = False
def n_queens(n: int) -> list[list[list[str]]]:
"""Solve n queens"""
# Initialize an n*n size chessboard, where 'Q' represents the queen and '#' represents an empty spot
"""Solve N queens"""
# Initialize an n*n chessboard, where 'Q' represents a queen and '#' represents an empty cell
state = [["#" for _ in range(n)] for _ in range(n)]
cols = [False] * n # Record columns with queens
diags1 = [False] * (2 * n - 1) # Record main diagonals with queens
diags2 = [False] * (2 * n - 1) # Record minor diagonals with queens
cols = [False] * n # Record whether there is a queen in the column
diags1 = [False] * (2 * n - 1) # Record whether there is a queen on the main diagonal
diags2 = [False] * (2 * n - 1) # Record whether there is a queen on the anti-diagonal
res = []
backtrack(0, n, state, res, cols, diags1, diags2)
@@ -105,7 +105,7 @@ Please note, in an $n$-dimensional square matrix, the range of $row - col$ is $[
=== "C++"
```cpp title="n_queens.cpp"
/* Backtracking algorithm: n queens */
/* Backtracking algorithm: N queens */
void backtrack(int row, int n, vector<vector<string>> &state, vector<vector<vector<string>>> &res, vector<bool> &cols,
vector<bool> &diags1, vector<bool> &diags2) {
// When all rows are placed, record the solution
@@ -115,30 +115,30 @@ Please note, in an $n$-dimensional square matrix, the range of $row - col$ is $[
}
// Traverse all columns
for (int col = 0; col < n; col++) {
// Calculate the main and minor diagonals corresponding to the cell
// Calculate the main diagonal and anti-diagonal corresponding to this cell
int diag1 = row - col + n - 1;
int diag2 = row + col;
// Pruning: do not allow queens on the column, main diagonal, or minor diagonal of the cell
// Pruning: do not allow queens to exist in the column, main diagonal, and anti-diagonal of this cell
if (!cols[col] && !diags1[diag1] && !diags2[diag2]) {
// Attempt: place the queen in the cell
// Attempt: place the queen in this cell
state[row][col] = "Q";
cols[col] = diags1[diag1] = diags2[diag2] = true;
// Place the next row
backtrack(row + 1, n, state, res, cols, diags1, diags2);
// Retract: restore the cell to an empty spot
// Backtrack: restore this cell to an empty cell
state[row][col] = "#";
cols[col] = diags1[diag1] = diags2[diag2] = false;
}
}
}
/* Solve n queens */
/* Solve N queens */
vector<vector<vector<string>>> nQueens(int n) {
// Initialize an n*n size chessboard, where 'Q' represents the queen and '#' represents an empty spot
// Initialize an n*n chessboard, where 'Q' represents a queen and '#' represents an empty cell
vector<vector<string>> state(n, vector<string>(n, "#"));
vector<bool> cols(n, false); // Record columns with queens
vector<bool> diags1(2 * n - 1, false); // Record main diagonals with queens
vector<bool> diags2(2 * n - 1, false); // Record minor diagonals with queens
vector<bool> cols(n, false); // Record whether there is a queen in the column
vector<bool> diags1(2 * n - 1, false); // Record whether there is a queen on the main diagonal
vector<bool> diags2(2 * n - 1, false); // Record whether there is a queen on the anti-diagonal
vector<vector<vector<string>>> res;
backtrack(0, n, state, res, cols, diags1, diags2);
@@ -150,7 +150,7 @@ Please note, in an $n$-dimensional square matrix, the range of $row - col$ is $[
=== "Java"
```java title="n_queens.java"
/* Backtracking algorithm: n queens */
/* Backtracking algorithm: N queens */
void backtrack(int row, int n, List<List<String>> state, List<List<List<String>>> res,
boolean[] cols, boolean[] diags1, boolean[] diags2) {
// When all rows are placed, record the solution
@@ -164,26 +164,26 @@ Please note, in an $n$-dimensional square matrix, the range of $row - col$ is $[
}
// Traverse all columns
for (int col = 0; col < n; col++) {
// Calculate the main and minor diagonals corresponding to the cell
// Calculate the main diagonal and anti-diagonal corresponding to this cell
int diag1 = row - col + n - 1;
int diag2 = row + col;
// Pruning: do not allow queens on the column, main diagonal, or minor diagonal of the cell
// Pruning: do not allow queens to exist in the column, main diagonal, and anti-diagonal of this cell
if (!cols[col] && !diags1[diag1] && !diags2[diag2]) {
// Attempt: place the queen in the cell
// Attempt: place the queen in this cell
state.get(row).set(col, "Q");
cols[col] = diags1[diag1] = diags2[diag2] = true;
// Place the next row
backtrack(row + 1, n, state, res, cols, diags1, diags2);
// Retract: restore the cell to an empty spot
// Backtrack: restore this cell to an empty cell
state.get(row).set(col, "#");
cols[col] = diags1[diag1] = diags2[diag2] = false;
}
}
}
/* Solve n queens */
/* Solve N queens */
List<List<List<String>>> nQueens(int n) {
// Initialize an n*n size chessboard, where 'Q' represents the queen and '#' represents an empty spot
// Initialize an n*n chessboard, where 'Q' represents a queen and '#' represents an empty cell
List<List<String>> state = new ArrayList<>();
for (int i = 0; i < n; i++) {
List<String> row = new ArrayList<>();
@@ -192,9 +192,9 @@ Please note, in an $n$-dimensional square matrix, the range of $row - col$ is $[
}
state.add(row);
}
boolean[] cols = new boolean[n]; // Record columns with queens
boolean[] diags1 = new boolean[2 * n - 1]; // Record main diagonals with queens
boolean[] diags2 = new boolean[2 * n - 1]; // Record minor diagonals with queens
boolean[] cols = new boolean[n]; // Record whether there is a queen in the column
boolean[] diags1 = new boolean[2 * n - 1]; // Record whether there is a queen on the main diagonal
boolean[] diags2 = new boolean[2 * n - 1]; // Record whether there is a queen on the anti-diagonal
List<List<List<String>>> res = new ArrayList<>();
backtrack(0, n, state, res, cols, diags1, diags2);
@@ -206,91 +206,544 @@ Please note, in an $n$-dimensional square matrix, the range of $row - col$ is $[
=== "C#"
```csharp title="n_queens.cs"
[class]{n_queens}-[func]{Backtrack}
/* Backtracking algorithm: N queens */
void Backtrack(int row, int n, List<List<string>> state, List<List<List<string>>> res,
bool[] cols, bool[] diags1, bool[] diags2) {
// When all rows are placed, record the solution
if (row == n) {
List<List<string>> copyState = [];
foreach (List<string> sRow in state) {
copyState.Add(new List<string>(sRow));
}
res.Add(copyState);
return;
}
// Traverse all columns
for (int col = 0; col < n; col++) {
// Calculate the main diagonal and anti-diagonal corresponding to this cell
int diag1 = row - col + n - 1;
int diag2 = row + col;
// Pruning: do not allow queens to exist in the column, main diagonal, and anti-diagonal of this cell
if (!cols[col] && !diags1[diag1] && !diags2[diag2]) {
// Attempt: place the queen in this cell
state[row][col] = "Q";
cols[col] = diags1[diag1] = diags2[diag2] = true;
// Place the next row
Backtrack(row + 1, n, state, res, cols, diags1, diags2);
// Backtrack: restore this cell to an empty cell
state[row][col] = "#";
cols[col] = diags1[diag1] = diags2[diag2] = false;
}
}
}
[class]{n_queens}-[func]{NQueens}
/* Solve N queens */
List<List<List<string>>> NQueens(int n) {
// Initialize an n*n chessboard, where 'Q' represents a queen and '#' represents an empty cell
List<List<string>> state = [];
for (int i = 0; i < n; i++) {
List<string> row = [];
for (int j = 0; j < n; j++) {
row.Add("#");
}
state.Add(row);
}
bool[] cols = new bool[n]; // Record whether there is a queen in the column
bool[] diags1 = new bool[2 * n - 1]; // Record whether there is a queen on the main diagonal
bool[] diags2 = new bool[2 * n - 1]; // Record whether there is a queen on the anti-diagonal
List<List<List<string>>> res = [];
Backtrack(0, n, state, res, cols, diags1, diags2);
return res;
}
```
=== "Go"
```go title="n_queens.go"
[class]{}-[func]{backtrack}
/* Backtracking algorithm: N queens */
func backtrack(row, n int, state *[][]string, res *[][][]string, cols, diags1, diags2 *[]bool) {
// When all rows are placed, record the solution
if row == n {
newState := make([][]string, len(*state))
for i, _ := range newState {
newState[i] = make([]string, len((*state)[0]))
copy(newState[i], (*state)[i])
[class]{}-[func]{nQueens}
}
*res = append(*res, newState)
return
}
// Traverse all columns
for col := 0; col < n; col++ {
// Calculate the main diagonal and anti-diagonal corresponding to this cell
diag1 := row - col + n - 1
diag2 := row + col
// Pruning: do not allow queens to exist in the column, main diagonal, and anti-diagonal of this cell
if !(*cols)[col] && !(*diags1)[diag1] && !(*diags2)[diag2] {
// Attempt: place the queen in this cell
(*state)[row][col] = "Q"
(*cols)[col], (*diags1)[diag1], (*diags2)[diag2] = true, true, true
// Place the next row
backtrack(row+1, n, state, res, cols, diags1, diags2)
// Backtrack: restore this cell to an empty cell
(*state)[row][col] = "#"
(*cols)[col], (*diags1)[diag1], (*diags2)[diag2] = false, false, false
}
}
}
/* Solve N queens */
func nQueens(n int) [][][]string {
// Initialize an n*n chessboard, where 'Q' represents a queen and '#' represents an empty cell
state := make([][]string, n)
for i := 0; i < n; i++ {
row := make([]string, n)
for i := 0; i < n; i++ {
row[i] = "#"
}
state[i] = row
}
// Record whether there is a queen in the column
cols := make([]bool, n)
diags1 := make([]bool, 2*n-1)
diags2 := make([]bool, 2*n-1)
res := make([][][]string, 0)
backtrack(0, n, &state, &res, &cols, &diags1, &diags2)
return res
}
```
=== "Swift"
```swift title="n_queens.swift"
[class]{}-[func]{backtrack}
/* Backtracking algorithm: N queens */
func backtrack(row: Int, n: Int, state: inout [[String]], res: inout [[[String]]], cols: inout [Bool], diags1: inout [Bool], diags2: inout [Bool]) {
// When all rows are placed, record the solution
if row == n {
res.append(state)
return
}
// Traverse all columns
for col in 0 ..< n {
// Calculate the main diagonal and anti-diagonal corresponding to this cell
let diag1 = row - col + n - 1
let diag2 = row + col
// Pruning: do not allow queens to exist in the column, main diagonal, and anti-diagonal of this cell
if !cols[col] && !diags1[diag1] && !diags2[diag2] {
// Attempt: place the queen in this cell
state[row][col] = "Q"
cols[col] = true
diags1[diag1] = true
diags2[diag2] = true
// Place the next row
backtrack(row: row + 1, n: n, state: &state, res: &res, cols: &cols, diags1: &diags1, diags2: &diags2)
// Backtrack: restore this cell to an empty cell
state[row][col] = "#"
cols[col] = false
diags1[diag1] = false
diags2[diag2] = false
}
}
}
[class]{}-[func]{nQueens}
/* Solve N queens */
func nQueens(n: Int) -> [[[String]]] {
// Initialize an n*n chessboard, where 'Q' represents a queen and '#' represents an empty cell
var state = Array(repeating: Array(repeating: "#", count: n), count: n)
var cols = Array(repeating: false, count: n) // Record whether there is a queen in the column
var diags1 = Array(repeating: false, count: 2 * n - 1) // Record whether there is a queen on the main diagonal
var diags2 = Array(repeating: false, count: 2 * n - 1) // Record whether there is a queen on the anti-diagonal
var res: [[[String]]] = []
backtrack(row: 0, n: n, state: &state, res: &res, cols: &cols, diags1: &diags1, diags2: &diags2)
return res
}
```
=== "JS"
```javascript title="n_queens.js"
[class]{}-[func]{backtrack}
/* Backtracking algorithm: N queens */
function backtrack(row, n, state, res, cols, diags1, diags2) {
// When all rows are placed, record the solution
if (row === n) {
res.push(state.map((row) => row.slice()));
return;
}
// Traverse all columns
for (let col = 0; col < n; col++) {
// Calculate the main diagonal and anti-diagonal corresponding to this cell
const diag1 = row - col + n - 1;
const diag2 = row + col;
// Pruning: do not allow queens to exist in the column, main diagonal, and anti-diagonal of this cell
if (!cols[col] && !diags1[diag1] && !diags2[diag2]) {
// Attempt: place the queen in this cell
state[row][col] = 'Q';
cols[col] = diags1[diag1] = diags2[diag2] = true;
// Place the next row
backtrack(row + 1, n, state, res, cols, diags1, diags2);
// Backtrack: restore this cell to an empty cell
state[row][col] = '#';
cols[col] = diags1[diag1] = diags2[diag2] = false;
}
}
}
[class]{}-[func]{nQueens}
/* Solve N queens */
function nQueens(n) {
// Initialize an n*n chessboard, where 'Q' represents a queen and '#' represents an empty cell
const state = Array.from({ length: n }, () => Array(n).fill('#'));
const cols = Array(n).fill(false); // Record whether there is a queen in the column
const diags1 = Array(2 * n - 1).fill(false); // Record whether there is a queen on the main diagonal
const diags2 = Array(2 * n - 1).fill(false); // Record whether there is a queen on the anti-diagonal
const res = [];
backtrack(0, n, state, res, cols, diags1, diags2);
return res;
}
```
=== "TS"
```typescript title="n_queens.ts"
[class]{}-[func]{backtrack}
/* Backtracking algorithm: N queens */
function backtrack(
row: number,
n: number,
state: string[][],
res: string[][][],
cols: boolean[],
diags1: boolean[],
diags2: boolean[]
): void {
// When all rows are placed, record the solution
if (row === n) {
res.push(state.map((row) => row.slice()));
return;
}
// Traverse all columns
for (let col = 0; col < n; col++) {
// Calculate the main diagonal and anti-diagonal corresponding to this cell
const diag1 = row - col + n - 1;
const diag2 = row + col;
// Pruning: do not allow queens to exist in the column, main diagonal, and anti-diagonal of this cell
if (!cols[col] && !diags1[diag1] && !diags2[diag2]) {
// Attempt: place the queen in this cell
state[row][col] = 'Q';
cols[col] = diags1[diag1] = diags2[diag2] = true;
// Place the next row
backtrack(row + 1, n, state, res, cols, diags1, diags2);
// Backtrack: restore this cell to an empty cell
state[row][col] = '#';
cols[col] = diags1[diag1] = diags2[diag2] = false;
}
}
}
[class]{}-[func]{nQueens}
/* Solve N queens */
function nQueens(n: number): string[][][] {
// Initialize an n*n chessboard, where 'Q' represents a queen and '#' represents an empty cell
const state = Array.from({ length: n }, () => Array(n).fill('#'));
const cols = Array(n).fill(false); // Record whether there is a queen in the column
const diags1 = Array(2 * n - 1).fill(false); // Record whether there is a queen on the main diagonal
const diags2 = Array(2 * n - 1).fill(false); // Record whether there is a queen on the anti-diagonal
const res: string[][][] = [];
backtrack(0, n, state, res, cols, diags1, diags2);
return res;
}
```
=== "Dart"
```dart title="n_queens.dart"
[class]{}-[func]{backtrack}
/* Backtracking algorithm: N queens */
void backtrack(
int row,
int n,
List<List<String>> state,
List<List<List<String>>> res,
List<bool> cols,
List<bool> diags1,
List<bool> diags2,
) {
// When all rows are placed, record the solution
if (row == n) {
List<List<String>> copyState = [];
for (List<String> sRow in state) {
copyState.add(List.from(sRow));
}
res.add(copyState);
return;
}
// Traverse all columns
for (int col = 0; col < n; col++) {
// Calculate the main diagonal and anti-diagonal corresponding to this cell
int diag1 = row - col + n - 1;
int diag2 = row + col;
// Pruning: do not allow queens to exist in the column, main diagonal, and anti-diagonal of this cell
if (!cols[col] && !diags1[diag1] && !diags2[diag2]) {
// Attempt: place the queen in this cell
state[row][col] = "Q";
cols[col] = true;
diags1[diag1] = true;
diags2[diag2] = true;
// Place the next row
backtrack(row + 1, n, state, res, cols, diags1, diags2);
// Backtrack: restore this cell to an empty cell
state[row][col] = "#";
cols[col] = false;
diags1[diag1] = false;
diags2[diag2] = false;
}
}
}
[class]{}-[func]{nQueens}
/* Solve N queens */
List<List<List<String>>> nQueens(int n) {
// Initialize an n*n chessboard, where 'Q' represents a queen and '#' represents an empty cell
List<List<String>> state = List.generate(n, (index) => List.filled(n, "#"));
List<bool> cols = List.filled(n, false); // Record whether there is a queen in the column
List<bool> diags1 = List.filled(2 * n - 1, false); // Record whether there is a queen on the main diagonal
List<bool> diags2 = List.filled(2 * n - 1, false); // Record whether there is a queen on the anti-diagonal
List<List<List<String>>> res = [];
backtrack(0, n, state, res, cols, diags1, diags2);
return res;
}
```
=== "Rust"
```rust title="n_queens.rs"
[class]{}-[func]{backtrack}
/* Backtracking algorithm: N queens */
fn backtrack(
row: usize,
n: usize,
state: &mut Vec<Vec<String>>,
res: &mut Vec<Vec<Vec<String>>>,
cols: &mut [bool],
diags1: &mut [bool],
diags2: &mut [bool],
) {
// When all rows are placed, record the solution
if row == n {
res.push(state.clone());
return;
}
// Traverse all columns
for col in 0..n {
// Calculate the main diagonal and anti-diagonal corresponding to this cell
let diag1 = row + n - 1 - col;
let diag2 = row + col;
// Pruning: do not allow queens to exist in the column, main diagonal, and anti-diagonal of this cell
if !cols[col] && !diags1[diag1] && !diags2[diag2] {
// Attempt: place the queen in this cell
state[row][col] = "Q".into();
(cols[col], diags1[diag1], diags2[diag2]) = (true, true, true);
// Place the next row
backtrack(row + 1, n, state, res, cols, diags1, diags2);
// Backtrack: restore this cell to an empty cell
state[row][col] = "#".into();
(cols[col], diags1[diag1], diags2[diag2]) = (false, false, false);
}
}
}
[class]{}-[func]{n_queens}
/* Solve N queens */
fn n_queens(n: usize) -> Vec<Vec<Vec<String>>> {
// Initialize an n*n chessboard, where 'Q' represents a queen and '#' represents an empty cell
let mut state: Vec<Vec<String>> = vec![vec!["#".to_string(); n]; n];
let mut cols = vec![false; n]; // Record whether there is a queen in the column
let mut diags1 = vec![false; 2 * n - 1]; // Record whether there is a queen on the main diagonal
let mut diags2 = vec![false; 2 * n - 1]; // Record whether there is a queen on the anti-diagonal
let mut res: Vec<Vec<Vec<String>>> = Vec::new();
backtrack(
0,
n,
&mut state,
&mut res,
&mut cols,
&mut diags1,
&mut diags2,
);
res
}
```
=== "C"
```c title="n_queens.c"
[class]{}-[func]{backtrack}
/* Backtracking algorithm: N queens */
void backtrack(int row, int n, char state[MAX_SIZE][MAX_SIZE], char ***res, int *resSize, bool cols[MAX_SIZE],
bool diags1[2 * MAX_SIZE - 1], bool diags2[2 * MAX_SIZE - 1]) {
// When all rows are placed, record the solution
if (row == n) {
res[*resSize] = (char **)malloc(sizeof(char *) * n);
for (int i = 0; i < n; ++i) {
res[*resSize][i] = (char *)malloc(sizeof(char) * (n + 1));
strcpy(res[*resSize][i], state[i]);
}
(*resSize)++;
return;
}
// Traverse all columns
for (int col = 0; col < n; col++) {
// Calculate the main diagonal and anti-diagonal corresponding to this cell
int diag1 = row - col + n - 1;
int diag2 = row + col;
// Pruning: do not allow queens to exist in the column, main diagonal, and anti-diagonal of this cell
if (!cols[col] && !diags1[diag1] && !diags2[diag2]) {
// Attempt: place the queen in this cell
state[row][col] = 'Q';
cols[col] = diags1[diag1] = diags2[diag2] = true;
// Place the next row
backtrack(row + 1, n, state, res, resSize, cols, diags1, diags2);
// Backtrack: restore this cell to an empty cell
state[row][col] = '#';
cols[col] = diags1[diag1] = diags2[diag2] = false;
}
}
}
[class]{}-[func]{nQueens}
/* Solve N queens */
char ***nQueens(int n, int *returnSize) {
char state[MAX_SIZE][MAX_SIZE];
// Initialize an n*n chessboard, where 'Q' represents a queen and '#' represents an empty cell
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
state[i][j] = '#';
}
state[i][n] = '\0';
}
bool cols[MAX_SIZE] = {false}; // Record whether there is a queen in the column
bool diags1[2 * MAX_SIZE - 1] = {false}; // Record whether there is a queen on the main diagonal
bool diags2[2 * MAX_SIZE - 1] = {false}; // Record whether there is a queen on the anti-diagonal
char ***res = (char ***)malloc(sizeof(char **) * MAX_SIZE);
*returnSize = 0;
backtrack(0, n, state, res, returnSize, cols, diags1, diags2);
return res;
}
```
=== "Kotlin"
```kotlin title="n_queens.kt"
[class]{}-[func]{backtrack}
/* Backtracking algorithm: N queens */
fun backtrack(
row: Int,
n: Int,
state: MutableList<MutableList<String>>,
res: MutableList<MutableList<MutableList<String>>?>,
cols: BooleanArray,
diags1: BooleanArray,
diags2: BooleanArray
) {
// When all rows are placed, record the solution
if (row == n) {
val copyState = mutableListOf<MutableList<String>>()
for (sRow in state) {
copyState.add(sRow.toMutableList())
}
res.add(copyState)
return
}
// Traverse all columns
for (col in 0..<n) {
// Calculate the main diagonal and anti-diagonal corresponding to this cell
val diag1 = row - col + n - 1
val diag2 = row + col
// Pruning: do not allow queens to exist in the column, main diagonal, and anti-diagonal of this cell
if (!cols[col] && !diags1[diag1] && !diags2[diag2]) {
// Attempt: place the queen in this cell
state[row][col] = "Q"
diags2[diag2] = true
diags1[diag1] = diags2[diag2]
cols[col] = diags1[diag1]
// Place the next row
backtrack(row + 1, n, state, res, cols, diags1, diags2)
// Backtrack: restore this cell to an empty cell
state[row][col] = "#"
diags2[diag2] = false
diags1[diag1] = diags2[diag2]
cols[col] = diags1[diag1]
}
}
}
[class]{}-[func]{nQueens}
/* Solve N queens */
fun nQueens(n: Int): MutableList<MutableList<MutableList<String>>?> {
// Initialize an n*n chessboard, where 'Q' represents a queen and '#' represents an empty cell
val state = mutableListOf<MutableList<String>>()
for (i in 0..<n) {
val row = mutableListOf<String>()
for (j in 0..<n) {
row.add("#")
}
state.add(row)
}
val cols = BooleanArray(n) // Record whether there is a queen in the column
val diags1 = BooleanArray(2 * n - 1) // Record whether there is a queen on the main diagonal
val diags2 = BooleanArray(2 * n - 1) // Record whether there is a queen on the anti-diagonal
val res = mutableListOf<MutableList<MutableList<String>>?>()
backtrack(0, n, state, res, cols, diags1, diags2)
return res
}
```
=== "Ruby"
```ruby title="n_queens.rb"
[class]{}-[func]{backtrack}
### Backtracking: n queens ###
def backtrack(row, n, state, res, cols, diags1, diags2)
# When all rows are placed, record the solution
if row == n
res << state.map { |row| row.dup }
return
end
[class]{}-[func]{n_queens}
# Traverse all columns
for col in 0...n
# Calculate the main diagonal and anti-diagonal corresponding to this cell
diag1 = row - col + n - 1
diag2 = row + col
# Pruning: do not allow queens to exist in the column, main diagonal, and anti-diagonal of this cell
if !cols[col] && !diags1[diag1] && !diags2[diag2]
# Attempt: place the queen in this cell
state[row][col] = "Q"
cols[col] = diags1[diag1] = diags2[diag2] = true
# Place the next row
backtrack(row + 1, n, state, res, cols, diags1, diags2)
# Backtrack: restore this cell to an empty cell
state[row][col] = "#"
cols[col] = diags1[diag1] = diags2[diag2] = false
end
end
end
### Solve n queens ###
def n_queens(n)
# Initialize an n*n chessboard, where 'Q' represents a queen and '#' represents an empty cell
state = Array.new(n) { Array.new(n, "#") }
cols = Array.new(n, false) # Record whether there is a queen in the column
diags1 = Array.new(2 * n - 1, false) # Record whether there is a queen on the main diagonal
diags2 = Array.new(2 * n - 1, false) # Record whether there is a queen on the anti-diagonal
res = []
backtrack(0, n, state, res, cols, diags1, diags2)
res
end
```
=== "Zig"
Placing $n$ queens row by row, considering the column constraint, from the first row to the last row there are $n$, $n-1$, $\dots$, $2$, $1$ choices, using $O(n!)$ time. When recording a solution, it is necessary to copy the matrix `state` and add it to `res`, and the copy operation uses $O(n^2)$ time. Therefore, **the overall time complexity is $O(n! \cdot n^2)$**. In practice, pruning based on diagonal constraints can also significantly reduce the search space, so the search efficiency is often better than the time complexity mentioned above.
```zig title="n_queens.zig"
[class]{}-[func]{backtrack}
[class]{}-[func]{nQueens}
```
Placing $n$ queens row-by-row, considering column constraints, from the first row to the last row, there are $n$, $n-1$, $\dots$, $2$, $1$ choices, using $O(n!)$ time. When recording a solution, it is necessary to copy the matrix `state` and add it to `res`, with the copying operation using $O(n^2)$ time. Therefore, **the overall time complexity is $O(n! \cdot n^2)$**. In practice, pruning based on diagonal constraints can significantly reduce the search space, thus often the search efficiency is better than the aforementioned time complexity.
Array `state` uses $O(n^2)$ space, and arrays `cols`, `diags1`, and `diags2` each use $O(n)$ space as well. The maximum recursion depth is $n$, using $O(n)$ stack frame space. Therefore, **the space complexity is $O(n^2)$**.
The array `state` uses $O(n^2)$ space, and the arrays `cols`, `diags1`, and `diags2` each use $O(n)$ space. The maximum recursion depth is $n$, using $O(n)$ stack frame space. Therefore, **the space complexity is $O(n^2)$**.
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# 13.5 &nbsp; Summary
### 1. &nbsp; Key review
### 1. &nbsp; Key Review
- The essence of the backtracking algorithm is exhaustive search. It seeks solutions that meet the conditions by performing a depth-first traversal of the solution space. During the search, if a satisfying solution is found, it is recorded, until all solutions are found or the traversal is completed.
- The search process of the backtracking algorithm includes trying and backtracking. It uses depth-first search to explore various choices, and when a choice does not meet the constraints, the previous choice is undone. Then it reverts to the previous state and continues to try other options. Trying and backtracking are operations in opposite directions.
- Backtracking problems usually contain multiple constraints. These constraints can be used to perform pruning operations. Pruning can terminate unnecessary search branches in advance, greatly enhancing search efficiency.
- The backtracking algorithm is mainly used to solve search problems and constraint satisfaction problems. Although combinatorial optimization problems can be solved using backtracking, there are often more efficient or effective solutions available.
- The permutation problem aims to search for all possible permutations of the elements in a given set. We use an array to record whether each element has been chosen, avoiding repeated selection of the same element. This ensures that each element is chosen only once.
- In permutation problems, if the set contains duplicate elements, the final result will include duplicate permutations. We need to restrict that identical elements can only be selected once in each round, which is usually implemented using a hash set.
- The subset-sum problem aims to find all subsets in a given set that sum to a target value. The set does not distinguish the order of elements, but the search process may generate duplicate subsets. This occurs because the algorithm explores different element orders as unique paths. Before backtracking, we sort the data and set a variable to indicate the starting point of the traversal for each round. This allows us to prune the search branches that generate duplicate subsets.
- For the subset-sum problem, equal elements in the array can produce duplicate sets. Using the precondition that the array is already sorted, we prune by determining if adjacent elements are equal. This ensures that equal elements are only selected once per round.
- The $n$ queens problem aims to find schemes to place $n$ queens on an $n \times n$ chessboard such that no two queens can attack each other. The constraints of the problem include row constraints, column constraints, and constraints on the main and secondary diagonals. To meet the row constraint, we adopt a strategy of placing one queen per row, ensuring each row has one queen placed.
- The handling of column constraints and diagonal constraints is similar. For column constraints, we use an array to record whether there is a queen in each column, thereby indicating whether the selected cell is legal. For diagonal constraints, we use two arrays to respectively record the presence of queens on the main and secondary diagonals. The challenge is to determine the relationship between row and column indices for cells on the same main or secondary diagonal.
- The backtracking algorithm is fundamentally an exhaustive search method. It finds solutions that meet specified conditions by performing a depth-first traversal of the solution space. During the search process, when a solution satisfying the conditions is found, it is recorded. The search ends either after finding all solutions or when the traversal is complete.
- The backtracking algorithm search process consists of two parts: attempting and backtracking. It tries various choices through depth-first search. When encountering situations that violate constraints, it reverts the previous choice, returns to the previous state, and continues exploring other options. Attempting and backtracking are operations in opposite directions.
- Backtracking problems typically contain multiple constraints, which can be utilized to implement pruning operations. Pruning can terminate unnecessary search branches early, significantly improving search efficiency.
- The backtracking algorithm is primarily used to solve search problems and constraint satisfaction problems. While combinatorial optimization problems can be solved with backtracking, there are often more efficient or better-performing solutions available.
- The permutation problem aims to find all possible permutations of elements in a given set. We use an array to record whether each element has been selected, thereby pruning search branches that attempt to select the same element repeatedly, ensuring each element is selected exactly once.
- In the permutation problem, if the set contains duplicate elements, the final result will contain duplicate permutations. We need to impose a constraint so that equal elements can only be selected once per round, which is typically achieved using a hash set.
- The subset-sum problem aims to find all subsets of a given set that sum to a target value. Since the set is unordered but the search process outputs results in all orders, duplicate subsets are generated. We sort the data before backtracking and use a variable to indicate the starting point of each round's traversal, thereby pruning search branches that generate duplicate subsets.
- For the subset-sum problem, equal elements in the array produce duplicate sets. We leverage the precondition that the array is sorted by checking whether adjacent elements are equal to implement pruning, ensuring that equal elements can only be selected once per round.
- The $n$ queens problem aims to find placements of $n$ queens on an $n \times n$ chessboard such that no two queens can attack each other. The constraints of this problem include row constraints, column constraints, and main and anti-diagonal constraints. To satisfy row constraints, we adopt a row-by-row placement strategy, ensuring exactly one queen is placed in each row.
- The handling of column constraints and diagonal constraints is similar. For column constraints, we use an array to record whether each column has a queen, thereby indicating whether a selected cell is valid. For diagonal constraints, we use two arrays to separately record whether queens exist on each main or anti-diagonal. The challenge lies in finding the row-column index pattern that characterizes cells on the same main (anti-)diagonal.
### 2. &nbsp; Q & A
**Q**: How can we understand the relationship between backtracking and recursion?
**Q**: How should we understand the relationship between backtracking and recursion?
Overall, backtracking is an "algorithmic strategy," while recursion is more of a "tool."
Overall, backtracking is an "algorithm strategy", while recursion is more like a "tool".
- Backtracking algorithms are typically based on recursion. However, backtracking is one of the application scenarios of recursion, specifically in search problems.
- The structure of recursion reflects the problem-solving paradigm of "sub-problem decomposition." It is commonly used in solving problems involving divide and conquer, backtracking, and dynamic programming (memoized recursion).
- The backtracking algorithm is typically implemented based on recursion. However, backtracking is one application scenario of recursion and represents the application of recursion in search problems.
- The structure of recursion embodies the "subproblem decomposition" problem-solving paradigm, commonly used to solve problems involving divide-and-conquer, backtracking, and dynamic programming (memoized recursion).
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icon: material/timer-sand
---
# Chapter 2. &nbsp; Complexity analysis
# Chapter 2. &nbsp; Complexity Analysis
![Complexity analysis](../assets/covers/chapter_complexity_analysis.jpg){ class="cover-image" }
!!! abstract
Complexity analysis is like a space-time navigator in the vast universe of algorithms.
Complexity analysis is like a space-time guide in the vast universe of algorithms.
It guides us in exploring deeper within the dimensions of time and space, seeking more elegant solutions.
It leads us to explore deeply within the two dimensions of time and space, seeking more elegant solutions.
## Chapter contents
- [2.1 &nbsp; Algorithm efficiency assessment](performance_evaluation.md)
- [2.2 &nbsp; Iteration and recursion](iteration_and_recursion.md)
- [2.3 &nbsp; Time complexity](time_complexity.md)
- [2.4 &nbsp; Space complexity](space_complexity.md)
- [2.1 &nbsp; Algorithm Efficiency Evaluation](performance_evaluation.md)
- [2.2 &nbsp; Iteration and Recursion](iteration_and_recursion.md)
- [2.3 &nbsp; Time Complexity](time_complexity.md)
- [2.4 &nbsp; Space Complexity](space_complexity.md)
- [2.5 &nbsp; Summary](summary.md)
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---
# 2.1 &nbsp; Algorithm efficiency assessment
# 2.1 &nbsp; Algorithm Efficiency Evaluation
In algorithm design, we pursue the following two objectives in sequence.
In algorithm design, we pursue the following two levels of objectives sequentially.
1. **Finding a Solution to the Problem**: The algorithm should reliably find the correct solution within the specified range of inputs.
2. **Seeking the Optimal Solution**: For the same problem, multiple solutions might exist, and we aim to find the most efficient algorithm possible.
1. **Finding a solution to the problem**: The algorithm must reliably obtain the correct solution within the specified input range.
2. **Seeking the optimal solution**: Multiple solutions may exist for the same problem, and we hope to find an algorithm that is as efficient as possible.
In other words, under the premise of being able to solve the problem, algorithm efficiency has become the main criterion for evaluating an algorithm, which includes the following two dimensions.
In other words, under the premise of being able to solve the problem, algorithm efficiency has become the primary evaluation criterion for measuring the quality of algorithms. It includes the following two dimensions.
- **Time efficiency**: The speed at which an algorithm runs.
- **Space efficiency**: The size of the memory space occupied by an algorithm.
- **Time efficiency**: The length of time the algorithm runs.
- **Space efficiency**: The size of memory space the algorithm occupies.
In short, **our goal is to design data structures and algorithms that are both fast and memory-efficient**. Effectively assessing algorithm efficiency is crucial because only then can we compare various algorithms and guide the process of algorithm design and optimization.
In short, **our goal is to design data structures and algorithms that are "both fast and memory-efficient"**. Effectively evaluating algorithm efficiency is crucial, because only in this way can we compare various algorithms and guide the algorithm design and optimization process.
There are mainly two methods of efficiency assessment: actual testing and theoretical estimation.
Efficiency evaluation methods are mainly divided into two types: actual testing and theoretical estimation.
## 2.1.1 &nbsp; Actual testing
## 2.1.1 &nbsp; Actual Testing
Suppose we have algorithms `A` and `B`, both capable of solving the same problem, and we need to compare their efficiencies. The most direct method is to use a computer to run these two algorithms, monitor and record their runtime and memory usage. This assessment method reflects the actual situation, but it has significant limitations.
Suppose we now have algorithm `A` and algorithm `B`, both of which can solve the same problem, and we need to compare the efficiency of these two algorithms. The most direct method is to find a computer, run these two algorithms, and monitor and record their running time and memory usage. This evaluation approach can reflect the real situation, but it also has considerable limitations.
On one hand, **it's difficult to eliminate interference from the testing environment**. Hardware configurations can affect algorithm performance. For example, an algorithm with a high degree of parallelism is better suited for running on multi-core CPUs, while an algorithm that involves intensive memory operations performs better with high-performance memory. The test results of an algorithm may vary across different machines. This means testing across multiple machines to calculate average efficiency becomes impractical.
On one hand, **it is difficult to eliminate interference factors from the testing environment**. Hardware configuration affects the performance of algorithms. For example, if an algorithm has a high degree of parallelism, it is more suitable for running on multi-core CPUs; if an algorithm has intensive memory operations, it will perform better on high-performance memory. In other words, the test results of an algorithm on different machines may be inconsistent. This means we need to test on various machines and calculate average efficiency, which is impractical.
On the other hand, **conducting a full test is very resource-intensive**. Algorithm efficiency varies with input data size. For example, with smaller data volumes, algorithm `A` might run faster than `B`, but with larger data volumes, the test results may be the opposite. Therefore, to draw convincing conclusions, we need to test a wide range of input data sizes, which requires excessive computational resources.
On the other hand, **conducting complete testing is very resource-intensive**. As the input data volume changes, the algorithm will exhibit different efficiencies. For example, when the input data volume is small, the running time of algorithm `A` is shorter than algorithm `B`; but when the input data volume is large, the test results may be exactly the opposite. Therefore, to obtain convincing conclusions, we need to test input data of various scales, which requires a large amount of computational resources.
## 2.1.2 &nbsp; Theoretical estimation
## 2.1.2 &nbsp; Theoretical Estimation
Due to the significant limitations of actual testing, we can consider evaluating algorithm efficiency solely through calculations. This estimation method is known as <u>asymptotic complexity analysis</u>, or simply <u>complexity analysis</u>.
Since actual testing has considerable limitations, we can consider evaluating algorithm efficiency through calculations alone. This estimation method is called <u>asymptotic complexity analysis</u>, or <u>complexity analysis</u> for short.
Complexity analysis reflects the relationship between the time and space resources required for algorithm execution and the size of the input data. **It describes the trend of growth in the time and space required by the algorithm as the size of the input data increases**. This definition might sound complex, but we can break it down into three key points to understand it better.
Complexity analysis can reflect the relationship between the time and space resources required for algorithm execution and the input data scale. **It describes the growth trend of the time and space required for algorithm execution as the input data scale increases**. This definition is somewhat convoluted, so we can break it down into three key points to understand.
- "Time and space resources" correspond to <u>time complexity</u> and <u>space complexity</u>, respectively.
- "As the size of input data increases" means that complexity reflects the relationship between algorithm efficiency and the volume of input data.
- "The trend of growth in time and space" indicates that complexity analysis focuses not on the specific values of runtime or space occupied, but on the "rate" at which time or space increases.
- "As the input data scale increases" means that complexity reflects the relationship between algorithm running efficiency and input data scale.
- "Growth trend of time and space" indicates that complexity analysis focuses not on the specific values of running time or occupied space, but on how "fast" time or space grows.
**Complexity analysis overcomes the disadvantages of actual testing methods**, reflected in the following aspects:
**Complexity analysis overcomes the drawbacks of the actual testing method**, reflected in the following aspects.
- It does not require actually running the code, making it more environmentally friendly and energy efficient.
- It is independent of the testing environment and applicable to all operating platforms.
- It can reflect algorithm efficiency under different data volumes, especially in the performance of algorithms with large data volumes.
- It does not need to actually run the code, making it more environmentally friendly and energy-efficient.
- It is independent of the testing environment, and the analysis results are applicable to all running platforms.
- It can reflect algorithm efficiency at different data volumes, especially algorithm performance at large data volumes.
!!! tip
If you're still confused about the concept of complexity, don't worry. We will cover it in detail in subsequent chapters.
If you are still confused about the concept of complexity, don't worry—we will introduce it in detail in subsequent chapters.
Complexity analysis provides us with a "ruler" to evaluate the efficiency of an algorithm, enabling us to measure the time and space resources required to execute it and compare the efficiency of different algorithms.
Complexity analysis provides us with a "ruler" for evaluating algorithm efficiency, allowing us to measure the time and space resources required to execute a certain algorithm and compare the efficiency between different algorithms.
Complexity is a mathematical concept that might be abstract and challenging for beginners. From this perspective, complexity analysis might not be the most suitable topic to introduce first. However, when discussing the characteristics of a particular data structure or algorithm, it's hard to avoid analyzing its speed and space usage.
Complexity is a mathematical concept that may be relatively abstract for beginners, with a relatively high learning difficulty. From this perspective, complexity analysis may not be very suitable as the first content to be introduced. However, when we discuss the characteristics of a certain data structure or algorithm, it is difficult to avoid analyzing its running speed and space usage.
In summary, it is recommended to develop a basic understanding of complexity analysis before diving deep into data structures and algorithms, **so that you can perform complexity analysis on simple algorithms**.
In summary, it is recommended that before diving deep into data structures and algorithms, **you first establish a preliminary understanding of complexity analysis so that you can complete complexity analysis of simple algorithms**.
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# 2.5 &nbsp; Summary
### 1. &nbsp; Key review
### 1. &nbsp; Key Review
**Algorithm Efficiency Assessment**
- Time efficiency and space efficiency are the two main criteria for assessing the merits of an algorithm.
- We can assess algorithm efficiency through actual testing, but it's challenging to eliminate the influence of the test environment, and it consumes substantial computational resources.
- Complexity analysis can overcome the disadvantages of actual testing. Its results are applicable across all operating platforms and can reveal the efficiency of algorithms at different data scales.
- Time efficiency and space efficiency are the two primary evaluation metrics for measuring algorithm performance.
- We can evaluate algorithm efficiency through actual testing, but it is difficult to eliminate the influence of the testing environment, and it consumes substantial computational resources.
- Complexity analysis can eliminate the drawbacks of actual testing, with results applicable to all running platforms, and it can reveal algorithm efficiency under different data scales.
**Time Complexity**
- Time complexity measures the trend of an algorithm's running time with the increase in data volume, effectively assessing algorithm efficiency. However, it can fail in certain cases, such as with small input data volumes or when time complexities are the same, making it challenging to precisely compare the efficiency of algorithms.
- Worst-case time complexity is denoted using big-$O$ notation, representing the asymptotic upper bound, reflecting the growth level of the number of operations $T(n)$ as $n$ approaches infinity.
- Calculating time complexity involves two steps: first counting the number of operations, then determining the asymptotic upper bound.
- Common time complexities, arranged from low to high, include $O(1)$, $O(\log n)$, $O(n)$, $O(n \log n)$, $O(n^2)$, $O(2^n)$, and $O(n!)$, among others.
- The time complexity of some algorithms is not fixed and depends on the distribution of input data. Time complexities are divided into worst, best, and average cases. The best case is rarely used because input data generally needs to meet strict conditions to achieve the best case.
- Average time complexity reflects the efficiency of an algorithm under random data inputs, closely resembling the algorithm's performance in actual applications. Calculating average time complexity requires accounting for the distribution of input data and the subsequent mathematical expectation.
- Time complexity is used to measure the trend of algorithm runtime as data volume increases. It can effectively evaluate algorithm efficiency, but may fail in certain situations, such as when the input data volume is small or when time complexities are identical, making it impossible to precisely compare algorithm efficiency.
- Worst-case time complexity is represented using Big $O$ notation, corresponding to the asymptotic upper bound of a function, reflecting the growth level of the number of operations $T(n)$ as $n$ approaches positive infinity.
- Deriving time complexity involves two steps: first, counting the number of operations, then determining the asymptotic upper bound.
- Common time complexities arranged from low to high include $O(1)$, $O(\log n)$, $O(n)$, $O(n \log n)$, $O(n^2)$, $O(2^n)$, and $O(n!)$.
- The time complexity of some algorithms is not fixed, but rather depends on the distribution of input data. Time complexity is divided into worst-case, best-case, and average-case time complexity. Best-case time complexity is rarely used because input data generally needs to satisfy strict conditions to achieve the best case.
- Average time complexity reflects the algorithm's runtime efficiency under random data input, and is closest to the algorithm's performance in practical applications. Calculating average time complexity requires statistical analysis of input data distribution and the combined mathematical expectation.
**Space Complexity**
- Space complexity, similar to time complexity, measures the trend of memory space occupied by an algorithm with the increase in data volume.
- The relevant memory space used during the algorithm's execution can be divided into input space, temporary space, and output space. Generally, input space is not included in space complexity calculations. Temporary space can be divided into temporary data, stack frame space, and instruction space, where stack frame space usually affects space complexity only in recursive functions.
- We usually focus only on the worst-case space complexity, which means calculating the space complexity of the algorithm under the worst input data and at the worst moment of operation.
- Common space complexities, arranged from low to high, include $O(1)$, $O(\log n)$, $O(n)$, $O(n^2)$, and $O(2^n)$, among others.
- Space complexity serves a similar purpose to time complexity, used to measure the trend of algorithm memory usage as data volume increases.
- The memory space related to algorithm execution can be divided into input space, temporary space, and output space. Typically, input space is not included in space complexity calculations. Temporary space can be divided into temporary data, stack frame space, and instruction space, where stack frame space usually affects space complexity only in recursive functions.
- We typically only focus on worst-case space complexity, which is the space complexity of an algorithm under worst-case input data and worst-case runtime.
- Common space complexities arranged from low to high include $O(1)$, $O(\log n)$, $O(n)$, $O(n^2)$, and $O(2^n)$.
### 2. &nbsp; Q & A
**Q**: Is the space complexity of tail recursion $O(1)$?
Theoretically, the space complexity of a tail-recursive function can be optimized to $O(1)$. However, most programming languages (such as Java, Python, C++, Go, C#) do not support automatic optimization of tail recursion, so it's generally considered to have a space complexity of $O(n)$.
Theoretically, the space complexity of tail recursive functions can be optimized to $O(1)$. However, most programming languages (such as Java, Python, C++, Go, C#, etc.) do not support automatic tail recursion optimization, so the space complexity is generally considered to be $O(n)$.
**Q**: What is the difference between the terms "function" and "method"?
**Q**: What is the difference between the terms function and method?
A <u>function</u> can be executed independently, with all parameters passed explicitly. A <u>method</u> is associated with an object and is implicitly passed to the object calling it, able to operate on the data contained within an instance of a class.
A <u>function</u> can be executed independently, with all parameters passed explicitly. A <u>method</u> is associated with an object, is implicitly passed to the object that invokes it, and can operate on data contained in class instances.
Here are some examples from common programming languages:
The following examples use several common programming languages for illustration.
- C is a procedural programming language without object-oriented concepts, so it only has functions. However, we can simulate object-oriented programming by creating structures (struct), and functions associated with these structures are equivalent to methods in other programming languages.
- C is a procedural programming language without object-oriented concepts, so it only has functions. However, we can simulate object-oriented programming by creating structures (struct), and functions associated with structures are equivalent to methods in other programming languages.
- Java and C# are object-oriented programming languages where code blocks (methods) are typically part of a class. Static methods behave like functions because they are bound to the class and cannot access specific instance variables.
- C++ and Python support both procedural programming (functions) and object-oriented programming (methods).
**Q**: Does the "Common Types of Space Complexity" figure reflect the absolute size of occupied space?
**Q**: Does the diagram for "common space complexity types" reflect the absolute size of occupied space?
No, the figure shows space complexities, which reflect growth trends, not the absolute size of the occupied space.
No, the diagram shows space complexity, which reflects growth trends rather than the absolute size of occupied space.
If you take $n = 8$, you might find that the values of each curve don't correspond to their functions. This is because each curve includes a constant term, intended to compress the value range into a visually comfortable range.
Assuming $n = 8$, you might find that the values of each curve do not correspond to the functions. This is because each curve contains a constant term used to compress the value range into a visually comfortable range.
In practice, since we usually don't know the "constant term" complexity of each method, it's generally not possible to choose the best solution for $n = 8$ based solely on complexity. However, for $n = 8^5$, it's much easier to choose, as the growth trend becomes dominant.
In practice, because we generally do not know what the "constant term" complexity of each method is, we usually cannot select the optimal solution for $n = 8$ based on complexity alone. But for $n = 8^5$, the choice is straightforward, as the growth trend already dominates.
**Q**: Are there situations where algorithms are designed to sacrifice time (or space) based on actual use cases?
In practical applications, most situations choose to sacrifice space for time. For example, with database indexes, we typically choose to build B+ trees or hash indexes, occupying substantial memory space in exchange for efficient queries of $O(\log n)$ or even $O(1)$.
In scenarios where space resources are precious, time may be sacrificed for space. For example, in embedded development, device memory is precious, and engineers may forgo using hash tables and choose to use array sequential search to save memory usage, at the cost of slower searches.
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---
# 3.2 &nbsp; Basic data types
# 3.2 &nbsp; Basic Data Types
When discussing data in computers, various forms like text, images, videos, voice and 3D models comes to mind. Despite their different organizational forms, they are all composed of various basic data types.
When we talk about data in computers, we think of various forms such as text, images, videos, audio, 3D models, and more. Although these data are organized in different ways, they are all composed of various basic data types.
**Basic data types are those that the CPU can directly operate on** and are directly used in algorithms, mainly including the following.
**Basic data types are types that the CPU can directly operate on**, and they are directly used in algorithms, mainly including the following.
- Integer types: `byte`, `short`, `int`, `long`.
- Floating-point types: `float`, `double`, used to represent decimals.
- Character type: `char`, used to represent letters, punctuation, and even emojis in various languages.
- Boolean type: `bool`, used to represent "yes" or "no" decisions.
- Integer types `byte`, `short`, `int`, `long`.
- Floating-point types `float`, `double`, used to represent decimal numbers.
- Character type `char`, used to represent letters, punctuation marks, and even emojis in various languages.
- Boolean type `bool`, used to represent "yes" and "no" judgments.
**Basic data types are stored in computers in binary form**. One binary digit is 1 bit. In most modern operating systems, 1 byte consists of 8 bits.
**Basic data types are stored in binary form in computers**. One binary bit is $1$ bit. In most modern operating systems, $1$ byte consists of $8$ bits.
The range of values for basic data types depends on the size of the space they occupy. Below, we take Java as an example.
The range of values for basic data types depends on the size of the space they occupy. Below is an example using Java.
- The integer type `byte` occupies 1 byte = 8 bits and can represent $2^8$ numbers.
- The integer type `int` occupies 4 bytes = 32 bits and can represent $2^{32}$ numbers.
- Integer type `byte` occupies $1$ byte = $8$ bits, and can represent $2^{8}$ numbers.
- Integer type `int` occupies $4$ bytes = $32$ bits, and can represent $2^{32}$ numbers.
The following table lists the space occupied, value range, and default values of various basic data types in Java. While memorizing this table isn't necessary, having a general understanding of it and referencing it when required is recommended.
The following table lists the space occupied, value ranges, and default values of various basic data types in Java. You don't need to memorize this table; a general understanding is sufficient, and you can refer to it when needed.
<p align="center"> Table 3-1 &nbsp; Space occupied and value range of basic data types </p>
<p align="center"> Table 3-1 &nbsp; Space occupied and value ranges of basic data types </p>
<div class="center-table" markdown>
| Type | Symbol | Space Occupied | Minimum Value | Maximum Value | Default Value |
| ------- | -------- | -------------- | ------------------------ | ----------------------- | -------------- |
| Integer | `byte` | 1 byte | $-2^7$ ($-128$) | $2^7 - 1$ ($127$) | 0 |
| | `short` | 2 bytes | $-2^{15}$ | $2^{15} - 1$ | 0 |
| | `int` | 4 bytes | $-2^{31}$ | $2^{31} - 1$ | 0 |
| | `long` | 8 bytes | $-2^{63}$ | $2^{63} - 1$ | 0 |
| Float | `float` | 4 bytes | $1.175 \times 10^{-38}$ | $3.403 \times 10^{38}$ | $0.0\text{f}$ |
| | `double` | 8 bytes | $2.225 \times 10^{-308}$ | $1.798 \times 10^{308}$ | 0.0 |
| Char | `char` | 2 bytes | 0 | $2^{16} - 1$ | 0 |
| Boolean | `bool` | 1 byte | $\text{false}$ | $\text{true}$ | $\text{false}$ |
| Type | Symbol | Space Occupied | Minimum Value | Maximum Value | Default Value |
| ---------- | -------- | -------------- | ------------------------ | ----------------------- | -------------- |
| Integer | `byte` | 1 byte | $-2^7$ ($-128$) | $2^7 - 1$ ($127$) | $0$ |
| | `short` | 2 bytes | $-2^{15}$ | $2^{15} - 1$ | $0$ |
| | `int` | 4 bytes | $-2^{31}$ | $2^{31} - 1$ | $0$ |
| | `long` | 8 bytes | $-2^{63}$ | $2^{63} - 1$ | $0$ |
| Float | `float` | 4 bytes | $1.175 \times 10^{-38}$ | $3.403 \times 10^{38}$ | $0.0\text{f}$ |
| | `double` | 8 bytes | $2.225 \times 10^{-308}$ | $1.798 \times 10^{308}$ | $0.0$ |
| Character | `char` | 2 bytes | $0$ | $2^{16} - 1$ | $0$ |
| Boolean | `bool` | 1 byte | $\text{false}$ | $\text{true}$ | $\text{false}$ |
</div>
Please note that the above table is specific to Java's basic data types. Every programming language has its own data type definitions, which might differ in space occupied, value ranges, and default values.
Please note that the above table is specific to Java's basic data types. Each programming language has its own data type definitions, and their space occupied, value ranges, and default values may vary.
- In Python, the integer type `int` can be of any size, limited only by available memory; the floating-point `float` is double precision 64-bit; there is no `char` type, as a single character is actually a string `str` of length 1.
- C and C++ do not specify the size of basic data types, it varies with implementation and platform. The above table follows the LP64 [data model](https://en.cppreference.com/w/cpp/language/types#Properties), used for Unix 64-bit operating systems including Linux and macOS.
- The size of `char` in C and C++ is 1 byte, while in most programming languages, it depends on the specific character encoding method, as detailed in the "Character Encoding" chapter.
- Even though representing a boolean only requires 1 bit (0 or 1), it is usually stored in memory as 1 byte. This is because modern computer CPUs typically use 1 byte as the smallest addressable memory unit.
- In Python, the integer type `int` can be of any size, limited only by available memory; the floating-point type `float` is double-precision 64-bit; there is no `char` type, a single character is actually a string `str` of length 1.
- C and C++ do not explicitly specify the size of basic data types, which varies by implementation and platform. The above table follows the LP64 [data model](https://en.cppreference.com/w/cpp/language/types#Properties), which is used in Unix 64-bit operating systems including Linux and macOS.
- The size of character `char` is 1 byte in C and C++, and in most programming languages it depends on the specific character encoding method, as detailed in the "Character Encoding" section.
- Even though representing a boolean value requires only 1 bit ($0$ or $1$), it is usually stored as 1 byte in memory. This is because modern computer CPUs typically use 1 byte as the minimum addressable memory unit.
So, what is the connection between basic data types and data structures? We know that data structures are ways to organize and store data in computers. The focus here is on "structure" rather than "data".
So, what is the relationship between basic data types and data structures? We know that data structures are ways of organizing and storing data in computers. The subject of this statement is "structure", not "data".
If we want to represent "a row of numbers", we naturally think of using an array. This is because the linear structure of an array can represent the adjacency and the ordering of the numbers, but whether the stored content is an integer `int`, a decimal `float`, or a character `char`, is irrelevant to the "data structure".
If we want to represent "a row of numbers", we naturally think of using an array. This is because the linear structure of an array can represent the adjacency and order relationships of numbers, but the content stored—whether integer `int`, floating-point `float`, or character `char`—is unrelated to the "data structure".
In other words, **basic data types provide the "content type" of data, while data structures provide the "way of organizing" data**. For example, in the following code, we use the same data structure (array) to store and represent different basic data types, including `int`, `float`, `char`, `bool`, etc.
In other words, **basic data types provide the "content type" of data, while data structures provide the "organization method" of data**. For example, in the following code, we use the same data structure (array) to store and represent different basic data types, including `int`, `float`, `char`, `bool`, etc.
=== "Python"
```python title=""
# Using various basic data types to initialize arrays
# Initialize arrays using various basic data types
numbers: list[int] = [0] * 5
decimals: list[float] = [0.0] * 5
# Python's characters are actually strings of length 1
# In Python, characters are actually strings of length 1
characters: list[str] = ['0'] * 5
bools: list[bool] = [False] * 5
# Python's lists can freely store various basic data types and object references
# Python lists can freely store various basic data types and object references
data = [0, 0.0, 'a', False, ListNode(0)]
```
=== "C++"
```cpp title=""
// Using various basic data types to initialize arrays
// Initialize arrays using various basic data types
int numbers[5];
float decimals[5];
char characters[5];
@@ -78,7 +78,7 @@ In other words, **basic data types provide the "content type" of data, while dat
=== "Java"
```java title=""
// Using various basic data types to initialize arrays
// Initialize arrays using various basic data types
int[] numbers = new int[5];
float[] decimals = new float[5];
char[] characters = new char[5];
@@ -88,7 +88,7 @@ In other words, **basic data types provide the "content type" of data, while dat
=== "C#"
```csharp title=""
// Using various basic data types to initialize arrays
// Initialize arrays using various basic data types
int[] numbers = new int[5];
float[] decimals = new float[5];
char[] characters = new char[5];
@@ -98,7 +98,7 @@ In other words, **basic data types provide the "content type" of data, while dat
=== "Go"
```go title=""
// Using various basic data types to initialize arrays
// Initialize arrays using various basic data types
var numbers = [5]int{}
var decimals = [5]float64{}
var characters = [5]byte{}
@@ -108,7 +108,7 @@ In other words, **basic data types provide the "content type" of data, while dat
=== "Swift"
```swift title=""
// Using various basic data types to initialize arrays
// Initialize arrays using various basic data types
let numbers = Array(repeating: 0, count: 5)
let decimals = Array(repeating: 0.0, count: 5)
let characters: [Character] = Array(repeating: "a", count: 5)
@@ -118,14 +118,14 @@ In other words, **basic data types provide the "content type" of data, while dat
=== "JS"
```javascript title=""
// JavaScript's arrays can freely store various basic data types and objects
// JavaScript arrays can freely store various basic data types and objects
const array = [0, 0.0, 'a', false];
```
=== "TS"
```typescript title=""
// Using various basic data types to initialize arrays
// Initialize arrays using various basic data types
const numbers: number[] = [];
const characters: string[] = [];
const bools: boolean[] = [];
@@ -134,7 +134,7 @@ In other words, **basic data types provide the "content type" of data, while dat
=== "Dart"
```dart title=""
// Using various basic data types to initialize arrays
// Initialize arrays using various basic data types
List<int> numbers = List.filled(5, 0);
List<double> decimals = List.filled(5, 0.0);
List<String> characters = List.filled(5, 'a');
@@ -144,9 +144,9 @@ In other words, **basic data types provide the "content type" of data, while dat
=== "Rust"
```rust title=""
// Using various basic data types to initialize arrays
// Initialize arrays using various basic data types
let numbers: Vec<i32> = vec![0; 5];
let decimals: Vec<f32> = vec![0.0, 5];
let decimals: Vec<f32> = vec![0.0; 5];
let characters: Vec<char> = vec!['0'; 5];
let bools: Vec<bool> = vec![false; 5];
```
@@ -154,7 +154,7 @@ In other words, **basic data types provide the "content type" of data, while dat
=== "C"
```c title=""
// Using various basic data types to initialize arrays
// Initialize arrays using various basic data types
int numbers[10];
float decimals[10];
char characters[10];
@@ -164,15 +164,20 @@ In other words, **basic data types provide the "content type" of data, while dat
=== "Kotlin"
```kotlin title=""
// Initialize arrays using various basic data types
val numbers = IntArray(5)
val decinals = FloatArray(5)
val characters = CharArray(5)
val bools = BooleanArray(5)
```
=== "Zig"
=== "Ruby"
```zig title=""
// Using various basic data types to initialize arrays
var numbers: [5]i32 = undefined;
var decimals: [5]f32 = undefined;
var characters: [5]u8 = undefined;
var bools: [5]bool = undefined;
```ruby title=""
# Ruby lists can freely store various basic data types and object references
data = [0, 0.0, 'a', false, ListNode(0)]
```
??? pythontutor "Visualized Execution"
https://pythontutor.com/render.html#code=class%20ListNode%3A%0A%20%20%20%20%22%22%22%E9%93%BE%E8%A1%A8%E8%8A%82%E7%82%B9%E7%B1%BB%22%22%22%0A%20%20%20%20def%20__init__%28self,%20val%3A%20int%29%3A%0A%20%20%20%20%20%20%20%20self.val%3A%20int%20%3D%20val%20%20%23%20%E8%8A%82%E7%82%B9%E5%80%BC%0A%20%20%20%20%20%20%20%20self.next%3A%20ListNode%20%7C%20None%20%3D%20None%20%20%23%20%E5%90%8E%E7%BB%A7%E8%8A%82%E7%82%B9%E5%BC%95%E7%94%A8%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20%23%20%E4%BD%BF%E7%94%A8%E5%A4%9A%E7%A7%8D%E5%9F%BA%E6%9C%AC%E6%95%B0%E6%8D%AE%E7%B1%BB%E5%9E%8B%E6%9D%A5%E5%88%9D%E5%A7%8B%E5%8C%96%E6%95%B0%E7%BB%84%0A%20%20%20%20numbers%20%3D%20%5B0%5D%20*%205%0A%20%20%20%20decimals%20%3D%20%5B0.0%5D%20*%205%0A%20%20%20%20%23%20Python%20%E7%9A%84%E5%AD%97%E7%AC%A6%E5%AE%9E%E9%99%85%E4%B8%8A%E6%98%AF%E9%95%BF%E5%BA%A6%E4%B8%BA%201%20%E7%9A%84%E5%AD%97%E7%AC%A6%E4%B8%B2%0A%20%20%20%20characters%20%3D%20%5B'0'%5D%20*%205%0A%20%20%20%20bools%20%3D%20%5BFalse%5D%20*%205%0A%20%20%20%20%23%20Python%20%E7%9A%84%E5%88%97%E8%A1%A8%E5%8F%AF%E4%BB%A5%E8%87%AA%E7%94%B1%E5%AD%98%E5%82%A8%E5%90%84%E7%A7%8D%E5%9F%BA%E6%9C%AC%E6%95%B0%E6%8D%AE%E7%B1%BB%E5%9E%8B%E5%92%8C%E5%AF%B9%E8%B1%A1%E5%BC%95%E7%94%A8%0A%20%20%20%20data%20%3D%20%5B0,%200.0,%20'a',%20False,%20ListNode%280%29%5D&cumulative=false&curInstr=12&heapPrimitives=nevernest&mode=display&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false
en/docs/chapter_data_structure/character_encoding.md
+45 -45
View File
@@ -2,96 +2,96 @@
comments: true
---
# 3.4 &nbsp; Character encoding *
# 3.4 &nbsp; Character Encoding *
In the computer system, all data is stored in binary form, and `char` is no exception. To represent characters, we need to develop a "character set" that defines a one-to-one mapping between each character and binary numbers. With the character set, computers can convert binary numbers to characters by looking up the table.
In computers, all data is stored in binary form, and character `char` is no exception. To represent characters, we need to establish a "character set" that defines a one-to-one correspondence between each character and binary numbers. With a character set, computers can convert binary numbers to characters by looking up the table.
## 3.4.1 &nbsp; ASCII character set
## 3.4.1 &nbsp; Ascii Character Set
The <u>ASCII code</u> is one of the earliest character sets, officially known as the American Standard Code for Information Interchange. It uses 7 binary digits (the lower 7 bits of a byte) to represent a character, allowing for a maximum of 128 different characters. As shown in Figure 3-6, ASCII includes uppercase and lowercase English letters, numbers 0 ~ 9, various punctuation marks, and certain control characters (such as newline and tab).
<u>ASCII code</u> is the earliest character set, with the full name American Standard Code for Information Interchange. It uses 7 binary bits (the lower 7 bits of one byte) to represent a character, and can represent a maximum of 128 different characters. As shown in Figure 3-6, ASCII code includes uppercase and lowercase English letters, numbers 0 ~ 9, some punctuation marks, and some control characters (such as newline and tab).
![ASCII code](character_encoding.assets/ascii_table.png){ class="animation-figure" }
<p align="center"> Figure 3-6 &nbsp; ASCII code </p>
However, **ASCII can only represent English characters**. With the globalization of computers, a character set called <u>EASCII</u> was developed to represent more languages. It expands from the 7-bit structure of ASCII to 8 bits, enabling the representation of 256 characters.
However, **ASCII code can only represent English**. With the globalization of computers, a character set called <u>EASCII</u> that can represent more languages emerged. It expands from the 7-bit basis of ASCII to 8 bits, and can represent 256 different characters.
Globally, various region-specific EASCII character sets have been introduced. The first 128 characters of these sets are consistent with the ASCII, while the remaining 128 characters are defined differently to accommodate the requirements of different languages.
Worldwide, a batch of EASCII character sets suitable for different regions have appeared successively. The first 128 characters of these character sets are unified as ASCII code, and the last 128 characters are defined differently to adapt to the needs of different languages.
## 3.4.2 &nbsp; GBK character set
## 3.4.2 &nbsp; Gbk Character Set
Later, it was found that **EASCII still could not meet the character requirements of many languages**. For instance, there are nearly a hundred thousand Chinese characters, with several thousand used regularly. In 1980, the Standardization Administration of China released the <u>GB2312</u> character set, which included 6763 Chinese characters, essentially fulfilling the computer processing needs for the Chinese language.
Later, people found that **EASCII code still cannot meet the character quantity requirements of many languages**. For example, there are nearly one hundred thousand Chinese characters, and several thousand are used daily. In 1980, the China National Standardization Administration released the <u>GB2312</u> character set, which included 6,763 Chinese characters, basically meeting the needs for computer processing of Chinese characters.
However, GB2312 could not handle some rare and traditional characters. The <u>GBK</u> character set expands GB2312 and includes 21886 Chinese characters. In the GBK encoding scheme, ASCII characters are represented with one byte, while Chinese characters use two bytes.
However, GB2312 cannot handle some rare characters and traditional Chinese characters. The <u>GBK</u> character set is an extension based on GB2312, which includes a total of 21,886 Chinese characters. In the GBK encoding scheme, ASCII characters are represented using one byte, and Chinese characters are represented using two bytes.
## 3.4.3 &nbsp; Unicode character set
## 3.4.3 &nbsp; Unicode Character Set
With the rapid evolution of computer technology and a plethora of character sets and encoding standards, numerous problems arose. On the one hand, these character sets generally only defined characters for specific languages and could not function properly in multilingual environments. On the other hand, the existence of multiple character set standards for the same language caused garbled text when information was exchanged between computers using different encoding standards.
With the vigorous development of computer technology, character sets and encoding standards flourished, which brought many problems. On the one hand, these character sets generally only define characters for specific languages and cannot work normally in multilingual environments. On the other hand, multiple character set standards exist for the same language, and if two computers use different encoding standards, garbled characters will appear during information transmission.
Researchers of that era thought: **What if a comprehensive character set encompassing all global languages and symbols was developed? Wouldn't this resolve the issues associated with cross-linguistic environments and garbled text?** Inspired by this idea, the extensive character set, Unicode, was born.
Researchers of that era thought: **If a sufficiently complete character set is released that includes all languages and symbols in the world, wouldn't it be possible to solve cross-language environment and garbled character problems**? Driven by this idea, a large and comprehensive character set, Unicode, was born.
<u>Unicode</u> is referred to as "统一码" (Unified Code) in Chinese, theoretically capable of accommodating over a million characters. It aims to incorporate characters from all over the world into a single set, providing a universal character set for processing and displaying various languages and reducing the issues of garbled text due to different encoding standards.
<u>Unicode</u> is called "统一码" (Unified Code) in Chinese and can theoretically accommodate over one million characters. It is committed to including characters from around the world into a unified character set, providing a universal character set to handle and display various language texts, reducing garbled character problems caused by different encoding standards.
Since its release in 1991, Unicode has continually expanded to include new languages and characters. As of September 2022, Unicode contains 149,186 characters, including characters, symbols, and even emojis from various languages. In the vast Unicode character set, commonly used characters occupy 2 bytes, while some rare characters may occupy 3 or even 4 bytes.
Since its release in 1991, Unicode has continuously expanded to include new languages and characters. As of September 2022, Unicode has included 149,186 characters, including characters, symbols, and even emojis from various languages. In the vast Unicode character set, commonly used characters occupy 2 bytes, and some rare characters occupy 3 bytes or even 4 bytes.
Unicode is a universal character set that assigns a number (called a "code point") to each character, **but it does not specify how these character code points should be stored in a computer system**. One might ask: How does a system interpret Unicode code points of varying lengths within a text? For example, given a 2-byte code, how does the system determine if it represents a single 2-byte character or two 1-byte characters?
Unicode is a universal character set that essentially assigns a number (called a "code point") to each character, **but it does not specify how to store these character code points in computers**. We can't help but ask: when Unicode code points of multiple lengths appear simultaneously in a text, how does the system parse the characters? For example, given an encoding with a length of 2 bytes, how does the system determine whether it is one 2-byte character or two 1-byte characters?
**A straightforward solution to this problem is to store all characters as equal-length encodings**. As shown in Figure 3-7, each character in "Hello" occupies 1 byte, while each character in "算法" (algorithm) occupies 2 bytes. We could encode all characters in "Hello 算法" as 2 bytes by padding the higher bits with zeros. This method would enable the system to interpret a character every 2 bytes, recovering the content of the phrase.
For the above problem, **a straightforward solution is to store all characters as equal-length encodings**. As shown in Figure 3-7, each character in "Hello" occupies 1 byte, and each character in "算法" (algorithm) occupies 2 bytes. We can encode all characters in "Hello 算法" as 2 bytes in length by padding the high bits with 0. In this way, the system can parse one character every 2 bytes and restore the content of this phrase.
![Unicode encoding example](character_encoding.assets/unicode_hello_algo.png){ class="animation-figure" }
<p align="center"> Figure 3-7 &nbsp; Unicode encoding example </p>
However, as ASCII has shown us, encoding English only requires 1 byte. Using the above approach would double the space occupied by English text compared to ASCII encoding, which is a waste of memory space. Therefore, a more efficient Unicode encoding method is needed.
However, ASCII code has already proven to us that encoding English only requires 1 byte. If the above scheme is adopted, the size of English text will be twice that under ASCII encoding, which is very wasteful of memory space. Therefore, we need a more efficient Unicode encoding method.
## 3.4.4 &nbsp; UTF-8 encoding
## 3.4.4 &nbsp; Utf-8 Encoding
Currently, UTF-8 has become the most widely used Unicode encoding method internationally. **It is a variable-length encoding**, using 1 to 4 bytes to represent a character, depending on the complexity of the character. ASCII characters need only 1 byte, Latin and Greek letters require 2 bytes, commonly used Chinese characters need 3 bytes, and some other rare characters need 4 bytes.
Currently, UTF-8 has become the most widely used Unicode encoding method internationally. **It is a variable-length encoding** that uses 1 to 4 bytes to represent a character, depending on the complexity of the character. ASCII characters only require 1 byte, Latin and Greek letters require 2 bytes, commonly used Chinese characters require 3 bytes, and some other rare characters require 4 bytes.
The encoding rules for UTF-8 are not complex and can be divided into two cases:
The encoding rules of UTF-8 are not complicated and can be divided into the following two cases.
- For 1-byte characters, set the highest bit to $0$, and the remaining 7 bits to the Unicode code point. Notably, ASCII characters occupy the first 128 code points in the Unicode set. This means that **UTF-8 encoding is backward compatible with ASCII**. This implies that UTF-8 can be used to parse ancient ASCII text.
- For characters of length $n$ bytes (where $n > 1$), set the highest $n$ bits of the first byte to $1$, and the $(n + 1)^{\text{th}}$ bit to $0$; starting from the second byte, set the highest 2 bits of each byte to $10$; the rest of the bits are used to fill the Unicode code point.
- For 1-byte characters, set the highest bit to $0$, and set the remaining 7 bits to the Unicode code point. It is worth noting that ASCII characters occupy the first 128 code points in the Unicode character set. That is to say, **UTF-8 encoding is backward compatible with ASCII code**. This means we can use UTF-8 to parse very old ASCII code text.
- For characters with a length of $n$ bytes (where $n > 1$), set the highest $n$ bits of the first byte to $1$, and set the $(n + 1)$-th bit to $0$; starting from the second byte, set the highest 2 bits of each byte to $10$; use all remaining bits to fill in the Unicode code point of the character.
Figure 3-8 shows the UTF-8 encoding for "Hello算法". It can be observed that since the highest $n$ bits are set to $1$, the system can determine the length of the character as $n$ by counting the number of highest bits set to $1$.
Figure 3-8 shows the UTF-8 encoding corresponding to "Hello算法". It can be observed that since the highest $n$ bits are all set to $1$, the system can parse the length of the character as $n$ by reading the number of highest bits that are $1$.
But why set the highest 2 bits of the remaining bytes to $10$? Actually, this $10$ serves as a kind of checksum. If the system starts parsing text from an incorrect byte, the $10$ at the beginning of the byte can help the system quickly detect anomalies.
But why set the highest 2 bits of all other bytes to $10$? In fact, this $10$ can serve as a check symbol. Assuming the system starts parsing text from an incorrect byte, the $10$ at the beginning of the byte can help the system quickly determine an anomaly.
The reason for using $10$ as a checksum is that, under UTF-8 encoding rules, it's impossible for the highest two bits of a character to be $10$. This can be proven by contradiction: If the highest two bits of a character are $10$, it indicates that the character's length is $1$, corresponding to ASCII. However, the highest bit of an ASCII character should be $0$, which contradicts the assumption.
The reason for using $10$ as a check symbol is that under UTF-8 encoding rules, it is impossible for a character's highest two bits to be $10$. This conclusion can be proven by contradiction: assuming the highest two bits of a character are $10$, it means the length of the character is $1$, corresponding to ASCII code. However, the highest bit of ASCII code should be $0$, which contradicts the assumption.
![UTF-8 encoding example](character_encoding.assets/utf-8_hello_algo.png){ class="animation-figure" }
<p align="center"> Figure 3-8 &nbsp; UTF-8 encoding example </p>
Apart from UTF-8, other common encoding methods include:
In addition to UTF-8, common encoding methods also include the following two.
- **UTF-16 encoding**: Uses 2 or 4 bytes to represent a character. All ASCII characters and commonly used non-English characters are represented with 2 bytes; a few characters require 4 bytes. For 2-byte characters, the UTF-16 encoding equals the Unicode code point.
- **UTF-32 encoding**: Every character uses 4 bytes. This means UTF-32 occupies more space than UTF-8 and UTF-16, especially for texts with a high proportion of ASCII characters.
- **UTF-16 encoding**: Uses 2 or 4 bytes to represent a character. All ASCII characters and commonly used non-English characters are represented with 2 bytes; a few characters need to use 4 bytes. For 2-byte characters, UTF-16 encoding is equal to the Unicode code point.
- **UTF-32 encoding**: Every character uses 4 bytes. This means that UTF-32 takes up more space than UTF-8 and UTF-16, especially for text with a high proportion of ASCII characters.
From the perspective of storage space, using UTF-8 to represent English characters is very efficient because it only requires 1 byte; using UTF-16 to encode some non-English characters (such as Chinese) can be more efficient because it only requires 2 bytes, while UTF-8 might need 3 bytes.
From the perspective of storage space occupation, using UTF-8 to represent English characters is very efficient because it only requires 1 byte; using UTF-16 encoding for some non-English characters (such as Chinese) will be more efficient because it only requires 2 bytes, while UTF-8 may require 3 bytes.
From a compatibility perspective, UTF-8 is the most versatile, with many tools and libraries supporting UTF-8 as a priority.
From a compatibility perspective, UTF-8 has the best universality, and many tools and libraries support UTF-8 first.
## 3.4.5 &nbsp; Character encoding in programming languages
## 3.4.5 &nbsp; Character Encoding in Programming Languages
Historically, many programming languages utilized fixed-length encodings such as UTF-16 or UTF-32 for processing strings during program execution. This allows strings to be handled as arrays, offering several advantages:
For most past programming languages, strings during program execution use fixed-length encodings such as UTF-16 or UTF-32. Under fixed-length encoding, we can treat strings as arrays for processing, and this approach has the following advantages.
- **Random access**: Strings encoded in UTF-16 can be accessed randomly with ease. For UTF-8, which is a variable-length encoding, locating the $i^{th}$ character requires traversing the string from the start to the $i^{th}$ position, taking $O(n)$ time.
- **Character counting**: Similar to random access, counting the number of characters in a UTF-16 encoded string is an $O(1)$ operation. However, counting characters in a UTF-8 encoded string requires traversing the entire string.
- **String operations**: Many string operations like splitting, concatenating, inserting, and deleting are easier on UTF-16 encoded strings. These operations generally require additional computation on UTF-8 encoded strings to ensure the validity of the UTF-8 encoding.
- **Random access**: UTF-16 encoded strings can be easily accessed randomly. UTF-8 is a variable-length encoding. To find the $i$-th character, we need to traverse from the beginning of the string to the $i$-th character, which requires $O(n)$ time.
- **Character counting**: Similar to random access, calculating the length of a UTF-16 encoded string is also an $O(1)$ operation. However, calculating the length of a UTF-8 encoded string requires traversing the entire string.
- **String operations**: Many string operations (such as splitting, joining, inserting, deleting, etc.) on UTF-16 encoded strings are easier to perform. Performing these operations on UTF-8 encoded strings usually requires additional calculations to ensure that invalid UTF-8 encoding is not generated.
The design of character encoding schemes in programming languages is an interesting topic involving various factors:
In fact, the design of character encoding schemes for programming languages is a very interesting topic involving many factors.
- Java’s `String` type uses UTF-16 encoding, with each character occupying 2 bytes. This was based on the initial belief that 16 bits were sufficient to represent all possible characters and proven incorrect later. As the Unicode standard expanded beyond 16 bits, characters in Java may now be represented by a pair of 16-bit values, known as “surrogate pairs.”
- JavaScript and TypeScript use UTF-16 encoding for similar reasons as Java. When JavaScript was first introduced by Netscape in 1995, Unicode was still in its early stages, and 16-bit encoding was sufficient to represent all Unicode characters.
- C# uses UTF-16 encoding, largely because the .NET platform, designed by Microsoft, and many Microsoft technologies, including the Windows operating system, extensively use UTF-16 encoding.
- Java's `String` type uses UTF-16 encoding, with each character occupying 2 bytes. This is because at the beginning of Java language design, people believed that 16 bits were sufficient to represent all possible characters. However, this was an incorrect judgment. Later, the Unicode specification expanded beyond 16 bits, so characters in Java may now be represented by a pair of 16-bit values (called "surrogate pairs").
- The strings of JavaScript and TypeScript use UTF-16 encoding for reasons similar to Java. When Netscape first introduced the JavaScript language in 1995, Unicode was still in its early stages of development, and at that time, using 16-bit encoding was sufficient to represent all Unicode characters.
- C# uses UTF-16 encoding mainly because the .NET platform was designed by Microsoft, and many of Microsoft's technologies (including the Windows operating system) extensively use UTF-16 encoding.
Due to the underestimation of character counts, these languages had to use "surrogate pairs" to represent Unicode characters exceeding 16 bits. This approach has its drawbacks: strings containing surrogate pairs may have characters occupying 2 or 4 bytes, losing the advantage of fixed-length encoding. Additionally, handling surrogate pairs adds complexity and debugging difficulty to programming.
Due to the underestimation of character quantities by the above programming languages, they had to adopt the "surrogate pair" method to represent Unicode characters with lengths exceeding 16 bits. This is a reluctant compromise. On the one hand, in strings containing surrogate pairs, one character may occupy 2 bytes or 4 bytes, thus losing the advantage of fixed-length encoding. On the other hand, handling surrogate pairs requires additional code, which increases the complexity and difficulty of debugging in programming.
Addressing these challenges, some languages have adopted alternative encoding strategies:
For the above reasons, some programming languages have proposed different encoding schemes.
- Python’s `str` type uses Unicode encoding with a flexible representation where the storage length of characters depends on the largest Unicode code point in the string. If all characters are ASCII, each character occupies 1 byte, 2 bytes for characters within the Basic Multilingual Plane (BMP), and 4 bytes for characters beyond the BMP.
- Go’s `string` type internally uses UTF-8 encoding. Go also provides the `rune` type for representing individual Unicode code points.
- Rust’s `str` and `String` types use UTF-8 encoding internally. Rust also offers the `char` type for individual Unicode code points.
- Python's `str` uses Unicode encoding and adopts a flexible string representation where the stored character length depends on the largest Unicode code point in the string. If all characters in the string are ASCII characters, each character occupies 1 byte; if there are characters exceeding the ASCII range but all within the Basic Multilingual Plane (BMP), each character occupies 2 bytes; if there are characters exceeding the BMP, each character occupies 4 bytes.
- Go language's `string` type uses UTF-8 encoding internally. Go language also provides the `rune` type, which is used to represent a single Unicode code point.
- Rust language's `str` and `String` types use UTF-8 encoding internally. Rust also provides the `char` type for representing a single Unicode code point.
It’s important to note that the above discussion pertains to how strings are stored in programming languages, **which is different from how strings are stored in files or transmitted over networks**. For file storage or network transmission, strings are usually encoded in UTF-8 format for optimal compatibility and space efficiency.
It should be noted that the above discussion is about how strings are stored in programming languages, **which is different from how strings are stored in files or transmitted over networks**. In file storage or network transmission, we usually encode strings into UTF-8 format to achieve optimal compatibility and space efficiency.
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---
# 3.1 &nbsp; Classification of data structures
# 3.1 &nbsp; Classification of Data Structures
Common data structures include arrays, linked lists, stacks, queues, hash tables, trees, heaps, and graphs. They can be classified into "logical structure" and "physical structure".
Common data structures include arrays, linked lists, stacks, queues, hash tables, trees, heaps, and graphs. They can be classified from two dimensions: "logical structure" and "physical structure".
## 3.1.1 &nbsp; Logical structure: linear and non-linear
## 3.1.1 &nbsp; Logical Structure: Linear and Non-Linear
**The logical structures reveal the logical relationships between data elements**. In arrays and linked lists, data are arranged in a specific sequence, demonstrating the linear relationship between data; while in trees, data are arranged hierarchically from the top down, showing the derived relationship between "ancestors" and "descendants"; and graphs are composed of nodes and edges, reflecting the intricate network relationship.
**Logical structure reveals the logical relationships between data elements**. In arrays and linked lists, data is arranged in a certain order, embodying the linear relationship between data; while in trees, data is arranged hierarchically from top to bottom, showing the derived relationship between "ancestors" and "descendants"; graphs are composed of nodes and edges, reflecting complex network relationships.
As shown in Figure 3-1, logical structures can be divided into two major categories: "linear" and "non-linear". Linear structures are more intuitive, indicating data is arranged linearly in logical relationships; non-linear structures, conversely, are arranged non-linearly.
As shown in Figure 3-1, logical structures can be divided into two major categories: "linear" and "non-linear". Linear structures are more intuitive, indicating that data is linearly arranged in logical relationships; non-linear structures are the opposite, arranged non-linearly.
- **Linear data structures**: Arrays, Linked Lists, Stacks, Queues, Hash Tables, where elements have a one-to-one sequential relationship.
- **Non-linear data structures**: Trees, Heaps, Graphs, Hash Tables.
- **Linear data structures**: Arrays, linked lists, stacks, queues, hash tables, where elements have a one-to-one sequential relationship.
- **Non-linear data structures**: Trees, heaps, graphs, hash tables.
Non-linear data structures can be further divided into tree structures and network structures.
- **Tree structures**: Trees, Heaps, Hash Tables, where elements have a one-to-many relationship.
- **Network structures**: Graphs, where elements have a many-to-many relationships.
- **Tree structures**: Trees, heaps, hash tables, where elements have a one-to-many relationship.
- **Network structures**: Graphs, where elements have a many-to-many relationship.
![Linear and non-linear data structures](classification_of_data_structure.assets/classification_logic_structure.png){ class="animation-figure" }
<p align="center"> Figure 3-1 &nbsp; Linear and non-linear data structures </p>
## 3.1.2 &nbsp; Physical structure: contiguous and dispersed
## 3.1.2 &nbsp; Physical Structure: Contiguous and Dispersed
**During the execution of an algorithm, the data being processed is stored in memory**. Figure 3-2 shows a computer memory stick where each black square is a physical memory space. We can think of memory as a vast Excel spreadsheet, with each cell capable of storing a certain amount of data.
**When an algorithm program runs, the data being processed is mainly stored in memory**. Figure 3-2 shows a computer memory stick, where each black square contains a memory space. We can imagine memory as a huge Excel spreadsheet, where each cell can store a certain amount of data.
**The system accesses the data at the target location by means of a memory address**. As shown in Figure 3-2, the computer assigns a unique identifier to each cell in the table according to specific rules, ensuring that each memory space has a unique memory address. With these addresses, the program can access the data stored in memory.
**The system accesses data at the target location through memory addresses**. As shown in Figure 3-2, the computer assigns a number to each cell in the spreadsheet according to specific rules, ensuring that each memory space has a unique memory address. With these addresses, the program can access data in memory.
![Memory stick, memory spaces, memory addresses](classification_of_data_structure.assets/computer_memory_location.png){ class="animation-figure" }
![Memory stick, memory space, memory address](classification_of_data_structure.assets/computer_memory_location.png){ class="animation-figure" }
<p align="center"> Figure 3-2 &nbsp; Memory stick, memory spaces, memory addresses </p>
<p align="center"> Figure 3-2 &nbsp; Memory stick, memory space, memory address </p>
!!! tip
It's worth noting that comparing memory to an Excel spreadsheet is a simplified analogy. The actual working mechanism of memory is more complex, involving concepts like address space, memory management, cache mechanisms, virtual memory, and physical memory.
It is worth noting that comparing memory to an Excel spreadsheet is a simplified analogy. The actual working mechanism of memory is quite complex, involving concepts such as address space, memory management, cache mechanisms, virtual memory, and physical memory.
Memory is a shared resource for all programs. When a block of memory is occupied by one program, it cannot be simultaneously used by other programs. **Therefore, memory resources are an important consideration in the design of data structures and algorithms**. For instance, the algorithm's peak memory usage should not exceed the remaining free memory of the system; if there is a lack of contiguous memory blocks, then the data structure chosen must be able to be stored in non-contiguous memory blocks.
Memory is a shared resource for all programs. When a block of memory is occupied by a program, it usually cannot be used by other programs at the same time. **Therefore, in the design of data structures and algorithms, memory resources are an important consideration**. For example, the peak memory occupied by an algorithm should not exceed the remaining free memory of the system; if there is a lack of contiguous large memory blocks, then the data structure chosen must be able to be stored in dispersed memory spaces.
As illustrated in Figure 3-3, **the physical structure reflects the way data is stored in computer memory** and it can be divided into contiguous space storage (arrays) and non-contiguous space storage (linked lists). The two types of physical structures exhibit complementary characteristics in terms of time efficiency and space efficiency.
As shown in Figure 3-3, **physical structure reflects the way data is stored in computer memory**, and can be divided into contiguous space storage (arrays) and dispersed space storage (linked lists). The two physical structures exhibit complementary characteristics in terms of time efficiency and space efficiency.
![Contiguous space storage and dispersed space storage](classification_of_data_structure.assets/classification_phisical_structure.png){ class="animation-figure" }
<p align="center"> Figure 3-3 &nbsp; Contiguous space storage and dispersed space storage </p>
**It is worth noting that all data structures are implemented based on arrays, linked lists, or a combination of both**. For example, stacks and queues can be implemented using either arrays or linked lists; while implementations of hash tables may involve both arrays and linked lists.
It is worth noting that **all data structures are implemented based on arrays, linked lists, or a combination of both**. For example, stacks and queues can be implemented using either arrays or linked lists; while the implementation of hash tables may include both arrays and linked lists.
- **Array-based implementations**: Stacks, Queues, Hash Tables, Trees, Heaps, Graphs, Matrices, Tensors (arrays with dimensions $\geq 3$).
- **Linked-list-based implementations**: Stacks, Queues, Hash Tables, Trees, Heaps, Graphs, etc.
- **Can be implemented based on arrays**: Stacks, queues, hash tables, trees, heaps, graphs, matrices, tensors (arrays with dimensions $\geq 3$), etc.
- **Can be implemented based on linked lists**: Stacks, queues, hash tables, trees, heaps, graphs, etc.
Data structures implemented based on arrays are also called “Static Data Structures,” meaning their length cannot be changed after initialization. Conversely, those based on linked lists are called “Dynamic Data Structures,” which can still adjust their size during program execution.
After initialization, linked lists can still adjust their length during program execution, so they are also called "dynamic data structures". After initialization, the length of arrays cannot be changed, so they are also called "static data structures". It is worth noting that arrays can achieve length changes by reallocating memory, thus possessing a certain degree of "dynamism".
!!! tip
If you find it challenging to comprehend the physical structure, it is recommended that you read the next chapter, "Arrays and Linked Lists," and revisit this section later.
If you find it difficult to understand physical structure, it is recommended to read the next chapter first, and then review this section.
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icon: material/shape-outline
---
# Chapter 3. &nbsp; Data structures
# Chapter 3. &nbsp; Data Structures
![Data structures](../assets/covers/chapter_data_structure.jpg){ class="cover-image" }
!!! abstract
Data structures serve as a robust and diverse framework.
Data structure is like a sturdy and diverse framework.
They offer a blueprint for the orderly organization of data, upon which algorithms come to life.
It provides a blueprint for the orderly organization of data, upon which algorithms come to life.
## Chapter contents
- [3.1 &nbsp; Classification of data structures](classification_of_data_structure.md)
- [3.2 &nbsp; Basic data types](basic_data_types.md)
- [3.3 &nbsp; Number encoding *](number_encoding.md)
- [3.4 &nbsp; Character encoding *](character_encoding.md)
- [3.1 &nbsp; Classification of Data Structures](classification_of_data_structure.md)
- [3.2 &nbsp; Basic Data Types](basic_data_types.md)
- [3.3 &nbsp; Number Encoding *](number_encoding.md)
- [3.4 &nbsp; Character Encoding *](character_encoding.md)
- [3.5 &nbsp; Summary](summary.md)
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---
# 3.3 &nbsp; Number encoding *
# 3.3 &nbsp; Number Encoding *
!!! tip
In this book, chapters marked with an asterisk '*' are optional readings. If you are short on time or find them challenging, you may skip these initially and return to them after completing the essential chapters.
In this book, chapters marked with an asterisk * are optional readings. If you are short on time or find them challenging, you may skip these initially and return to them after completing the essential chapters.
## 3.3.1 &nbsp; Integer encoding
## 3.3.1 &nbsp; Sign-Magnitude, 1's Complement, and 2's Complement
In the table from the previous section, we observed that all integer types can represent one more negative number than positive numbers, such as the `byte` range of $[-128, 127]$. This phenomenon seems counterintuitive, and its underlying reason involves knowledge of sign-magnitude, one's complement, and two's complement encoding.
In the table from the previous section, we found that all integer types can represent one more negative number than positive numbers. For example, the `byte` range is $[-128, 127]$. This phenomenon is counterintuitive, and its underlying reason involves knowledge of sign-magnitude, 1's complement, and 2's complement.
Firstly, it's important to note that **numbers are stored in computers using the two's complement form**. Before analyzing why this is the case, let's define these three encoding methods:
First, it should be noted that **numbers are stored in computers in the form of "2's complement"**. Before analyzing the reasons for this, let's first define these three concepts.
- **Sign-magnitude**: The highest bit of a binary representation of a number is considered the sign bit, where $0$ represents a positive number and $1$ represents a negative number. The remaining bits represent the value of the number.
- **One's complement**: The one's complement of a positive number is the same as its sign-magnitude. For negative numbers, it's obtained by inverting all bits except the sign bit.
- **Two's complement**: The two's complement of a positive number is the same as its sign-magnitude. For negative numbers, it's obtained by adding $1$ to their one's complement.
- **Sign-magnitude**: We treat the highest bit of the binary representation of a number as the sign bit, where $0$ represents a positive number and $1$ represents a negative number, and the remaining bits represent the value of the number.
- **1's complement**: The 1's complement of a positive number is the same as its sign-magnitude. For a negative number, the 1's complement is obtained by inverting all bits except the sign bit of its sign-magnitude.
- **2's complement**: The 2's complement of a positive number is the same as its sign-magnitude. For a negative number, the 2's complement is obtained by adding $1$ to its 1's complement.
Figure 3-4 illustrates the conversions among sign-magnitude, one's complement, and two's complement:
Figure 3-4 shows the conversion methods among sign-magnitude, 1's complement, and 2's complement.
![Conversions between sign-magnitude, one's complement, and two's complement](number_encoding.assets/1s_2s_complement.png){ class="animation-figure" }
![Conversions among sign-magnitude, 1's complement, and 2's complement](number_encoding.assets/1s_2s_complement.png){ class="animation-figure" }
<p align="center"> Figure 3-4 &nbsp; Conversions between sign-magnitude, one's complement, and two's complement </p>
<p align="center"> Figure 3-4 &nbsp; Conversions among sign-magnitude, 1's complement, and 2's complement </p>
Although <u>sign-magnitude</u> is the most intuitive, it has limitations. For one, **negative numbers in sign-magnitude cannot be directly used in calculations**. For example, in sign-magnitude, calculating $1 + (-2)$ results in $-3$, which is incorrect.
<u>Sign-magnitude</u>, although the most intuitive, has some limitations. On one hand, **the sign-magnitude of negative numbers cannot be directly used in operations**. For example, calculating $1 + (-2)$ in sign-magnitude yields $-3$, which is clearly incorrect.
$$
\begin{aligned}
@@ -35,20 +35,20 @@ $$
\end{aligned}
$$
To address this, computers introduced the <u>one's complement</u>. If we convert to one's complement and calculate $1 + (-2)$, then convert the result back to sign-magnitude, we get the correct result of $-1$.
To solve this problem, computers introduced <u>1's complement</u>. If we first convert sign-magnitude to 1's complement and calculate $1 + (-2)$ in 1's complement, then convert the result back to sign-magnitude, we can obtain the correct result of $-1$.
$$
\begin{aligned}
& 1 + (-2) \newline
& \rightarrow 0000 \; 0001 \; \text{(Sign-magnitude)} + 1000 \; 0010 \; \text{(Sign-magnitude)} \newline
& = 0000 \; 0001 \; \text{(One's complement)} + 1111 \; 1101 \; \text{(One's complement)} \newline
& = 1111 \; 1110 \; \text{(One's complement)} \newline
& = 0000 \; 0001 \; \text{(1's complement)} + 1111 \; 1101 \; \text{(1's complement)} \newline
& = 1111 \; 1110 \; \text{(1's complement)} \newline
& = 1000 \; 0001 \; \text{(Sign-magnitude)} \newline
& \rightarrow -1
\end{aligned}
$$
Additionally, **there are two representations of zero in sign-magnitude**: $+0$ and $-0$. This means two different binary encodings for zero, which could lead to ambiguity. For example, in conditional checks, not differentiating between positive and negative zero might result in incorrect outcomes. Addressing this ambiguity would require additional checks, potentially reducing computational efficiency.
On the other hand, **the sign-magnitude of the number zero has two representations, $+0$ and $-0$**. This means that the number zero corresponds to two different binary encodings, which may cause ambiguity. For example, in conditional judgments, if we don't distinguish between positive zero and negative zero, it may lead to incorrect judgment results. If we want to handle the ambiguity of positive and negative zero, we need to introduce additional judgment operations, which may reduce the computational efficiency of the computer.
$$
\begin{aligned}
@@ -57,67 +57,67 @@ $$
\end{aligned}
$$
Like sign-magnitude, one's complement also suffers from the positive and negative zero ambiguity. Therefore, computers further introduced the <u>two's complement</u>. Let's observe the conversion process for negative zero in sign-magnitude, one's complement, and two's complement:
Like sign-magnitude, 1's complement also has the problem of positive and negative zero ambiguity. Therefore, computers further introduced <u>2's complement</u>. Let's first observe the conversion process of negative zero from sign-magnitude to 1's complement to 2's complement:
$$
\begin{aligned}
-0 \rightarrow \; & 1000 \; 0000 \; \text{(Sign-magnitude)} \newline
= \; & 1111 \; 1111 \; \text{(One's complement)} \newline
= 1 \; & 0000 \; 0000 \; \text{(Two's complement)} \newline
= \; & 1111 \; 1111 \; \text{(1's complement)} \newline
= 1 \; & 0000 \; 0000 \; \text{(2's complement)} \newline
\end{aligned}
$$
Adding $1$ to the one's complement of negative zero produces a carry, but with `byte` length being only 8 bits, the carried-over $1$ to the 9th bit is discarded. Therefore, **the two's complement of negative zero is $0000 \; 0000$**, the same as positive zero, thus resolving the ambiguity.
Adding $1$ to the 1's complement of negative zero produces a carry, but since the `byte` type has a length of only 8 bits, the $1$ that overflows to the 9th bit is discarded. That is to say, **the 2's complement of negative zero is $0000 \; 0000$, which is the same as the 2's complement of positive zero**. This means that in 2's complement representation, there is only one zero, and the positive and negative zero ambiguity is thus resolved.
One last puzzle is the $[-128, 127]$ range for `byte`, with an additional negative number, $-128$. We observe that for the interval $[-127, +127]$, all integers have corresponding sign-magnitude, one's complement, and two's complement, allowing for mutual conversion between them.
One last question remains: the range of the `byte` type is $[-128, 127]$, and how is the extra negative number $-128$ obtained? We notice that all integers in the interval $[-127, +127]$ have corresponding sign-magnitude, 1's complement, and 2's complement, and sign-magnitude and 2's complement can be converted to each other.
However, **the two's complement $1000 \; 0000$ is an exception without a corresponding sign-magnitude**. According to the conversion method, its sign-magnitude would be $0000 \; 0000$, indicating zero. This presents a contradiction because its two's complement should represent itself. Computers designate this special two's complement $1000 \; 0000$ as representing $-128$. In fact, the calculation of $(-1) + (-127)$ in two's complement results in $-128$.
However, **the 2's complement $1000 \; 0000$ is an exception, and it does not have a corresponding sign-magnitude**. According to the conversion method, we get that the sign-magnitude of this 2's complement is $0000 \; 0000$. This is clearly contradictory because this sign-magnitude represents the number $0$, and its 2's complement should be itself. The computer specifies that this special 2's complement $1000 \; 0000$ represents $-128$. In fact, the result of calculating $(-1) + (-127)$ in 2's complement is $-128$.
$$
\begin{aligned}
& (-127) + (-1) \newline
& \rightarrow 1111 \; 1111 \; \text{(Sign-magnitude)} + 1000 \; 0001 \; \text{(Sign-magnitude)} \newline
& = 1000 \; 0000 \; \text{(One's complement)} + 1111 \; 1110 \; \text{(One's complement)} \newline
& = 1000 \; 0001 \; \text{(Two's complement)} + 1111 \; 1111 \; \text{(Two's complement)} \newline
& = 1000 \; 0000 \; \text{(Two's complement)} \newline
& = 1000 \; 0000 \; \text{(1's complement)} + 1111 \; 1110 \; \text{(1's complement)} \newline
& = 1000 \; 0001 \; \text{(2's complement)} + 1111 \; 1111 \; \text{(2's complement)} \newline
& = 1000 \; 0000 \; \text{(2's complement)} \newline
& \rightarrow -128
\end{aligned}
$$
As you might have noticed, all these calculations are additions, hinting at an important fact: **computers' internal hardware circuits are primarily designed around addition operations**. This is because addition is simpler to implement in hardware compared to other operations like multiplication, division, and subtraction, allowing for easier parallelization and faster computation.
You may have noticed that all the above calculations are addition operations. This hints at an important fact: **the hardware circuits inside computers are mainly designed based on addition operations**. This is because addition operations are simpler to implement in hardware compared to other operations (such as multiplication, division, and subtraction), easier to parallelize, and have faster operation speeds.
It's important to note that this doesn't mean computers can only perform addition. **By combining addition with basic logical operations, computers can execute a variety of other mathematical operations**. For example, the subtraction $a - b$ can be translated into $a + (-b)$; multiplication and division can be translated into multiple additions or subtractions.
Please note that this does not mean that computers can only perform addition. **By combining addition with some basic logical operations, computers can implement various other mathematical operations**. For example, calculating the subtraction $a - b$ can be converted to calculating the addition $a + (-b)$; calculating multiplication and division can be converted to calculating multiple additions or subtractions.
We can now summarize the reason for using two's complement in computers: with two's complement representation, computers can use the same circuits and operations to handle both positive and negative number addition, eliminating the need for special hardware circuits for subtraction and avoiding the ambiguity of positive and negative zero. This greatly simplifies hardware design and enhances computational efficiency.
Now we can summarize the reasons why computers use 2's complement: based on 2's complement representation, computers can use the same circuits and operations to handle the addition of positive and negative numbers, without the need to design special hardware circuits to handle subtraction, and without the need to specially handle the ambiguity problem of positive and negative zero. This greatly simplifies hardware design and improves operational efficiency.
The design of two's complement is quite ingenious, and due to space constraints, we'll stop here. Interested readers are encouraged to explore further.
The design of 2's complement is very ingenious. Due to space limitations, we will stop here. Interested readers are encouraged to explore further.
## 3.3.2 &nbsp; Floating-point number encoding
## 3.3.2 &nbsp; Floating-Point Number Encoding
You might have noticed something intriguing: despite having the same length of 4 bytes, why does a `float` have a much larger range of values compared to an `int`? This seems counterintuitive, as one would expect the range to shrink for `float` since it needs to represent fractions.
Careful readers may have noticed: `int` and `float` have the same length, both are 4 bytes, but why does `float` have a much larger range than `int`? This is very counterintuitive because it stands to reason that `float` needs to represent decimals, so the range should be smaller.
In fact, **this is due to the different representation method used by floating-point numbers (`float`)**. Let's consider a 32-bit binary number as:
In fact, **this is because floating-point number `float` uses a different representation method**. Let's denote a 32-bit binary number as:
$$
b_{31} b_{30} b_{29} \ldots b_2 b_1 b_0
$$
According to the IEEE 754 standard, a 32-bit `float` consists of the following three parts:
According to the IEEE 754 standard, a 32-bit `float` consists of the following three parts.
- Sign bit $\mathrm{S}$: Occupies 1 bit, corresponding to $b_{31}$.
- Exponent bit $\mathrm{E}$: Occupies 8 bits, corresponding to $b_{30} b_{29} \ldots b_{23}$.
- Fraction bit $\mathrm{N}$: Occupies 23 bits, corresponding to $b_{22} b_{21} \ldots b_0$.
- Sign bit $\mathrm{S}$: occupies 1 bit, corresponding to $b_{31}$.
- Exponent bit $\mathrm{E}$: occupies 8 bits, corresponding to $b_{30} b_{29} \ldots b_{23}$.
- Fraction bit $\mathrm{N}$: occupies 23 bits, corresponding to $b_{22} b_{21} \ldots b_0$.
The value of a binary `float` number is calculated as:
The calculation method for the value corresponding to the binary `float` is:
$$
\text{val} = (-1)^{b_{31}} \times 2^{\left(b_{30} b_{29} \ldots b_{23}\right)_2 - 127} \times \left(1 . b_{22} b_{21} \ldots b_0\right)_2
\text {val} = (-1)^{b_{31}} \times 2^{\left(b_{30} b_{29} \ldots b_{23}\right)_2-127} \times\left(1 . b_{22} b_{21} \ldots b_0\right)_2
$$
Converted to a decimal formula, this becomes:
Converted to decimal, the calculation formula is:
$$
\text{val} = (-1)^{\mathrm{S}} \times 2^{\mathrm{E} - 127} \times (1 + \mathrm{N})
\text {val}=(-1)^{\mathrm{S}} \times 2^{\mathrm{E} -127} \times (1 + \mathrm{N})
$$
The range of each component is:
@@ -125,23 +125,23 @@ The range of each component is:
$$
\begin{aligned}
\mathrm{S} \in & \{ 0, 1\}, \quad \mathrm{E} \in \{ 1, 2, \dots, 254 \} \newline
(1 + \mathrm{N}) = & (1 + \sum_{i=1}^{23} b_{23-i} \times 2^{-i}) \subset [1, 2 - 2^{-23}]
(1 + \mathrm{N}) = & (1 + \sum_{i=1}^{23} b_{23-i} 2^{-i}) \subset [1, 2 - 2^{-23}]
\end{aligned}
$$
![Example calculation of a float in IEEE 754 standard](number_encoding.assets/ieee_754_float.png){ class="animation-figure" }
![Calculation example of float under IEEE 754 standard](number_encoding.assets/ieee_754_float.png){ class="animation-figure" }
<p align="center"> Figure 3-5 &nbsp; Example calculation of a float in IEEE 754 standard </p>
<p align="center"> Figure 3-5 &nbsp; Calculation example of float under IEEE 754 standard </p>
Observing Figure 3-5, given an example data $\mathrm{S} = 0$, $\mathrm{E} = 124$, $\mathrm{N} = 2^{-2} + 2^{-3} = 0.375$, we have:
Observing Figure 3-5, given example data $\mathrm{S} = 0$, $\mathrm{E} = 124$, $\mathrm{N} = 2^{-2} + 2^{-3} = 0.375$, we have:
$$
\text{val} = (-1)^0 \times 2^{124 - 127} \times (1 + 0.375) = 0.171875
\text { val } = (-1)^0 \times 2^{124 - 127} \times (1 + 0.375) = 0.171875
$$
Now we can answer the initial question: **The representation of `float` includes an exponent bit, leading to a much larger range than `int`**. Based on the above calculation, the maximum positive number representable by `float` is approximately $2^{254 - 127} \times (2 - 2^{-23}) \approx 3.4 \times 10^{38}$, and the minimum negative number is obtained by switching the sign bit.
Now we can answer the initial question: **the representation of `float` includes an exponent bit, resulting in a range far greater than `int`**. According to the above calculation, the maximum positive number that `float` can represent is $2^{254 - 127} \times (2 - 2^{-23}) \approx 3.4 \times 10^{38}$, and the minimum negative number can be obtained by switching the sign bit.
**However, the trade-off for `float`'s expanded range is a sacrifice in precision**. The integer type `int` uses all 32 bits to represent the number, with values evenly distributed; but due to the exponent bit, the larger the value of a `float`, the greater the difference between adjacent numbers.
**Although floating-point number `float` expands the range, its side effect is sacrificing precision**. The integer type `int` uses all 32 bits to represent numbers, and the numbers are evenly distributed; however, due to the existence of the exponent bit, the larger the value of floating-point number `float`, the larger the difference between two adjacent numbers tends to be.
As shown in Table 3-2, exponent bits $\mathrm{E} = 0$ and $\mathrm{E} = 255$ have special meanings, **used to represent zero, infinity, $\mathrm{NaN}$, etc.**
@@ -151,12 +151,12 @@ As shown in Table 3-2, exponent bits $\mathrm{E} = 0$ and $\mathrm{E} = 255$ hav
| Exponent Bit E | Fraction Bit $\mathrm{N} = 0$ | Fraction Bit $\mathrm{N} \ne 0$ | Calculation Formula |
| ------------------ | ----------------------------- | ------------------------------- | ---------------------------------------------------------------------- |
| $0$ | $\pm 0$ | Subnormal Numbers | $(-1)^{\mathrm{S}} \times 2^{-126} \times (0.\mathrm{N})$ |
| $1, 2, \dots, 254$ | Normal Numbers | Normal Numbers | $(-1)^{\mathrm{S}} \times 2^{(\mathrm{E} -127)} \times (1.\mathrm{N})$ |
| $0$ | $\pm 0$ | Subnormal Number | $(-1)^{\mathrm{S}} \times 2^{-126} \times (0.\mathrm{N})$ |
| $1, 2, \dots, 254$ | Normal Number | Normal Number | $(-1)^{\mathrm{S}} \times 2^{(\mathrm{E} -127)} \times (1.\mathrm{N})$ |
| $255$ | $\pm \infty$ | $\mathrm{NaN}$ | |
</div>
It's worth noting that subnormal numbers significantly improve the precision of floating-point numbers. The smallest positive normal number is $2^{-126}$, and the smallest positive subnormal number is $2^{-126} \times 2^{-23}$.
It is worth noting that subnormal numbers significantly improve the precision of floating-point numbers. The smallest positive normal number is $2^{-126}$, and the smallest positive subnormal number is $2^{-126} \times 2^{-23}$.
Double-precision `double` also uses a similar representation method to `float`, which is not elaborated here for brevity.
Double-precision `double` also uses a representation method similar to `float`, which will not be elaborated here.
en/docs/chapter_data_structure/summary.md
+30 -30
View File
@@ -4,67 +4,67 @@ comments: true
# 3.5 &nbsp; Summary
### 1. &nbsp; Key review
### 1. &nbsp; Key Review
- Data structures can be categorized from two perspectives: logical structure and physical structure. Logical structure describes the logical relationships between data, while physical structure describes how data is stored in memory.
- Frequently used logical structures include linear structures, trees, and networks. We usually divide data structures into linear (arrays, linked lists, stacks, queues) and non-linear (trees, graphs, heaps) based on their logical structure. The implementation of hash tables may involve both linear and non-linear data structures.
- When a program is running, data is stored in memory. Each memory space has a corresponding address, and the program accesses data through these addresses.
- Physical structures can be divided into continuous space storage (arrays) and discrete space storage (linked lists). All data structures are implemented using arrays, linked lists, or a combination of both.
- The basic data types in computers include integers (`byte`, `short`, `int`, `long`), floating-point numbers (`float`, `double`), characters (`char`), and booleans (`bool`). The value range of a data type depends on its size and representation.
- Sign-magnitude, 1's complement, 2's complement are three methods of encoding integers in computers, and they can be converted into each other. The most significant bit of the sign-magnitude is the sign bit, and the remaining bits represent the value of the number.
- Integers are encoded by 2's complement in computers. The benefits of this representation include (i) the computer can unify the addition of positive and negative integers, (ii) no need to design special hardware circuits for subtraction, and (iii) no ambiguity of positive and negative zero.
- The encoding of floating-point numbers consists of 1 sign bit, 8 exponent bits, and 23 fraction bits. Due to the exponent bit, the range of floating-point numbers is much greater than that of integers, but at the cost of precision.
- ASCII is the earliest English character set, with 1 byte in length and a total of 127 characters. GBK is a popular Chinese character set, which includes more than 20,000 Chinese characters. Unicode aims to provide a complete character set standard that includes characters from various languages in the world, thus solving the garbled character problem caused by inconsistent character encoding methods.
- UTF-8 is the most popular and general Unicode encoding method. It is a variable-length encoding method with good scalability and space efficiency. UTF-16 and UTF-32 are fixed-length encoding methods. When encoding Chinese characters, UTF-16 takes up less space than UTF-8. Programming languages like Java and C# use UTF-16 encoding by default.
- Data structures can be classified from two perspectives: logical structure and physical structure. Logical structure describes the logical relationships between data elements, while physical structure describes how data is stored in computer memory.
- Common logical structures include linear, tree, and network structures. We typically classify data structures as linear (arrays, linked lists, stacks, queues) and non-linear (trees, graphs, heaps) based on their logical structure. The implementation of hash tables may involve both linear and non-linear data structures.
- When a program runs, data is stored in computer memory. Each memory space has a corresponding memory address, and the program accesses data through these memory addresses.
- Physical structures are primarily divided into contiguous space storage (arrays) and dispersed space storage (linked lists). All data structures are implemented using arrays, linked lists, or a combination of both.
- Basic data types in computers include integers `byte`, `short`, `int`, `long`, floating-point numbers `float`, `double`, characters `char`, and booleans `bool`. Their value ranges depend on the size of space they occupy and their representation method.
- Sign-magnitude, 1's complement, and 2's complement are three methods for encoding numbers in computers, and they can be converted into each other. The most significant bit of sign-magnitude is the sign bit, and the remaining bits represent the value of the number.
- Integers are stored in computers in 2's complement form. Under 2's complement representation, computers can treat the addition of positive and negative numbers uniformly, without needing to design special hardware circuits for subtraction, and there is no ambiguity of positive and negative zero.
- The encoding of floating-point numbers consists of 1 sign bit, 8 exponent bits, and 23 fraction bits. Due to the exponent bits, the range of floating-point numbers is much larger than that of integers, at the cost of sacrificing precision.
- ASCII is the earliest English character set, with a length of 1 byte, containing a total of 127 characters. GBK is a commonly used Chinese character set, containing over 20,000 Chinese characters. Unicode is committed to providing a complete character set standard, collecting characters from various languages around the world, thereby solving the garbled text problem caused by inconsistent character encoding methods.
- UTF-8 is the most popular Unicode encoding method, with excellent universality. It is a variable-length encoding method with good scalability, effectively improving storage space efficiency. UTF-16 and UTF-32 are fixed-length encoding methods. When encoding Chinese characters, UTF-16 occupies less space than UTF-8. Programming languages such as Java and C# use UTF-16 encoding by default.
### 2. &nbsp; Q & A
**Q**: Why does a hash table contain both linear and non-linear data structures?
**Q**: Why do hash tables contain both linear and non-linear data structures?
The underlying structure of a hash table is an array. To resolve hash collisions, we may use "chaining" (discussed in a later section, "Hash collision"): each bucket in the array points to a linked list, which may transform into a tree (usually a red-black tree) when its length is larger than a certain threshold.
From a storage perspective, the underlying structure of a hash table is an array, where each bucket might contain a value, a linked list, or a tree. Therefore, hash tables may contain both linear data structures (arrays, linked lists) and non-linear data structures (trees).
The underlying structure of a hash table is an array. To resolve hash collisions, we may use "chaining" (discussed in the subsequent "Hash Collision" section): each bucket in the array points to a linked list, which may be converted to a tree (usually a red-black tree) when the list length exceeds a certain threshold.
From a storage perspective, the underlying structure of a hash table is an array, where each bucket slot may contain a value, a linked list, or a tree. Therefore, hash tables may contain both linear data structures (arrays, linked lists) and non-linear data structures (trees).
**Q**: Is the length of the `char` type 1 byte?
The length of the `char` type is determined by the encoding method of the programming language. For example, Java, JavaScript, TypeScript, and C# all use UTF-16 encoding (to save Unicode code points), so the length of the `char` type is 2 bytes.
The length of the `char` type is determined by the encoding method used by the programming language. For example, Java, JavaScript, TypeScript, and C# all use UTF-16 encoding (to store Unicode code points), so the `char` type has a length of 2 bytes.
**Q**: Is there any ambiguity when we refer to array-based data structures as "static data structures"? The stack can also perform "dynamic" operations such as popping and pushing.
**Q**: Is there ambiguity in referring to array-based data structures as "static data structures"? Stacks can also perform "dynamic" operations such as push and pop.
The stack can implement dynamic data operations, but the data structure is still "static" (the length is fixed). Although array-based data structures can dynamically add or remove elements, their capacity is fixed. If the stack size exceeds the pre-allocated size, then the old array will be copied into a newly created and larger array.
Stacks can indeed implement dynamic data operations, but the data structure is still "static" (fixed length). Although array-based data structures can dynamically add or remove elements, their capacity is fixed. If the data volume exceeds the pre-allocated size, a new larger array needs to be created, and the contents of the old array must be copied to the new array.
**Q**: When building a stack (queue), its size is not specified, so why are they "static data structures"?
**Q**: When constructing a stack (queue), its size is not specified. Why are they "static data structures"?
In high-level programming languages, we do not need to manually specify the initial capacity of stacks (queues); this task is automatically completed within the class. For example, the initial capacity of Java's `ArrayList` is usually 10. Furthermore, the expansion operation is also completed automatically. See the subsequent "List" chapter for details.
In high-level programming languages, we do not need to manually specify the initial capacity of a stack (queue); this work is automatically completed within the class. For example, the initial capacity of Java's `ArrayList` is typically 10. Additionally, the expansion operation is also automatically implemented. See the subsequent "List" section for details.
**Q**：The method of converting the sign-magnitude to the 2's complement is "first negate and then add 1", so converting the 2's complement to the sign-magnitude should be its inverse operation "first subtract 1 and then negate".
However, the 2's complement can also be converted to the sign-magnitude through "first negate and then add 1", why is this?
**Q**: The method of converting sign-magnitude to 2's complement is "first negate then add 1". So converting 2's complement to sign-magnitude should be the inverse operation "first subtract 1 then negate". However, 2's complement can also be converted to sign-magnitude through "first negate then add 1". Why is this?
**A**：This is because the mutual conversion between the sign-magnitude and the 2's complement is equivalent to computing the "complement". We first define the complement: assuming $a + b = c$, then we say that $a$ is the complement of $b$ to $c$, and vice versa, $b$ is the complement of $a$ to $c$.
This is because the mutual conversion between sign-magnitude and 2's complement is actually the process of computing the "complement". Let us first define the complement: assuming $a + b = c$, then we say that $a$ is the complement of $b$ to $c$, and conversely, $b$ is the complement of $a$ to $c$.
Given a binary number $0010$ with length $n = 4$, if this number is the sign-magnitude (ignoring the sign bit), then its 2's complement can be obtained by "first negating and then adding 1":
Given an $n = 4$ bit binary number $0010$, if we treat this number as sign-magnitude (ignoring the sign bit), then its 2's complement can be obtained through "first negate then add 1":
$$
0010 \rightarrow 1101 \rightarrow 1110
$$
Observe that the sum of the sign-magnitude and the 2's complement is $0010 + 1110 = 10000$, i.e., the 2's complement $1110$ is the "complement" of the sign-magnitude $0010$ to $10000$. **This means that the above "first negate and then add 1" is equivalent to computing the complement to $10000$**.
We find that the sum of sign-magnitude and 2's complement is $0010 + 1110 = 10000$, which means the 2's complement $1110$ is the "complement" of sign-magnitude $0010$ to $10000$. **This means the above "first negate then add 1" is actually the process of computing the complement to $10000$**.
So, what is the "complement" of $1110$ to $10000$? We can still compute it by "negating first and then adding 1":
So, what is the "complement" of 2's complement $1110$ to $10000$? We can still use "first negate then add 1" to obtain it:
$$
1110 \rightarrow 0001 \rightarrow 0010
$$
In other words, the sign-magnitude and the 2's complement are each other's "complement" to $10000$, so "sign-magnitude to 2's complement" and "2's complement to sign-magnitude" can be implemented with the same operation (first negate and then add 1).
In other words, sign-magnitude and 2's complement are each other's "complement" to $10000$, so "sign-magnitude to 2's complement" and "2's complement to sign-magnitude" can be implemented using the same operation (first negate then add 1).
Of course, we can also use the inverse operation of "first negate and then add 1" to find the sign-magnitude of the 2's complement $1110$, that is, "first subtract 1 and then negate":
Of course, we can also use the inverse operation to find the sign-magnitude of 2's complement $1110$, that is, "first subtract 1 then negate":
$$
1110 \rightarrow 1101 \rightarrow 0010
$$
To sum up, "first negate and then add 1" and "first subtract 1 and then negate" are both computing the complement to $10000$, and they are equivalent.
In summary, both "first negate then add 1" and "first subtract 1 then negate" are computing the complement to $10000$, and they are equivalent.
Essentially, the "negate" operation is actually to find the complement to $1111$ (because `sign-magnitude + 1's complement = 1111` always holds); and the 1's complement plus 1 is equal to the 2's complement to $10000$.
Essentially, the "negate" operation is actually finding the complement to $1111$ (because `sign-magnitude + 1's complement = 1111` always holds); and adding 1 to the 1's complement yields the 2's complement, which is the complement to $10000$.
We take $n = 4$ as an example in the above, and it can be generalized to any binary number with any number of digits.
The above uses $n = 4$ as an example, and it can be generalized to binary numbers of any number of bits.
en/docs/chapter_divide_and_conquer/binary_search_recur.md
+295 -68
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@@ -2,47 +2,47 @@
comments: true
---
# 12.2 &nbsp; Divide and conquer search strategy
# 12.2 &nbsp; Divide and Conquer Search Strategy
We have learned that search algorithms fall into two main categories.
We have already learned that search algorithms are divided into two major categories.
- **Brute-force search**: It is implemented by traversing the data structure, with a time complexity of $O(n)$.
- **Adaptive search**: It utilizes a unique data organization form or prior information, and its time complexity can reach $O(\log n)$ or even $O(1)$.
- **Brute-force search**: Implemented by traversing the data structure, with a time complexity of $O(n)$.
- **Adaptive search**: Utilizes unique data organization forms or prior information, with time complexity reaching $O(\log n)$ or even $O(1)$.
In fact, **search algorithms with a time complexity of $O(\log n)$ are usually based on the divide-and-conquer strategy**, such as binary search and trees.
In fact, **search algorithms with time complexity of $O(\log n)$ are typically implemented based on the divide and conquer strategy**, such as binary search and trees.
- Each step of binary search divides the problem (searching for a target element in an array) into a smaller problem (searching for the target element in half of the array), continuing until the array is empty or the target element is found.
- Trees represent the divide-and-conquer idea, where in data structures like binary search trees, AVL trees, and heaps, the time complexity of various operations is $O(\log n)$.
- Trees are representative of the divide and conquer idea. In data structures such as binary search trees, AVL trees, and heaps, the time complexity of various operations is $O(\log n)$.
The divide-and-conquer strategy of binary search is as follows.
The divide and conquer strategy of binary search is as follows.
- **The problem can be divided**: Binary search recursively divides the original problem (searching in an array) into subproblems (searching in half of the array), achieved by comparing the middle element with the target element.
- **Subproblems are independent**: In binary search, each round handles one subproblem, unaffected by other subproblems.
- **The solutions of subproblems do not need to be merged**: Binary search aims to find a specific element, so there is no need to merge the solutions of subproblems. When a subproblem is solved, the original problem is also solved.
- **The problem can be decomposed**: Binary search recursively decomposes the original problem (searching in an array) into subproblems (searching in half of the array), achieved by comparing the middle element with the target element.
- **Subproblems are independent**: In binary search, each round only processes one subproblem, which is not affected by other subproblems.
- **Solutions of subproblems do not need to be merged**: Binary search aims to find a specific element, so there is no need to merge the solutions of subproblems. When a subproblem is solved, the original problem is also solved.
Divide-and-conquer can enhance search efficiency because brute-force search can only eliminate one option per round, **whereas divide-and-conquer can eliminate half of the options**.
Divide and conquer can improve search efficiency because brute-force search can only eliminate one option per round, **while divide and conquer search can eliminate half of the options per round**.
### 1. &nbsp; Implementing binary search based on divide-and-conquer
### 1. &nbsp; Implementing Binary Search Based on Divide and Conquer
In previous chapters, binary search was implemented based on iteration. Now, we implement it based on divide-and-conquer (recursion).
In previous sections, binary search was implemented based on iteration. Now we implement it based on divide and conquer (recursion).
!!! question
Given an ordered array `nums` of length $n$, where all elements are unique, please find the element `target`.
Given a sorted array `nums` of length $n$, where all elements are unique, find the element `target`.
From a divide-and-conquer perspective, we denote the subproblem corresponding to the search interval $[i, j]$ as $f(i, j)$.
From a divide and conquer perspective, we denote the subproblem corresponding to the search interval $[i, j]$ as $f(i, j)$.
Starting from the original problem $f(0, n-1)$, perform the binary search through the following steps.
Starting from the original problem $f(0, n-1)$, perform binary search through the following steps.
1. Calculate the midpoint $m$ of the search interval $[i, j]$, and use it to eliminate half of the search interval.
2. Recursively solve the subproblem reduced by half in size, which could be $f(i, m-1)$ or $f(m+1, j)$.
3. Repeat steps `1.` and `2.`, until `target` is found or the interval is empty and returns.
3. Repeat steps `1.` and `2.` until `target` is found or the interval is empty and return.
Figure 12-4 shows the divide-and-conquer process of binary search for element $6$ in an array.
Figure 12-4 shows the divide and conquer process of binary search for element $6$ in an array.
![The divide-and-conquer process of binary search](binary_search_recur.assets/binary_search_recur.png){ class="animation-figure" }
![Divide and conquer process of binary search](binary_search_recur.assets/binary_search_recur.png){ class="animation-figure" }
<p align="center"> Figure 12-4 &nbsp; The divide-and-conquer process of binary search </p>
<p align="center"> Figure 12-4 &nbsp; Divide and conquer process of binary search </p>
In the implementation code, we declare a recursive function `dfs()` to solve the problem $f(i, j)$:
@@ -51,25 +51,25 @@ In the implementation code, we declare a recursive function `dfs()` to solve the
```python title="binary_search_recur.py"
def dfs(nums: list[int], target: int, i: int, j: int) -> int:
"""Binary search: problem f(i, j)"""
# If the interval is empty, indicating no target element, return -1
# If the interval is empty, it means there is no target element, return -1
if i > j:
return -1
# Calculate midpoint index m
# Calculate the midpoint index m
m = (i + j) // 2
if nums[m] < target:
# Recursive subproblem f(m+1, j)
# Recursion subproblem f(m+1, j)
return dfs(nums, target, m + 1, j)
elif nums[m] > target:
# Recursive subproblem f(i, m-1)
# Recursion subproblem f(i, m-1)
return dfs(nums, target, i, m - 1)
else:
# Found the target element, thus return its index
# Found the target element, return its index
return m
def binary_search(nums: list[int], target: int) -> int:
"""Binary search"""
n = len(nums)
# Solve problem f(0, n-1)
# Solve the problem f(0, n-1)
return dfs(nums, target, 0, n - 1)
```
@@ -78,20 +78,20 @@ In the implementation code, we declare a recursive function `dfs()` to solve the
```cpp title="binary_search_recur.cpp"
/* Binary search: problem f(i, j) */
int dfs(vector<int> &nums, int target, int i, int j) {
// If the interval is empty, indicating no target element, return -1
// If the interval is empty, it means there is no target element, return -1
if (i > j) {
return -1;
}
// Calculate midpoint index m
int m = i + (j - i) / 2;
// Calculate the midpoint index m
int m = (i + j) / 2;
if (nums[m] < target) {
// Recursive subproblem f(m+1, j)
// Recursion subproblem f(m+1, j)
return dfs(nums, target, m + 1, j);
} else if (nums[m] > target) {
// Recursive subproblem f(i, m-1)
// Recursion subproblem f(i, m-1)
return dfs(nums, target, i, m - 1);
} else {
// Found the target element, thus return its index
// Found the target element, return its index
return m;
}
}
@@ -99,7 +99,7 @@ In the implementation code, we declare a recursive function `dfs()` to solve the
/* Binary search */
int binarySearch(vector<int> &nums, int target) {
int n = nums.size();
// Solve problem f(0, n-1)
// Solve the problem f(0, n-1)
return dfs(nums, target, 0, n - 1);
}
```
@@ -109,20 +109,20 @@ In the implementation code, we declare a recursive function `dfs()` to solve the
```java title="binary_search_recur.java"
/* Binary search: problem f(i, j) */
int dfs(int[] nums, int target, int i, int j) {
// If the interval is empty, indicating no target element, return -1
// If the interval is empty, it means there is no target element, return -1
if (i > j) {
return -1;
}
// Calculate midpoint index m
int m = i + (j - i) / 2;
// Calculate the midpoint index m
int m = (i + j) / 2;
if (nums[m] < target) {
// Recursive subproblem f(m+1, j)
// Recursion subproblem f(m+1, j)
return dfs(nums, target, m + 1, j);
} else if (nums[m] > target) {
// Recursive subproblem f(i, m-1)
// Recursion subproblem f(i, m-1)
return dfs(nums, target, i, m - 1);
} else {
// Found the target element, thus return its index
// Found the target element, return its index
return m;
}
}
@@ -130,7 +130,7 @@ In the implementation code, we declare a recursive function `dfs()` to solve the
/* Binary search */
int binarySearch(int[] nums, int target) {
int n = nums.length;
// Solve problem f(0, n-1)
// Solve the problem f(0, n-1)
return dfs(nums, target, 0, n - 1);
}
```
@@ -138,87 +138,314 @@ In the implementation code, we declare a recursive function `dfs()` to solve the
=== "C#"
```csharp title="binary_search_recur.cs"
[class]{binary_search_recur}-[func]{DFS}
/* Binary search: problem f(i, j) */
int DFS(int[] nums, int target, int i, int j) {
// If the interval is empty, it means there is no target element, return -1
if (i > j) {
return -1;
}
// Calculate the midpoint index m
int m = (i + j) / 2;
if (nums[m] < target) {
// Recursion subproblem f(m+1, j)
return DFS(nums, target, m + 1, j);
} else if (nums[m] > target) {
// Recursion subproblem f(i, m-1)
return DFS(nums, target, i, m - 1);
} else {
// Found the target element, return its index
return m;
}
}
[class]{binary_search_recur}-[func]{BinarySearch}
/* Binary search */
int BinarySearch(int[] nums, int target) {
int n = nums.Length;
// Solve the problem f(0, n-1)
return DFS(nums, target, 0, n - 1);
}
```
=== "Go"
```go title="binary_search_recur.go"
[class]{}-[func]{dfs}
/* Binary search: problem f(i, j) */
func dfs(nums []int, target, i, j int) int {
// If interval is empty, indicating no target element, return -1
if i > j {
return -1
}
// Calculate midpoint index
m := i + ((j - i) >> 1)
// Compare midpoint with target element
if nums[m] < target {
// If smaller, recurse on right half of array
// Recursion subproblem f(m+1, j)
return dfs(nums, target, m+1, j)
} else if nums[m] > target {
// If larger, recurse on left half of array
// Recursion subproblem f(i, m-1)
return dfs(nums, target, i, m-1)
} else {
// Found the target element, return its index
return m
}
}
[class]{}-[func]{binarySearch}
/* Binary search */
func binarySearch(nums []int, target int) int {
n := len(nums)
return dfs(nums, target, 0, n-1)
}
```
=== "Swift"
```swift title="binary_search_recur.swift"
[class]{}-[func]{dfs}
/* Binary search: problem f(i, j) */
func dfs(nums: [Int], target: Int, i: Int, j: Int) -> Int {
// If the interval is empty, it means there is no target element, return -1
if i > j {
return -1
}
// Calculate the midpoint index m
let m = (i + j) / 2
if nums[m] < target {
// Recursion subproblem f(m+1, j)
return dfs(nums: nums, target: target, i: m + 1, j: j)
} else if nums[m] > target {
// Recursion subproblem f(i, m-1)
return dfs(nums: nums, target: target, i: i, j: m - 1)
} else {
// Found the target element, return its index
return m
}
}
[class]{}-[func]{binarySearch}
/* Binary search */
func binarySearch(nums: [Int], target: Int) -> Int {
// Solve the problem f(0, n-1)
dfs(nums: nums, target: target, i: nums.startIndex, j: nums.endIndex - 1)
}
```
=== "JS"
```javascript title="binary_search_recur.js"
[class]{}-[func]{dfs}
/* Binary search: problem f(i, j) */
function dfs(nums, target, i, j) {
// If the interval is empty, it means there is no target element, return -1
if (i > j) {
return -1;
}
// Calculate the midpoint index m
const m = i + ((j - i) >> 1);
if (nums[m] < target) {
// Recursion subproblem f(m+1, j)
return dfs(nums, target, m + 1, j);
} else if (nums[m] > target) {
// Recursion subproblem f(i, m-1)
return dfs(nums, target, i, m - 1);
} else {
// Found the target element, return its index
return m;
}
}
[class]{}-[func]{binarySearch}
/* Binary search */
function binarySearch(nums, target) {
const n = nums.length;
// Solve the problem f(0, n-1)
return dfs(nums, target, 0, n - 1);
}
```
=== "TS"
```typescript title="binary_search_recur.ts"
[class]{}-[func]{dfs}
/* Binary search: problem f(i, j) */
function dfs(nums: number[], target: number, i: number, j: number): number {
// If the interval is empty, it means there is no target element, return -1
if (i > j) {
return -1;
}
// Calculate the midpoint index m
const m = i + ((j - i) >> 1);
if (nums[m] < target) {
// Recursion subproblem f(m+1, j)
return dfs(nums, target, m + 1, j);
} else if (nums[m] > target) {
// Recursion subproblem f(i, m-1)
return dfs(nums, target, i, m - 1);
} else {
// Found the target element, return its index
return m;
}
}
[class]{}-[func]{binarySearch}
/* Binary search */
function binarySearch(nums: number[], target: number): number {
const n = nums.length;
// Solve the problem f(0, n-1)
return dfs(nums, target, 0, n - 1);
}
```
=== "Dart"
```dart title="binary_search_recur.dart"
[class]{}-[func]{dfs}
/* Binary search: problem f(i, j) */
int dfs(List<int> nums, int target, int i, int j) {
// If the interval is empty, it means there is no target element, return -1
if (i > j) {
return -1;
}
// Calculate the midpoint index m
int m = (i + j) ~/ 2;
if (nums[m] < target) {
// Recursion subproblem f(m+1, j)
return dfs(nums, target, m + 1, j);
} else if (nums[m] > target) {
// Recursion subproblem f(i, m-1)
return dfs(nums, target, i, m - 1);
} else {
// Found the target element, return its index
return m;
}
}
[class]{}-[func]{binarySearch}
/* Binary search */
int binarySearch(List<int> nums, int target) {
int n = nums.length;
// Solve the problem f(0, n-1)
return dfs(nums, target, 0, n - 1);
}
```
=== "Rust"
```rust title="binary_search_recur.rs"
[class]{}-[func]{dfs}
/* Binary search: problem f(i, j) */
fn dfs(nums: &[i32], target: i32, i: i32, j: i32) -> i32 {
// If the interval is empty, it means there is no target element, return -1
if i > j {
return -1;
}
let m: i32 = i + (j - i) / 2;
if nums[m as usize] < target {
// Recursion subproblem f(m+1, j)
return dfs(nums, target, m + 1, j);
} else if nums[m as usize] > target {
// Recursion subproblem f(i, m-1)
return dfs(nums, target, i, m - 1);
} else {
// Found the target element, return its index
return m;
}
}
[class]{}-[func]{binary_search}
/* Binary search */
fn binary_search(nums: &[i32], target: i32) -> i32 {
let n = nums.len() as i32;
// Solve the problem f(0, n-1)
dfs(nums, target, 0, n - 1)
}
```
=== "C"
```c title="binary_search_recur.c"
[class]{}-[func]{dfs}
/* Binary search: problem f(i, j) */
int dfs(int nums[], int target, int i, int j) {
// If the interval is empty, it means there is no target element, return -1
if (i > j) {
return -1;
}
// Calculate the midpoint index m
int m = (i + j) / 2;
if (nums[m] < target) {
// Recursion subproblem f(m+1, j)
return dfs(nums, target, m + 1, j);
} else if (nums[m] > target) {
// Recursion subproblem f(i, m-1)
return dfs(nums, target, i, m - 1);
} else {
// Found the target element, return its index
return m;
}
}
[class]{}-[func]{binarySearch}
/* Binary search */
int binarySearch(int nums[], int target, int numsSize) {
int n = numsSize;
// Solve the problem f(0, n-1)
return dfs(nums, target, 0, n - 1);
}
```
=== "Kotlin"
```kotlin title="binary_search_recur.kt"
[class]{}-[func]{dfs}
/* Binary search: problem f(i, j) */
fun dfs(
nums: IntArray,
target: Int,
i: Int,
j: Int
): Int {
// If the interval is empty, it means there is no target element, return -1
if (i > j) {
return -1
}
// Calculate the midpoint index m
val m = (i + j) / 2
return if (nums[m] < target) {
// Recursion subproblem f(m+1, j)
dfs(nums, target, m + 1, j)
} else if (nums[m] > target) {
// Recursion subproblem f(i, m-1)
dfs(nums, target, i, m - 1)
} else {
// Found the target element, return its index
m
}
}
[class]{}-[func]{binarySearch}
/* Binary search */
fun binarySearch(nums: IntArray, target: Int): Int {
val n = nums.size
// Solve the problem f(0, n-1)
return dfs(nums, target, 0, n - 1)
}
```
=== "Ruby"
```ruby title="binary_search_recur.rb"
[class]{}-[func]{dfs}
### Binary search: problem f(i, j) ###
def dfs(nums, target, i, j)
# If the interval is empty, it means there is no target element, return -1
return -1 if i > j
# Calculate the midpoint index m
m = (i + j) / 2
[class]{}-[func]{binary_search}
```
=== "Zig"
```zig title="binary_search_recur.zig"
[class]{}-[func]{dfs}
[class]{}-[func]{binarySearch}
if nums[m] < target
# Recursion subproblem f(m+1, j)
return dfs(nums, target, m + 1, j)
elsif nums[m] > target
# Recursion subproblem f(i, m-1)
return dfs(nums, target, i, m - 1)
else
# Found the target element, return its index
return m
end
end
### Binary search ###
def binary_search(nums, target)
n = nums.length
# Solve the problem f(0, n-1)
dfs(nums, target, 0, n - 1)
end
```
en/docs/chapter_divide_and_conquer/build_binary_tree_problem.md
+349 -93
View File
@@ -2,74 +2,74 @@
comments: true
---
# 12.3 &nbsp; Building a binary tree problem
# 12.3 &nbsp; Building a Binary Tree Problem
!!! question
Given the pre-order traversal `preorder` sequence and the in-order traversal `inorder` sequence of a binary tree, construct the binary tree and return its root node. Assume there are no duplicate node values in the binary tree (as shown in Figure 12-5).
Given the preorder traversal `preorder` and inorder traversal `inorder` of a binary tree, construct the binary tree and return the root node of the binary tree. Assume there are no duplicate node values in the binary tree (as shown in Figure 12-5).
![Example data for building a binary tree](build_binary_tree_problem.assets/build_tree_example.png){ class="animation-figure" }
<p align="center"> Figure 12-5 &nbsp; Example data for building a binary tree </p>
### 1. &nbsp; Determining if it is a divide-and-conquer problem
### 1. &nbsp; Determining If It Is a Divide and Conquer Problem
The original problem of building a binary tree from the `preorder` and the `inorder` sequences is a typical divide-and-conquer problem.
The original problem is defined as constructing a binary tree from `preorder` and `inorder`, which is a typical divide and conquer problem.
- **The problem can be decomposed**: From the perspective of divide-and-conquer, we can divide the original problem into two subproblems—building the left subtree and building the right subtree—plus one operation of initializing the root node. For each subtree (subproblem), we continue applying the same approach, partitioning it into smaller subtrees (subproblems), until reaching the smallest subproblem (an empty subtree).
- **The subproblems are independent**: The left and right subtrees do not overlap. When building the left subtree, we only need the segments of the in-order and pre-order traversals that correspond to the left subtree. The same approach applies to the right subtree.
- **Solutions to subproblems can be combined**: Once we have constructed the left and right subtrees (the subproblem solutions), we can attach them to the root node to obtain the solution to the original problem.
- **The problem can be decomposed**: From a divide and conquer perspective, we can divide the original problem into two subproblems: constructing the left subtree and constructing the right subtree, plus one operation: initializing the root node. For each subtree (subproblem), we can still reuse the above division method, dividing it into smaller subtrees (subproblems) until the smallest subproblem (empty subtree) is reached.
- **Subproblems are independent**: The left and right subtrees are independent of each other; there is no overlap between them. When constructing the left subtree, we only need to focus on the parts of the inorder and preorder traversals corresponding to the left subtree. The same applies to the right subtree.
- **Solutions of subproblems can be merged**: Once we have the left and right subtrees (solutions of subproblems), we can link them to the root node to obtain the solution to the original problem.
### 2. &nbsp; How to divide the subtrees
### 2. &nbsp; How to Divide Subtrees
Based on the above analysis, this problem can be solved using divide-and-conquer. **However, how do we use the pre-order traversal `preorder` sequence and the in-order traversal `inorder` sequence to divide the left and right subtrees?**
Based on the above analysis, this problem can be solved using divide and conquer, **but how do we divide the left and right subtrees through the preorder traversal `preorder` and inorder traversal `inorder`**?
By definition, both the `preorder` and `inorder` sequences can be divided into three parts:
According to the definition, both `preorder` and `inorder` can be divided into three parts.
- Pre-order traversal: `[ Root | Left Subtree | Right Subtree ]`. For example, in the figure, the tree corresponds to `[ 3 | 9 | 2 1 7 ]`.
- In-order traversal: `[ Left Subtree | Root | Right Subtree ]`. For example, in the figure, the tree corresponds to `[ 9 | 3 | 1 2 7 ]`.
- Preorder traversal: `[ Root Node | Left Subtree | Right Subtree ]`, for example, the tree in Figure 12-5 corresponds to `[ 3 | 9 | 2 1 7 ]`.
- Inorder traversal: `[ Left Subtree | Root Node ｜ Right Subtree ]`, for example, the tree in Figure 12-5 corresponds to `[ 9 | 3 | 1 2 7 ]`.
Using the data from the preceding figure, we can follow the steps shown in the next figure to obtain the division results:
Using the data from the figure above as an example, we can obtain the division results through the steps shown in Figure 12-6.
1. The first element 3 in the pre-order traversal is the value of the root node.
2. Find the index of the root node 3 in the `inorder` sequence, and use this index to split `inorder` into `[ 9 | 3 ｜ 1 2 7 ]`.
3. According to the split of the `inorder` sequence, it is straightforward to determine that the left and right subtrees contain 1 and 3 nodes, respectively, so we can split the `preorder` sequence into `[ 3 | 9 | 2 1 7 ]` accordingly.
1. The first element 3 in the preorder traversal is the value of the root node.
2. Find the index of root node 3 in `inorder`, and use this index to divide `inorder` into `[ 9 | 3 ｜ 1 2 7 ]`.
3. Based on the division result of `inorder`, it is easy to determine that the left and right subtrees have 1 and 3 nodes respectively, allowing us to divide `preorder` into `[ 3 | 9 | 2 1 7 ]`.
![Dividing the subtrees in pre-order and in-order traversals](build_binary_tree_problem.assets/build_tree_preorder_inorder_division.png){ class="animation-figure" }
![Dividing subtrees in preorder and inorder traversals](build_binary_tree_problem.assets/build_tree_preorder_inorder_division.png){ class="animation-figure" }
<p align="center"> Figure 12-6 &nbsp; Dividing the subtrees in pre-order and in-order traversals </p>
<p align="center"> Figure 12-6 &nbsp; Dividing subtrees in preorder and inorder traversals </p>
### 3. &nbsp; Describing subtree ranges based on variables
### 3. &nbsp; Describing Subtree Intervals Based on Variables
Based on the above division method, **we have now obtained the index ranges of the root, left subtree, and right subtree in the `preorder` and `inorder` sequences**. To describe these index ranges, we use several pointer variables.
Based on the above division method, **we have obtained the index intervals of the root node, left subtree, and right subtree in `preorder` and `inorder`**. To describe these index intervals, we need to use several pointer variables.
- Let the index of the current tree's root node in the `preorder` sequence be denoted as $i$.
- Let the index of the current tree's root node in the `inorder` sequence be denoted as $m$.
- Let the index range of the current tree in the `inorder` sequence be denoted as $[l, r]$.
- Denote the index of the current tree's root node in `preorder` as $i$.
- Denote the index of the current tree's root node in `inorder` as $m$.
- Denote the index interval of the current tree in `inorder` as $[l, r]$.
As shown in Table 12-1, these variables represent the root node’s index in the `preorder` sequence and the index ranges of the subtrees in the `inorder` sequence.
As shown in Table 12-1, through these variables we can represent the index of the root node in `preorder` and the index intervals of the subtrees in `inorder`.
<p align="center"> Table 12-1 &nbsp; Indexes of the root node and subtrees in pre-order and in-order traversals </p>
<p align="center"> Table 12-1 &nbsp; Indices of root node and subtrees in preorder and inorder traversals </p>
<div class="center-table" markdown>
| | Root node index in `preorder` | Subtree index range in `inorder` |
| ------------- | ----------------------------- | ----------------------------------- |
| Current tree | $i$ | $[l, r]$ |
| Left subtree | $i + 1$ | $[l, m-1]$ |
| Right subtree | $i + 1 + (m - l)$ | $[m+1, r]$ |
| | Root node index in `preorder` | Subtree index interval in `inorder` |
| ------------ | ----------------------------- | ----------------------------------- |
| Current tree | $i$ | $[l, r]$ |
| Left subtree | $i + 1$ | $[l, m-1]$ |
| Right subtree| $i + 1 + (m - l)$ | $[m+1, r]$ |
</div>
Please note that $(m-l)$ in the right subtree root index represents "the number of nodes in the left subtree." It may help to consult Figure 12-7 for a clearer understanding.
Please note that $(m-l)$ in the right subtree root node index means "the number of nodes in the left subtree". It is recommended to understand this in conjunction with Figure 12-7.
![Indexes of the root node and left and right subtrees](build_binary_tree_problem.assets/build_tree_division_pointers.png){ class="animation-figure" }
![Index interval representation of root node and left and right subtrees](build_binary_tree_problem.assets/build_tree_division_pointers.png){ class="animation-figure" }
<p align="center"> Figure 12-7 &nbsp; Indexes of the root node and left and right subtrees </p>
<p align="center"> Figure 12-7 &nbsp; Index interval representation of root node and left and right subtrees </p>
### 4. &nbsp; Code implementation
### 4. &nbsp; Code Implementation
To improve the efficiency of querying $m$, we use a hash table `hmap` to store the mapping from elements in the `inorder` sequence to their indexes:
To improve the efficiency of querying $m$, we use a hash table `hmap` to store the mapping from elements in the `inorder` array to their indices:
=== "Python"
@@ -81,24 +81,24 @@ To improve the efficiency of querying $m$, we use a hash table `hmap` to store t
l: int,
r: int,
) -> TreeNode | None:
"""Build binary tree: Divide and conquer"""
# Terminate when subtree interval is empty
"""Build binary tree: divide and conquer"""
# Terminate when the subtree interval is empty
if r - l < 0:
return None
# Initialize root node
# Initialize the root node
root = TreeNode(preorder[i])
# Query m to divide left and right subtrees
# Query m to divide the left and right subtrees
m = inorder_map[preorder[i]]
# Subproblem: build left subtree
# Subproblem: build the left subtree
root.left = dfs(preorder, inorder_map, i + 1, l, m - 1)
# Subproblem: build right subtree
# Subproblem: build the right subtree
root.right = dfs(preorder, inorder_map, i + 1 + m - l, m + 1, r)
# Return root node
# Return the root node
return root
def build_tree(preorder: list[int], inorder: list[int]) -> TreeNode | None:
"""Build binary tree"""
# Initialize hash table, storing in-order elements to indices mapping
# Initialize hash map, storing the mapping from inorder elements to indices
inorder_map = {val: i for i, val in enumerate(inorder)}
root = dfs(preorder, inorder_map, 0, 0, len(inorder) - 1)
return root
@@ -107,26 +107,26 @@ To improve the efficiency of querying $m$, we use a hash table `hmap` to store t
=== "C++"
```cpp title="build_tree.cpp"
/* Build binary tree: Divide and conquer */
/* Build binary tree: divide and conquer */
TreeNode *dfs(vector<int> &preorder, unordered_map<int, int> &inorderMap, int i, int l, int r) {
// Terminate when subtree interval is empty
// Terminate when the subtree interval is empty
if (r - l < 0)
return NULL;
// Initialize root node
// Initialize the root node
TreeNode *root = new TreeNode(preorder[i]);
// Query m to divide left and right subtrees
// Query m to divide the left and right subtrees
int m = inorderMap[preorder[i]];
// Subproblem: build left subtree
// Subproblem: build the left subtree
root->left = dfs(preorder, inorderMap, i + 1, l, m - 1);
// Subproblem: build right subtree
// Subproblem: build the right subtree
root->right = dfs(preorder, inorderMap, i + 1 + m - l, m + 1, r);
// Return root node
// Return the root node
return root;
}
/* Build binary tree */
TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {
// Initialize hash table, storing in-order elements to indices mapping
// Initialize hash map, storing the mapping from inorder elements to indices
unordered_map<int, int> inorderMap;
for (int i = 0; i < inorder.size(); i++) {
inorderMap[inorder[i]] = i;
@@ -139,26 +139,26 @@ To improve the efficiency of querying $m$, we use a hash table `hmap` to store t
=== "Java"
```java title="build_tree.java"
/* Build binary tree: Divide and conquer */
/* Build binary tree: divide and conquer */
TreeNode dfs(int[] preorder, Map<Integer, Integer> inorderMap, int i, int l, int r) {
// Terminate when subtree interval is empty
// Terminate when the subtree interval is empty
if (r - l < 0)
return null;
// Initialize root node
// Initialize the root node
TreeNode root = new TreeNode(preorder[i]);
// Query m to divide left and right subtrees
// Query m to divide the left and right subtrees
int m = inorderMap.get(preorder[i]);
// Subproblem: build left subtree
// Subproblem: build the left subtree
root.left = dfs(preorder, inorderMap, i + 1, l, m - 1);
// Subproblem: build right subtree
// Subproblem: build the right subtree
root.right = dfs(preorder, inorderMap, i + 1 + m - l, m + 1, r);
// Return root node
// Return the root node
return root;
}
/* Build binary tree */
TreeNode buildTree(int[] preorder, int[] inorder) {
// Initialize hash table, storing in-order elements to indices mapping
// Initialize hash map, storing the mapping from inorder elements to indices
Map<Integer, Integer> inorderMap = new HashMap<>();
for (int i = 0; i < inorder.length; i++) {
inorderMap.put(inorder[i], i);
@@ -171,92 +171,348 @@ To improve the efficiency of querying $m$, we use a hash table `hmap` to store t
=== "C#"
```csharp title="build_tree.cs"
[class]{build_tree}-[func]{DFS}
/* Build binary tree: divide and conquer */
TreeNode? DFS(int[] preorder, Dictionary<int, int> inorderMap, int i, int l, int r) {
// Terminate when the subtree interval is empty
if (r - l < 0)
return null;
// Initialize the root node
TreeNode root = new(preorder[i]);
// Query m to divide the left and right subtrees
int m = inorderMap[preorder[i]];
// Subproblem: build the left subtree
root.left = DFS(preorder, inorderMap, i + 1, l, m - 1);
// Subproblem: build the right subtree
root.right = DFS(preorder, inorderMap, i + 1 + m - l, m + 1, r);
// Return the root node
return root;
}
[class]{build_tree}-[func]{BuildTree}
/* Build binary tree */
TreeNode? BuildTree(int[] preorder, int[] inorder) {
// Initialize hash map, storing the mapping from inorder elements to indices
Dictionary<int, int> inorderMap = [];
for (int i = 0; i < inorder.Length; i++) {
inorderMap.TryAdd(inorder[i], i);
}
TreeNode? root = DFS(preorder, inorderMap, 0, 0, inorder.Length - 1);
return root;
}
```
=== "Go"
```go title="build_tree.go"
[class]{}-[func]{dfsBuildTree}
/* Build binary tree: divide and conquer */
func dfsBuildTree(preorder []int, inorderMap map[int]int, i, l, r int) *TreeNode {
// Terminate when the subtree interval is empty
if r-l < 0 {
return nil
}
// Initialize the root node
root := NewTreeNode(preorder[i])
// Query m to divide the left and right subtrees
m := inorderMap[preorder[i]]
// Subproblem: build the left subtree
root.Left = dfsBuildTree(preorder, inorderMap, i+1, l, m-1)
// Subproblem: build the right subtree
root.Right = dfsBuildTree(preorder, inorderMap, i+1+m-l, m+1, r)
// Return the root node
return root
}
[class]{}-[func]{buildTree}
/* Build binary tree */
func buildTree(preorder, inorder []int) *TreeNode {
// Initialize hash map, storing the mapping from inorder elements to indices
inorderMap := make(map[int]int, len(inorder))
for i := 0; i < len(inorder); i++ {
inorderMap[inorder[i]] = i
}
root := dfsBuildTree(preorder, inorderMap, 0, 0, len(inorder)-1)
return root
}
```
=== "Swift"
```swift title="build_tree.swift"
[class]{}-[func]{dfs}
/* Build binary tree: divide and conquer */
func dfs(preorder: [Int], inorderMap: [Int: Int], i: Int, l: Int, r: Int) -> TreeNode? {
// Terminate when the subtree interval is empty
if r - l < 0 {
return nil
}
// Initialize the root node
let root = TreeNode(x: preorder[i])
// Query m to divide the left and right subtrees
let m = inorderMap[preorder[i]]!
// Subproblem: build the left subtree
root.left = dfs(preorder: preorder, inorderMap: inorderMap, i: i + 1, l: l, r: m - 1)
// Subproblem: build the right subtree
root.right = dfs(preorder: preorder, inorderMap: inorderMap, i: i + 1 + m - l, l: m + 1, r: r)
// Return the root node
return root
}
[class]{}-[func]{buildTree}
/* Build binary tree */
func buildTree(preorder: [Int], inorder: [Int]) -> TreeNode? {
// Initialize hash map, storing the mapping from inorder elements to indices
let inorderMap = inorder.enumerated().reduce(into: [:]) { $0[$1.element] = $1.offset }
return dfs(preorder: preorder, inorderMap: inorderMap, i: inorder.startIndex, l: inorder.startIndex, r: inorder.endIndex - 1)
}
```
=== "JS"
```javascript title="build_tree.js"
[class]{}-[func]{dfs}
/* Build binary tree: divide and conquer */
function dfs(preorder, inorderMap, i, l, r) {
// Terminate when the subtree interval is empty
if (r - l < 0) return null;
// Initialize the root node
const root = new TreeNode(preorder[i]);
// Query m to divide the left and right subtrees
const m = inorderMap.get(preorder[i]);
// Subproblem: build the left subtree
root.left = dfs(preorder, inorderMap, i + 1, l, m - 1);
// Subproblem: build the right subtree
root.right = dfs(preorder, inorderMap, i + 1 + m - l, m + 1, r);
// Return the root node
return root;
}
[class]{}-[func]{buildTree}
/* Build binary tree */
function buildTree(preorder, inorder) {
// Initialize hash map, storing the mapping from inorder elements to indices
let inorderMap = new Map();
for (let i = 0; i < inorder.length; i++) {
inorderMap.set(inorder[i], i);
}
const root = dfs(preorder, inorderMap, 0, 0, inorder.length - 1);
return root;
}
```
=== "TS"
```typescript title="build_tree.ts"
[class]{}-[func]{dfs}
/* Build binary tree: divide and conquer */
function dfs(
preorder: number[],
inorderMap: Map<number, number>,
i: number,
l: number,
r: number
): TreeNode | null {
// Terminate when the subtree interval is empty
if (r - l < 0) return null;
// Initialize the root node
const root: TreeNode = new TreeNode(preorder[i]);
// Query m to divide the left and right subtrees
const m = inorderMap.get(preorder[i]);
// Subproblem: build the left subtree
root.left = dfs(preorder, inorderMap, i + 1, l, m - 1);
// Subproblem: build the right subtree
root.right = dfs(preorder, inorderMap, i + 1 + m - l, m + 1, r);
// Return the root node
return root;
}
[class]{}-[func]{buildTree}
/* Build binary tree */
function buildTree(preorder: number[], inorder: number[]): TreeNode | null {
// Initialize hash map, storing the mapping from inorder elements to indices
let inorderMap = new Map<number, number>();
for (let i = 0; i < inorder.length; i++) {
inorderMap.set(inorder[i], i);
}
const root = dfs(preorder, inorderMap, 0, 0, inorder.length - 1);
return root;
}
```
=== "Dart"
```dart title="build_tree.dart"
[class]{}-[func]{dfs}
/* Build binary tree: divide and conquer */
TreeNode? dfs(
List<int> preorder,
Map<int, int> inorderMap,
int i,
int l,
int r,
) {
// Terminate when the subtree interval is empty
if (r - l < 0) {
return null;
}
// Initialize the root node
TreeNode? root = TreeNode(preorder[i]);
// Query m to divide the left and right subtrees
int m = inorderMap[preorder[i]]!;
// Subproblem: build the left subtree
root.left = dfs(preorder, inorderMap, i + 1, l, m - 1);
// Subproblem: build the right subtree
root.right = dfs(preorder, inorderMap, i + 1 + m - l, m + 1, r);
// Return the root node
return root;
}
[class]{}-[func]{buildTree}
/* Build binary tree */
TreeNode? buildTree(List<int> preorder, List<int> inorder) {
// Initialize hash map, storing the mapping from inorder elements to indices
Map<int, int> inorderMap = {};
for (int i = 0; i < inorder.length; i++) {
inorderMap[inorder[i]] = i;
}
TreeNode? root = dfs(preorder, inorderMap, 0, 0, inorder.length - 1);
return root;
}
```
=== "Rust"
```rust title="build_tree.rs"
[class]{}-[func]{dfs}
/* Build binary tree: divide and conquer */
fn dfs(
preorder: &[i32],
inorder_map: &HashMap<i32, i32>,
i: i32,
l: i32,
r: i32,
) -> Option<Rc<RefCell<TreeNode>>> {
// Terminate when the subtree interval is empty
if r - l < 0 {
return None;
}
// Initialize the root node
let root = TreeNode::new(preorder[i as usize]);
// Query m to divide the left and right subtrees
let m = inorder_map.get(&preorder[i as usize]).unwrap();
// Subproblem: build the left subtree
root.borrow_mut().left = dfs(preorder, inorder_map, i + 1, l, m - 1);
// Subproblem: build the right subtree
root.borrow_mut().right = dfs(preorder, inorder_map, i + 1 + m - l, m + 1, r);
// Return the root node
Some(root)
}
[class]{}-[func]{build_tree}
/* Build binary tree */
fn build_tree(preorder: &[i32], inorder: &[i32]) -> Option<Rc<RefCell<TreeNode>>> {
// Initialize hash map, storing the mapping from inorder elements to indices
let mut inorder_map: HashMap<i32, i32> = HashMap::new();
for i in 0..inorder.len() {
inorder_map.insert(inorder[i], i as i32);
}
let root = dfs(preorder, &inorder_map, 0, 0, inorder.len() as i32 - 1);
root
}
```
=== "C"
```c title="build_tree.c"
[class]{}-[func]{dfs}
/* Build binary tree: divide and conquer */
TreeNode *dfs(int *preorder, int *inorderMap, int i, int l, int r, int size) {
// Terminate when the subtree interval is empty
if (r - l < 0)
return NULL;
// Initialize the root node
TreeNode *root = (TreeNode *)malloc(sizeof(TreeNode));
root->val = preorder[i];
root->left = NULL;
root->right = NULL;
// Query m to divide the left and right subtrees
int m = inorderMap[preorder[i]];
// Subproblem: build the left subtree
root->left = dfs(preorder, inorderMap, i + 1, l, m - 1, size);
// Subproblem: build the right subtree
root->right = dfs(preorder, inorderMap, i + 1 + m - l, m + 1, r, size);
// Return the root node
return root;
}
[class]{}-[func]{buildTree}
/* Build binary tree */
TreeNode *buildTree(int *preorder, int preorderSize, int *inorder, int inorderSize) {
// Initialize hash map, storing the mapping from inorder elements to indices
int *inorderMap = (int *)malloc(sizeof(int) * MAX_SIZE);
for (int i = 0; i < inorderSize; i++) {
inorderMap[inorder[i]] = i;
}
TreeNode *root = dfs(preorder, inorderMap, 0, 0, inorderSize - 1, inorderSize);
free(inorderMap);
return root;
}
```
=== "Kotlin"
```kotlin title="build_tree.kt"
[class]{}-[func]{dfs}
/* Build binary tree: divide and conquer */
fun dfs(
preorder: IntArray,
inorderMap: Map<Int?, Int?>,
i: Int,
l: Int,
r: Int
): TreeNode? {
// Terminate when the subtree interval is empty
if (r - l < 0) return null
// Initialize the root node
val root = TreeNode(preorder[i])
// Query m to divide the left and right subtrees
val m = inorderMap[preorder[i]]!!
// Subproblem: build the left subtree
root.left = dfs(preorder, inorderMap, i + 1, l, m - 1)
// Subproblem: build the right subtree
root.right = dfs(preorder, inorderMap, i + 1 + m - l, m + 1, r)
// Return the root node
return root
}
[class]{}-[func]{buildTree}
/* Build binary tree */
fun buildTree(preorder: IntArray, inorder: IntArray): TreeNode? {
// Initialize hash map, storing the mapping from inorder elements to indices
val inorderMap = HashMap<Int?, Int?>()
for (i in inorder.indices) {
inorderMap[inorder[i]] = i
}
val root = dfs(preorder, inorderMap, 0, 0, inorder.size - 1)
return root
}
```
=== "Ruby"
```ruby title="build_tree.rb"
[class]{}-[func]{dfs}
### Build binary tree: divide and conquer ###
def dfs(preorder, inorder_map, i, l, r)
# Terminate when the subtree interval is empty
return if r - l < 0
[class]{}-[func]{build_tree}
# Initialize the root node
root = TreeNode.new(preorder[i])
# Query m to divide the left and right subtrees
m = inorder_map[preorder[i]]
# Subproblem: build the left subtree
root.left = dfs(preorder, inorder_map, i + 1, l, m - 1)
# Subproblem: build the right subtree
root.right = dfs(preorder, inorder_map, i + 1 + m - l, m + 1, r)
# Return the root node
root
end
### Build binary tree ###
def build_tree(preorder, inorder)
# Initialize hash map, storing the mapping from inorder elements to indices
inorder_map = {}
inorder.each_with_index { |val, i| inorder_map[val] = i }
dfs(preorder, inorder_map, 0, 0, inorder.length - 1)
end
```
=== "Zig"
```zig title="build_tree.zig"
[class]{}-[func]{dfs}
[class]{}-[func]{buildTree}
```
Figure 12-8 shows the recursive process of building the binary tree. Each node is created during the "descending" phase of the recursion, and each edge (reference) is formed during the "ascending" phase.
Figure 12-8 shows the recursive process of building the binary tree. Each node is established during the downward "recursion" process, while each edge (reference) is established during the upward "return" process.
=== "<1>"
![Recursive process of building a binary tree](build_binary_tree_problem.assets/built_tree_step1.png){ class="animation-figure" }
@@ -287,12 +543,12 @@ Figure 12-8 shows the recursive process of building the binary tree. Each node i
<p align="center"> Figure 12-8 &nbsp; Recursive process of building a binary tree </p>
Each recursive function's division of the `preorder` and `inorder` sequences is illustrated in Figure 12-9.
The division results of the preorder traversal `preorder` and inorder traversal `inorder` within each recursive function are shown in Figure 12-9.
![Division in each recursive function](build_binary_tree_problem.assets/built_tree_overall.png){ class="animation-figure" }
![Division results in each recursive function](build_binary_tree_problem.assets/built_tree_overall.png){ class="animation-figure" }
<p align="center"> Figure 12-9 &nbsp; Division in each recursive function </p>
<p align="center"> Figure 12-9 &nbsp; Division results in each recursive function </p>
Assuming the binary tree has $n$ nodes, initializing each node (calling the recursive function `dfs()`) takes $O(1)$ time. **Therefore, the overall time complexity is $O(n)$**.
Let the number of nodes in the tree be $n$. Initializing each node (executing one recursive function `dfs()`) takes $O(1)$ time. **Therefore, the overall time complexity is $O(n)$**.
Because the hash table stores the mapping from `inorder` elements to their indexes, it requires $O(n)$ space. In the worst case, if the binary tree degenerates into a linked list, the recursive depth can reach $n$, consuming $O(n)$ stack space. **Hence, the overall space complexity is $O(n)$**.
The hash table stores the mapping from `inorder` elements to their indices, with a space complexity of $O(n)$. In the worst case, when the binary tree degenerates into a linked list, the recursion depth reaches $n$, using $O(n)$ stack frame space. **Therefore, the overall space complexity is $O(n)$**.
en/docs/chapter_divide_and_conquer/divide_and_conquer.md
+50 -50
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comments: true
---
# 12.1 &nbsp; Divide and conquer algorithms
# 12.1 &nbsp; Divide and Conquer Algorithms
<u>Divide and conquer</u> is an important and popular algorithm strategy. As the name suggests, the algorithm is typically implemented recursively and consists of two steps: "divide" and "conquer".
<u>Divide and conquer</u> is a very important and common algorithm strategy. Divide and conquer is typically implemented based on recursion, consisting of two steps: "divide" and "conquer".
1. **Divide (partition phase)**: Recursively break down the original problem into two or more smaller sub-problems until the smallest sub-problem is reached.
2. **Conquer (merge phase)**: Starting from the smallest sub-problem with known solution, we construct the solution to the original problem by merging the solutions of sub-problems in a bottom-up manner.
1. **Divide (partition phase)**: Recursively divide the original problem into two or more subproblems until the smallest subproblem is reached.
2. **Conquer (merge phase)**: Starting from the smallest subproblems with known solutions, merge the solutions of subproblems from bottom to top to construct the solution to the original problem.
As shown in Figure 12-1, "merge sort" is one of the typical applications of the divide and conquer strategy.
1. **Divide**: Recursively divide the original array (original problem) into two sub-arrays (sub-problems), until the sub-array has only one element (smallest sub-problem).
2. **Conquer**: Merge the ordered sub-arrays (solutions to the sub-problems) from bottom to top to obtain an ordered original array (solution to the original problem).
1. **Divide**: Recursively divide the original array (original problem) into two subarrays (subproblems) until the subarray has only one element (smallest subproblem).
2. **Conquer**: Merge the sorted subarrays (solutions to subproblems) from bottom to top to obtain a sorted original array (solution to the original problem).
![Merge sort's divide and conquer strategy](divide_and_conquer.assets/divide_and_conquer_merge_sort.png){ class="animation-figure" }
![Divide and conquer strategy of merge sort](divide_and_conquer.assets/divide_and_conquer_merge_sort.png){ class="animation-figure" }
<p align="center"> Figure 12-1 &nbsp; Merge sort's divide and conquer strategy </p>
<p align="center"> Figure 12-1 &nbsp; Divide and conquer strategy of merge sort </p>
## 12.1.1 &nbsp; How to identify divide and conquer problems
## 12.1.1 &nbsp; How to Determine Divide and Conquer Problems
Whether a problem is suitable for a divide-and-conquer solution can usually be decided based on the following criteria.
Whether a problem is suitable for solving with divide and conquer can usually be determined based on the following criteria.
1. **The problem can be broken down into smaller ones**: The original problem can be divided into smaller, similar sub-problems and such process can be recursively done in the same manner.
2. **Sub-problems are independent**: There is no overlap between sub-problems, and they are independent and can be solved separately.
3. **Solutions to sub-problems can be merged**: The solution to the original problem is derived by combining the solutions of the sub-problems.
1. **The problem can be decomposed**: The original problem can be divided into smaller, similar subproblems, and can be recursively divided in the same way.
2. **Subproblems are independent**: There is no overlap between subproblems, they are independent of each other and can be solved independently.
3. **Solutions of subproblems can be merged**: The solution to the original problem is obtained by merging the solutions of subproblems.
Clearly, merge sort meets these three criteria.
Clearly, merge sort satisfies these three criteria.
1. **The problem can be broken down into smaller ones**: Recursively divide the array (original problem) into two sub-arrays (sub-problems).
2. **Sub-problems are independent**: Each sub-array can be sorted independently (sub-problems can be solved independently).
3. **Solutions to sub-problems can be merged**: Two ordered sub-arrays (solutions to the sub-problems) can be merged into one ordered array (solution to the original problem).
1. **The problem can be decomposed**: Recursively divide the array (original problem) into two subarrays (subproblems).
2. **Subproblems are independent**: Each subarray can be sorted independently (subproblems can be solved independently).
3. **Solutions of subproblems can be merged**: Two sorted subarrays (solutions of subproblems) can be merged into one sorted array (solution of the original problem).
## 12.1.2 &nbsp; Improve efficiency through divide and conquer
## 12.1.2 &nbsp; Improving Efficiency Through Divide and Conquer
The **divide-and-conquer strategy not only effectively solves algorithm problems but also often enhances efficiency**. In sorting algorithms, quick sort, merge sort, and heap sort are faster than selection sort, bubble sort, and insertion sort because they apply the divide-and-conquer strategy.
**Divide and conquer can not only effectively solve algorithmic problems but often also improve algorithm efficiency**. In sorting algorithms, quick sort, merge sort, and heap sort are faster than selection, bubble, and insertion sort because they apply the divide and conquer strategy.
We may have a question in mind: **Why can divide and conquer improve algorithm efficiency, and what is the underlying logic?** In other words, why is breaking a problem into sub-problems, solving them, and combining their solutions to address the original problem offer more efficiency than directly solving the original problem? This question can be analyzed from two aspects: operation count and parallel computation.
This raises the question: **Why can divide and conquer improve algorithm efficiency, and what is the underlying logic**? In other words, why is dividing a large problem into multiple subproblems, solving the subproblems, and merging their solutions more efficient than directly solving the original problem? This question can be discussed from two aspects: operation count and parallel computation.
### 1. &nbsp; Optimization of operation count
### 1. &nbsp; Operation Count Optimization
Taking "bubble sort" as an example, it requires $O(n^2)$ time to process an array of length $n$. Suppose we divide the array from the midpoint into two sub-arrays as shown in Figure 12-2, such division requires $O(n)$ time. Sorting each sub-array requires $O((n / 2)^2)$ time. And merging the two sub-arrays requires $O(n)$ time. Thus, the overall time complexity is:
Taking "bubble sort" as an example, processing an array of length $n$ requires $O(n^2)$ time. Suppose we divide the array into two subarrays from the midpoint as shown in Figure 12-2, the division requires $O(n)$ time, sorting each subarray requires $O((n / 2)^2)$ time, and merging the two subarrays requires $O(n)$ time, resulting in an overall time complexity of:
$$
O(n + (\frac{n}{2})^2 \times 2 + n) = O(\frac{n^2}{2} + 2n)
$$
![Bubble sort before and after array partition](divide_and_conquer.assets/divide_and_conquer_bubble_sort.png){ class="animation-figure" }
![Bubble sort before and after array division](divide_and_conquer.assets/divide_and_conquer_bubble_sort.png){ class="animation-figure" }
<p align="center"> Figure 12-2 &nbsp; Bubble sort before and after array partition </p>
<p align="center"> Figure 12-2 &nbsp; Bubble sort before and after array division </p>
Let's calculate the following inequality, where the left side represents the total number of operations before division and the right side represents the total number of operations after division, respectively:
Next, we compute the following inequality, where the left and right sides represent the total number of operations before and after division, respectively:
$$
\begin{aligned}
@@ -60,42 +60,42 @@ n(n - 4) & > 0
\end{aligned}
$$
**This means that when $n > 4$, the number of operations after partitioning is fewer, leading to better performance**. Please note that the time complexity after partitioning is still quadratic $O(n^2)$, but the constant factor in the complexity has decreased.
**This means that when $n > 4$, the number of operations after division is smaller, and sorting efficiency should be higher**. Note that the time complexity after division is still quadratic $O(n^2)$, but the constant term in the complexity has become smaller.
We can go even further. **How about keeping dividing the sub-arrays from their midpoints into two sub-arrays** until the sub-arrays have only one element left? This idea is actually "merge sort," with a time complexity of $O(n \log n)$.
Going further, **what if we continuously divide the subarrays from their midpoints into two subarrays** until the subarrays have only one element? This approach is actually "merge sort", with a time complexity of $O(n \log n)$.
Let's try something a bit different again. **How about splitting into more partitions instead of just two?** For example, we evenly divide the original array into $k$ sub-arrays? This approach is very similar to "bucket sort," which is very suitable for sorting massive data. Theoretically, the time complexity can reach $O(n + k)$.
Thinking further, **what if we set multiple division points** and evenly divide the original array into $k$ subarrays? This situation is very similar to "bucket sort", which is well-suited for sorting massive amounts of data, with a theoretical time complexity of $O(n + k)$.
### 2. &nbsp; Optimization through parallel computation
### 2. &nbsp; Parallel Computation Optimization
We know that the sub-problems generated by divide and conquer are independent of each other, **which means that they can be solved in parallel.** As a result, divide and conquer not only reduces the algorithm's time complexity, **but also facilitates parallel optimization by modern operating systems.**
We know that the subproblems generated by divide and conquer are independent of each other, **so they can typically be solved in parallel**. This means divide and conquer can not only reduce the time complexity of algorithms, **but also benefits from parallel optimization by operating systems**.
Parallel optimization is particularly effective in environments with multiple cores or processors. As the system can process multiple sub-problems simultaneously, fully utilizing computing resources, the overall runtime is significantly reduced.
Parallel optimization is particularly effective in multi-core or multi-processor environments, as the system can simultaneously handle multiple subproblems, making fuller use of computing resources and significantly reducing overall runtime.
For example, in the "bucket sort" shown in Figure 12-3, we break massive data evenly into various buckets. The jobs of sorting each bucket can be allocated to available computing units. Once all jobs are done, all sorted buckets are merged to produce the final result.
For example, in the "bucket sort" shown in Figure 12-3, we evenly distribute massive data into various buckets, and the sorting tasks for all buckets can be distributed to various computing units. After completion, the results are merged.
![Bucket sort's parallel computation](divide_and_conquer.assets/divide_and_conquer_parallel_computing.png){ class="animation-figure" }
![Parallel computation in bucket sort](divide_and_conquer.assets/divide_and_conquer_parallel_computing.png){ class="animation-figure" }
<p align="center"> Figure 12-3 &nbsp; Bucket sort's parallel computation </p>
<p align="center"> Figure 12-3 &nbsp; Parallel computation in bucket sort </p>
## 12.1.3 &nbsp; Common applications of divide and conquer
## 12.1.3 &nbsp; Common Applications of Divide and Conquer
Divide and conquer can be used to solve many classic algorithm problems.
On one hand, divide and conquer can be used to solve many classic algorithmic problems.
- **Finding the closest pair of points**: This algorithm works by dividing the set of points into two halves. Then it recursively finds the closest pair in each half. Finally it considers pairs that span the two halves to find the overall closest pair.
- **Large integer multiplication**: One algorithm is called Karatsuba. It breaks down large integer multiplication into several smaller integer multiplications and additions.
- **Matrix multiplication**: One example is the Strassen algorithm. It breaks down a large matrix multiplication into multiple small matrix multiplications and additions.
- **Tower of Hanoi problem**: The Tower of Hanoi problem can be solved recursively, a typical application of the divide-and-conquer strategy.
- **Solving inversion pairs**: In a sequence, if a preceding number is greater than a following number, then these two numbers constitute an inversion pair. Solving inversion pair problem can utilize the idea of divide and conquer, with the aid of merge sort.
- **Finding the closest pair of points**: This algorithm first divides the point set into two parts, then finds the closest pair of points in each part separately, and finally finds the closest pair of points that spans both parts.
- **Large integer multiplication**: For example, the Karatsuba algorithm, which decomposes large integer multiplication into several smaller integer multiplications and additions.
- **Matrix multiplication**: For example, the Strassen algorithm, which decomposes large matrix multiplication into multiple small matrix multiplications and additions.
- **Hanota problem**: The hanota problem can be solved through recursion, which is a typical application of the divide and conquer strategy.
- **Solving inversion pairs**: In a sequence, if a preceding number is greater than a following number, these two numbers form an inversion pair. Solving the inversion pair problem can utilize the divide and conquer approach with the help of merge sort.
Divide and conquer is also widely applied in the design of algorithms and data structures.
On the other hand, divide and conquer is widely applied in the design of algorithms and data structures.
- **Binary search**: Binary search divides a sorted array into two halves from the midpoint index. And then based on the comparison result between the target value and the middle element value, one half is discarded. The search continues on the remaining half with the same process until the target is found or there is no remaining element.
- **Merge sort**: Already introduced at the beginning of this section, no further elaboration is needed.
- **Quicksort**: Quicksort picks a pivot value to divide the array into two sub-arrays, one with elements smaller than the pivot and the other with elements larger than the pivot. Such process goes on against each of these two sub-arrays until they hold only one element.
- **Bucket sort**: The basic idea of bucket sort is to distribute data to multiple buckets. After sorting the elements within each bucket, retrieve the elements from the buckets in order to obtain an ordered array.
- **Trees**: For example, binary search trees, AVL trees, red-black trees, B-trees, and B+ trees, etc. Their operations, such as search, insertion, and deletion, can all be regarded as applications of the divide-and-conquer strategy.
- **Heap**: A heap is a special type of complete binary tree. Its various operations, such as insertion, deletion, and heapify, actually imply the idea of divide and conquer.
- **Hash table**: Although hash tables do not directly apply divide and conquer, some hash collision resolution solutions indirectly apply the strategy. For example, long lists in chained addressing may be converted to red-black trees to improve query efficiency.
- **Binary search**: Binary search divides a sorted array into two parts from the midpoint index, then decides which half to eliminate based on the comparison result between the target value and the middle element value, and performs the same binary operation on the remaining interval.
- **Merge sort**: Already introduced at the beginning of this section, no further elaboration needed.
- **Quick sort**: Quick sort selects a pivot value, then divides the array into two subarrays, one with elements smaller than the pivot and the other with elements larger than the pivot, then performs the same division operation on these two parts until the subarrays have only one element.
- **Bucket sort**: The basic idea of bucket sort is to scatter data into multiple buckets, then sort the elements within each bucket, and finally extract the elements from each bucket in sequence to obtain a sorted array.
- **Trees**: For example, binary search trees, AVL trees, red-black trees, B-trees, B+ trees, etc. Their search, insertion, and deletion operations can all be viewed as applications of the divide and conquer strategy.
- **Heaps**: A heap is a special complete binary tree, and its various operations, such as insertion, deletion, and heapify, actually imply the divide and conquer idea.
- **Hash tables**: Although hash tables do not directly apply divide and conquer, some hash collision resolution solutions indirectly apply the divide and conquer strategy. For example, long linked lists in chaining may be converted to red-black trees to improve query efficiency.
It can be seen that **divide and conquer is a subtly pervasive algorithmic idea**, embedded within various algorithms and data structures.
It can be seen that **divide and conquer is a "subtly pervasive" algorithmic idea**, embedded in various algorithms and data structures.
en/docs/chapter_divide_and_conquer/hanota_problem.md
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---
# 12.4 &nbsp; Tower of Hanoi Problem
# 12.4 &nbsp; Hanota Problem
In both merge sort and binary tree construction, we break the original problem into two subproblems, each half the size of the original problem. However, for the Tower of Hanoi, we adopt a different decomposition strategy.
In merge sort and building binary trees, we decompose the original problem into two subproblems, each half the size of the original problem. However, for the hanota problem, we adopt a different decomposition strategy.
!!! question
We are given three pillars, denoted as `A`, `B`, and `C`. Initially, pillar `A` has $n$ discs, arranged from top to bottom in ascending size. Our task is to move these $n$ discs to pillar `C`, maintaining their original order (as shown in Figure 12-10). The following rules apply during the movement:
1. A disc can be removed only from the top of a pillar and must be placed on the top of another pillar.
Given three pillars, denoted as `A`, `B`, and `C`. Initially, pillar `A` has $n$ discs stacked on it, arranged from top to bottom in ascending order of size. Our task is to move these $n$ discs to pillar `C` while maintaining their original order (as shown in Figure 12-10). The following rules must be followed when moving the discs.
1. A disc can only be taken from the top of one pillar and placed on top of another pillar.
2. Only one disc can be moved at a time.
3. A smaller disc must always be on top of a larger disc.
![Example of the Tower of Hanoi](hanota_problem.assets/hanota_example.png){ class="animation-figure" }
![Example of the hanota problem](hanota_problem.assets/hanota_example.png){ class="animation-figure" }
<p align="center"> Figure 12-10 &nbsp; Example of the Tower of Hanoi </p>
<p align="center"> Figure 12-10 &nbsp; Example of the hanota problem </p>
**We denote the Tower of Hanoi problem of size $i$ as $f(i)$**. For example, $f(3)$ represents moving $3$ discs from pillar `A` to pillar `C`.
**We denote the hanota problem of size $i$ as $f(i)$**. For example, $f(3)$ represents moving $3$ discs from `A` to `C`.
### 1. &nbsp; Consider the base cases
### 1. &nbsp; Considering the Base Cases
As shown in Figure 12-11, for the problem $f(1)$—which has only one disc—we can directly move it from `A` to `C`.
As shown in Figure 12-11, for problem $f(1)$, when there is only one disc, we can move it directly from `A` to `C`.
=== "<1>"
![Solution for a problem of size 1](hanota_problem.assets/hanota_f1_step1.png){ class="animation-figure" }
@@ -32,7 +32,7 @@ As shown in Figure 12-11, for the problem $f(1)$—which has only one disc—we
<p align="center"> Figure 12-11 &nbsp; Solution for a problem of size 1 </p>
For $f(2)$—which has two discs—**we rely on pillar `B` to help keep the smaller disc above the larger disc**, as illustrated in the following figure:
As shown in Figure 12-12, for problem $f(2)$, when there are two discs, **since we must always keep the smaller disc on top of the larger disc, we need to use `B` to assist in the move**.
1. First, move the smaller disc from `A` to `B`.
2. Then move the larger disc from `A` to `C`.
@@ -52,17 +52,17 @@ For $f(2)$—which has two discs—**we rely on pillar `B` to help keep the smal
<p align="center"> Figure 12-12 &nbsp; Solution for a problem of size 2 </p>
The process of solving $f(2)$ can be summarized as: **moving two discs from `A` to `C` with the help of `B`**. Here, `C` is called the target pillar, and `B` is called the buffer pillar.
The process of solving problem $f(2)$ can be summarized as: **moving two discs from `A` to `C` with the help of `B`**. Here, `C` is called the target pillar, and `B` is called the buffer pillar.
### 2. &nbsp; Decomposition of subproblems
### 2. &nbsp; Subproblem Decomposition
For the problem $f(3)$—that is, when there are three discs—the situation becomes slightly more complicated.
For problem $f(3)$, when there are three discs, the situation becomes slightly more complex.
Since we already know the solutions to $f(1)$ and $f(2)$, we can adopt a divide-and-conquer perspective and **treat the top two discs on `A` as a single unit**, performing the steps shown in Figure 12-13. This allows the three discs to be successfully moved from `A` to `C`.
Since we already know the solutions to $f(1)$ and $f(2)$, we can think from a divide and conquer perspective, **treating the top two discs on `A` as a whole**, and execute the steps shown in Figure 12-13. This successfully moves the three discs from `A` to `C`.
1. Let `B` be the target pillar and `C` the buffer pillar, then move the two discs from `A` to `B`.
1. Let `B` be the target pillar and `C` be the buffer pillar, and move two discs from `A` to `B`.
2. Move the remaining disc from `A` directly to `C`.
3. Let `C` be the target pillar and `A` the buffer pillar, then move the two discs from `B` to `C`.
3. Let `C` be the target pillar and `A` be the buffer pillar, and move two discs from `B` to `C`.
=== "<1>"
![Solution for a problem of size 3](hanota_problem.assets/hanota_f3_step1.png){ class="animation-figure" }
@@ -78,85 +78,85 @@ Since we already know the solutions to $f(1)$ and $f(2)$, we can adopt a divide-
<p align="center"> Figure 12-13 &nbsp; Solution for a problem of size 3 </p>
Essentially, **we decompose $f(3)$ into two $f(2)$ subproblems and one $f(1)$ subproblem**. By solving these three subproblems in sequence, the original problem is solved, indicating that the subproblems are independent and their solutions can be merged.
Essentially, **we divide problem $f(3)$ into two subproblems $f(2)$ and one subproblem $f(1)$**. By solving these three subproblems in order, the original problem is solved. This shows that the subproblems are independent and their solutions can be merged.
From this, we can summarize the divide-and-conquer strategy for the Tower of Hanoi, illustrated in Figure 12-14. We divide the original problem $f(n)$ into two subproblems $f(n-1)$ and one subproblem $f(1)$, and solve these three subproblems in the following order:
From this, we can summarize the divide and conquer strategy for solving the hanota problem shown in Figure 12-14: divide the original problem $f(n)$ into two subproblems $f(n-1)$ and one subproblem $f(1)$, and solve these three subproblems in the following order.
1. Move $n-1$ discs from `A` to `B`, using `C` as a buffer.
2. Move the remaining disc directly from `A` to `C`.
3. Move $n-1$ discs from `B` to `C`, using `A` as a buffer.
1. Move $n-1$ discs from `A` to `B` with the help of `C`.
2. Move the remaining $1$ disc directly from `A` to `C`.
3. Move $n-1$ discs from `B` to `C` with the help of `A`.
For each $f(n-1)$ subproblem, **we can apply the same recursive partition** until we reach the smallest subproblem $f(1)$. Because $f(1)$ is already known to require just a single move, it is trivial to solve.
For these two subproblems $f(n-1)$, **we can recursively divide them in the same way** until reaching the smallest subproblem $f(1)$. The solution to $f(1)$ is known and requires only one move operation.
![Divide-and-conquer strategy for solving the Tower of Hanoi](hanota_problem.assets/hanota_divide_and_conquer.png){ class="animation-figure" }
![Divide and conquer strategy for solving the hanota problem](hanota_problem.assets/hanota_divide_and_conquer.png){ class="animation-figure" }
<p align="center"> Figure 12-14 &nbsp; Divide-and-conquer strategy for solving the Tower of Hanoi </p>
<p align="center"> Figure 12-14 &nbsp; Divide and conquer strategy for solving the hanota problem </p>
### 3. &nbsp; Code implementation
### 3. &nbsp; Code Implementation
In the code, we define a recursive function `dfs(i, src, buf, tar)` which moves the top $i$ discs from pillar `src` to pillar `tar`, using pillar `buf` as a buffer:
In the code, we declare a recursive function `dfs(i, src, buf, tar)`, whose purpose is to move the top $i$ discs from pillar `src` to target pillar `tar` with the help of buffer pillar `buf`:
=== "Python"
```python title="hanota.py"
def move(src: list[int], tar: list[int]):
"""Move a disc"""
# Take out a disc from the top of src
"""Move a disk"""
# Take out a disk from the top of src
pan = src.pop()
# Place the disc on top of tar
# Place the disk on top of tar
tar.append(pan)
def dfs(i: int, src: list[int], buf: list[int], tar: list[int]):
"""Solve the Tower of Hanoi problem f(i)"""
# If only one disc remains on src, move it to tar
# If there is only one disk left in src, move it directly to tar
if i == 1:
move(src, tar)
return
# Subproblem f(i-1): move the top i-1 discs from src with the help of tar to buf
# Subproblem f(i-1): move the top i-1 disks from src to buf using tar
dfs(i - 1, src, tar, buf)
# Subproblem f(1): move the remaining one disc from src to tar
# Subproblem f(1): move the remaining disk from src to tar
move(src, tar)
# Subproblem f(i-1): move the top i-1 discs from buf with the help of src to tar
# Subproblem f(i-1): move the top i-1 disks from buf to tar using src
dfs(i - 1, buf, src, tar)
def solve_hanota(A: list[int], B: list[int], C: list[int]):
"""Solve the Tower of Hanoi problem"""
n = len(A)
# Move the top n discs from A with the help of B to C
# Move the top n disks from A to C using B
dfs(n, A, B, C)
```
=== "C++"
```cpp title="hanota.cpp"
/* Move a disc */
/* Move a disk */
void move(vector<int> &src, vector<int> &tar) {
// Take out a disc from the top of src
// Take out a disk from the top of src
int pan = src.back();
src.pop_back();
// Place the disc on top of tar
// Place the disk on top of tar
tar.push_back(pan);
}
/* Solve the Tower of Hanoi problem f(i) */
void dfs(int i, vector<int> &src, vector<int> &buf, vector<int> &tar) {
// If only one disc remains on src, move it to tar
// If there is only one disk left in src, move it directly to tar
if (i == 1) {
move(src, tar);
return;
}
// Subproblem f(i-1): move the top i-1 discs from src with the help of tar to buf
// Subproblem f(i-1): move the top i-1 disks from src to buf using tar
dfs(i - 1, src, tar, buf);
// Subproblem f(1): move the remaining one disc from src to tar
// Subproblem f(1): move the remaining disk from src to tar
move(src, tar);
// Subproblem f(i-1): move the top i-1 discs from buf with the help of src to tar
// Subproblem f(i-1): move the top i-1 disks from buf to tar using src
dfs(i - 1, buf, src, tar);
}
/* Solve the Tower of Hanoi problem */
void solveHanota(vector<int> &A, vector<int> &B, vector<int> &C) {
int n = A.size();
// Move the top n discs from A with the help of B to C
// Move the top n disks from A to C using B
dfs(n, A, B, C);
}
```
@@ -164,33 +164,33 @@ In the code, we define a recursive function `dfs(i, src, buf, tar)` which moves
=== "Java"
```java title="hanota.java"
/* Move a disc */
/* Move a disk */
void move(List<Integer> src, List<Integer> tar) {
// Take out a disc from the top of src
// Take out a disk from the top of src
Integer pan = src.remove(src.size() - 1);
// Place the disc on top of tar
// Place the disk on top of tar
tar.add(pan);
}
/* Solve the Tower of Hanoi problem f(i) */
void dfs(int i, List<Integer> src, List<Integer> buf, List<Integer> tar) {
// If only one disc remains on src, move it to tar
// If there is only one disk left in src, move it directly to tar
if (i == 1) {
move(src, tar);
return;
}
// Subproblem f(i-1): move the top i-1 discs from src with the help of tar to buf
// Subproblem f(i-1): move the top i-1 disks from src to buf using tar
dfs(i - 1, src, tar, buf);
// Subproblem f(1): move the remaining one disc from src to tar
// Subproblem f(1): move the remaining disk from src to tar
move(src, tar);
// Subproblem f(i-1): move the top i-1 discs from buf with the help of src to tar
// Subproblem f(i-1): move the top i-1 disks from buf to tar using src
dfs(i - 1, buf, src, tar);
}
/* Solve the Tower of Hanoi problem */
void solveHanota(List<Integer> A, List<Integer> B, List<Integer> C) {
int n = A.size();
// Move the top n discs from A with the help of B to C
// Move the top n disks from A to C using B
dfs(n, A, B, C);
}
```
@@ -198,121 +198,358 @@ In the code, we define a recursive function `dfs(i, src, buf, tar)` which moves
=== "C#"
```csharp title="hanota.cs"
[class]{hanota}-[func]{Move}
/* Move a disk */
void Move(List<int> src, List<int> tar) {
// Take out a disk from the top of src
int pan = src[^1];
src.RemoveAt(src.Count - 1);
// Place the disk on top of tar
tar.Add(pan);
}
[class]{hanota}-[func]{DFS}
/* Solve the Tower of Hanoi problem f(i) */
void DFS(int i, List<int> src, List<int> buf, List<int> tar) {
// If there is only one disk left in src, move it directly to tar
if (i == 1) {
Move(src, tar);
return;
}
// Subproblem f(i-1): move the top i-1 disks from src to buf using tar
DFS(i - 1, src, tar, buf);
// Subproblem f(1): move the remaining disk from src to tar
Move(src, tar);
// Subproblem f(i-1): move the top i-1 disks from buf to tar using src
DFS(i - 1, buf, src, tar);
}
[class]{hanota}-[func]{SolveHanota}
/* Solve the Tower of Hanoi problem */
void SolveHanota(List<int> A, List<int> B, List<int> C) {
int n = A.Count;
// Move the top n disks from A to C using B
DFS(n, A, B, C);
}
```
=== "Go"
```go title="hanota.go"
[class]{}-[func]{move}
/* Move a disk */
func move(src, tar *list.List) {
// Take out a disk from the top of src
pan := src.Back()
// Place the disk on top of tar
tar.PushBack(pan.Value)
// Remove top disk from src
src.Remove(pan)
}
[class]{}-[func]{dfsHanota}
/* Solve the Tower of Hanoi problem f(i) */
func dfsHanota(i int, src, buf, tar *list.List) {
// If there is only one disk left in src, move it directly to tar
if i == 1 {
move(src, tar)
return
}
// Subproblem f(i-1): move the top i-1 disks from src to buf using tar
dfsHanota(i-1, src, tar, buf)
// Subproblem f(1): move the remaining disk from src to tar
move(src, tar)
// Subproblem f(i-1): move the top i-1 disks from buf to tar using src
dfsHanota(i-1, buf, src, tar)
}
[class]{}-[func]{solveHanota}
/* Solve the Tower of Hanoi problem */
func solveHanota(A, B, C *list.List) {
n := A.Len()
// Move the top n disks from A to C using B
dfsHanota(n, A, B, C)
}
```
=== "Swift"
```swift title="hanota.swift"
[class]{}-[func]{move}
/* Move a disk */
func move(src: inout [Int], tar: inout [Int]) {
// Take out a disk from the top of src
let pan = src.popLast()!
// Place the disk on top of tar
tar.append(pan)
}
[class]{}-[func]{dfs}
/* Solve the Tower of Hanoi problem f(i) */
func dfs(i: Int, src: inout [Int], buf: inout [Int], tar: inout [Int]) {
// If there is only one disk left in src, move it directly to tar
if i == 1 {
move(src: &src, tar: &tar)
return
}
// Subproblem f(i-1): move the top i-1 disks from src to buf using tar
dfs(i: i - 1, src: &src, buf: &tar, tar: &buf)
// Subproblem f(1): move the remaining disk from src to tar
move(src: &src, tar: &tar)
// Subproblem f(i-1): move the top i-1 disks from buf to tar using src
dfs(i: i - 1, src: &buf, buf: &src, tar: &tar)
}
[class]{}-[func]{solveHanota}
/* Solve the Tower of Hanoi problem */
func solveHanota(A: inout [Int], B: inout [Int], C: inout [Int]) {
let n = A.count
// The tail of the list is the top of the rod
// Move top n disks from src to C using B
dfs(i: n, src: &A, buf: &B, tar: &C)
}
```
=== "JS"
```javascript title="hanota.js"
[class]{}-[func]{move}
/* Move a disk */
function move(src, tar) {
// Take out a disk from the top of src
const pan = src.pop();
// Place the disk on top of tar
tar.push(pan);
}
[class]{}-[func]{dfs}
/* Solve the Tower of Hanoi problem f(i) */
function dfs(i, src, buf, tar) {
// If there is only one disk left in src, move it directly to tar
if (i === 1) {
move(src, tar);
return;
}
// Subproblem f(i-1): move the top i-1 disks from src to buf using tar
dfs(i - 1, src, tar, buf);
// Subproblem f(1): move the remaining disk from src to tar
move(src, tar);
// Subproblem f(i-1): move the top i-1 disks from buf to tar using src
dfs(i - 1, buf, src, tar);
}
[class]{}-[func]{solveHanota}
/* Solve the Tower of Hanoi problem */
function solveHanota(A, B, C) {
const n = A.length;
// Move the top n disks from A to C using B
dfs(n, A, B, C);
}
```
=== "TS"
```typescript title="hanota.ts"
[class]{}-[func]{move}
/* Move a disk */
function move(src: number[], tar: number[]): void {
// Take out a disk from the top of src
const pan = src.pop();
// Place the disk on top of tar
tar.push(pan);
}
[class]{}-[func]{dfs}
/* Solve the Tower of Hanoi problem f(i) */
function dfs(i: number, src: number[], buf: number[], tar: number[]): void {
// If there is only one disk left in src, move it directly to tar
if (i === 1) {
move(src, tar);
return;
}
// Subproblem f(i-1): move the top i-1 disks from src to buf using tar
dfs(i - 1, src, tar, buf);
// Subproblem f(1): move the remaining disk from src to tar
move(src, tar);
// Subproblem f(i-1): move the top i-1 disks from buf to tar using src
dfs(i - 1, buf, src, tar);
}
[class]{}-[func]{solveHanota}
/* Solve the Tower of Hanoi problem */
function solveHanota(A: number[], B: number[], C: number[]): void {
const n = A.length;
// Move the top n disks from A to C using B
dfs(n, A, B, C);
}
```
=== "Dart"
```dart title="hanota.dart"
[class]{}-[func]{move}
/* Move a disk */
void move(List<int> src, List<int> tar) {
// Take out a disk from the top of src
int pan = src.removeLast();
// Place the disk on top of tar
tar.add(pan);
}
[class]{}-[func]{dfs}
/* Solve the Tower of Hanoi problem f(i) */
void dfs(int i, List<int> src, List<int> buf, List<int> tar) {
// If there is only one disk left in src, move it directly to tar
if (i == 1) {
move(src, tar);
return;
}
// Subproblem f(i-1): move the top i-1 disks from src to buf using tar
dfs(i - 1, src, tar, buf);
// Subproblem f(1): move the remaining disk from src to tar
move(src, tar);
// Subproblem f(i-1): move the top i-1 disks from buf to tar using src
dfs(i - 1, buf, src, tar);
}
[class]{}-[func]{solveHanota}
/* Solve the Tower of Hanoi problem */
void solveHanota(List<int> A, List<int> B, List<int> C) {
int n = A.length;
// Move the top n disks from A to C using B
dfs(n, A, B, C);
}
```
=== "Rust"
```rust title="hanota.rs"
[class]{}-[func]{move_pan}
/* Move a disk */
fn move_pan(src: &mut Vec<i32>, tar: &mut Vec<i32>) {
// Take out a disk from the top of src
let pan = src.pop().unwrap();
// Place the disk on top of tar
tar.push(pan);
}
[class]{}-[func]{dfs}
/* Solve the Tower of Hanoi problem f(i) */
fn dfs(i: i32, src: &mut Vec<i32>, buf: &mut Vec<i32>, tar: &mut Vec<i32>) {
// If there is only one disk left in src, move it directly to tar
if i == 1 {
move_pan(src, tar);
return;
}
// Subproblem f(i-1): move the top i-1 disks from src to buf using tar
dfs(i - 1, src, tar, buf);
// Subproblem f(1): move the remaining disk from src to tar
move_pan(src, tar);
// Subproblem f(i-1): move the top i-1 disks from buf to tar using src
dfs(i - 1, buf, src, tar);
}
[class]{}-[func]{solve_hanota}
/* Solve the Tower of Hanoi problem */
fn solve_hanota(A: &mut Vec<i32>, B: &mut Vec<i32>, C: &mut Vec<i32>) {
let n = A.len() as i32;
// Move the top n disks from A to C using B
dfs(n, A, B, C);
}
```
=== "C"
```c title="hanota.c"
[class]{}-[func]{move}
/* Move a disk */
void move(int *src, int *srcSize, int *tar, int *tarSize) {
// Take out a disk from the top of src
int pan = src[*srcSize - 1];
src[*srcSize - 1] = 0;
(*srcSize)--;
// Place the disk on top of tar
tar[*tarSize] = pan;
(*tarSize)++;
}
[class]{}-[func]{dfs}
/* Solve the Tower of Hanoi problem f(i) */
void dfs(int i, int *src, int *srcSize, int *buf, int *bufSize, int *tar, int *tarSize) {
// If there is only one disk left in src, move it directly to tar
if (i == 1) {
move(src, srcSize, tar, tarSize);
return;
}
// Subproblem f(i-1): move the top i-1 disks from src to buf using tar
dfs(i - 1, src, srcSize, tar, tarSize, buf, bufSize);
// Subproblem f(1): move the remaining disk from src to tar
move(src, srcSize, tar, tarSize);
// Subproblem f(i-1): move the top i-1 disks from buf to tar using src
dfs(i - 1, buf, bufSize, src, srcSize, tar, tarSize);
}
[class]{}-[func]{solveHanota}
/* Solve the Tower of Hanoi problem */
void solveHanota(int *A, int *ASize, int *B, int *BSize, int *C, int *CSize) {
// Move the top n disks from A to C using B
dfs(*ASize, A, ASize, B, BSize, C, CSize);
}
```
=== "Kotlin"
```kotlin title="hanota.kt"
[class]{}-[func]{move}
/* Move a disk */
fun move(src: MutableList<Int>, tar: MutableList<Int>) {
// Take out a disk from the top of src
val pan = src.removeAt(src.size - 1)
// Place the disk on top of tar
tar.add(pan)
}
[class]{}-[func]{dfs}
/* Solve the Tower of Hanoi problem f(i) */
fun dfs(i: Int, src: MutableList<Int>, buf: MutableList<Int>, tar: MutableList<Int>) {
// If there is only one disk left in src, move it directly to tar
if (i == 1) {
move(src, tar)
return
}
// Subproblem f(i-1): move the top i-1 disks from src to buf using tar
dfs(i - 1, src, tar, buf)
// Subproblem f(1): move the remaining disk from src to tar
move(src, tar)
// Subproblem f(i-1): move the top i-1 disks from buf to tar using src
dfs(i - 1, buf, src, tar)
}
[class]{}-[func]{solveHanota}
/* Solve the Tower of Hanoi problem */
fun solveHanota(A: MutableList<Int>, B: MutableList<Int>, C: MutableList<Int>) {
val n = A.size
// Move the top n disks from A to C using B
dfs(n, A, B, C)
}
```
=== "Ruby"
```ruby title="hanota.rb"
[class]{}-[func]{move}
### Move one disk ###
def move(src, tar)
# Take out a disk from the top of src
pan = src.pop
# Place the disk on top of tar
tar << pan
end
[class]{}-[func]{dfs}
### Solve Tower of Hanoi f(i) ###
def dfs(i, src, buf, tar)
# If there is only one disk left in src, move it directly to tar
if i == 1
move(src, tar)
return
end
[class]{}-[func]{solve_hanota}
# Subproblem f(i-1): move the top i-1 disks from src to buf using tar
dfs(i - 1, src, tar, buf)
# Subproblem f(1): move the remaining disk from src to tar
move(src, tar)
# Subproblem f(i-1): move the top i-1 disks from buf to tar using src
dfs(i - 1, buf, src, tar)
end
### Solve Tower of Hanoi ###
def solve_hanota(_A, _B, _C)
n = _A.length
# Move the top n disks from A to C using B
dfs(n, _A, _B, _C)
end
```
=== "Zig"
As shown in Figure 12-15, the hanota problem forms a recursion tree of height $n$, where each node represents a subproblem corresponding to an invocation of the `dfs()` function, **therefore the time complexity is $O(2^n)$ and the space complexity is $O(n)$**.
```zig title="hanota.zig"
[class]{}-[func]{move}
![Recursion tree of the hanota problem](hanota_problem.assets/hanota_recursive_tree.png){ class="animation-figure" }
[class]{}-[func]{dfs}
[class]{}-[func]{solveHanota}
```
As shown in Figure 12-15, the Tower of Hanoi problem can be visualized as a recursive tree of height $n$. Each node represents a subproblem, corresponding to a call to `dfs()`, **Hence, the time complexity is $O(2^n)$, and the space complexity is $O(n)$.**
![Recursive tree of the Tower of Hanoi](hanota_problem.assets/hanota_recursive_tree.png){ class="animation-figure" }
<p align="center"> Figure 12-15 &nbsp; Recursive tree of the Tower of Hanoi </p>
<p align="center"> Figure 12-15 &nbsp; Recursion tree of the hanota problem </p>
!!! quote
The Tower of Hanoi originates from an ancient legend. In a temple in ancient India, monks had three tall diamond pillars and $64$ differently sized golden discs. They believed that when the last disc was correctly placed, the world would end.
The hanota problem originates from an ancient legend. In a temple in ancient India, monks had three tall diamond pillars and $64$ golden discs of different sizes. The monks continuously moved the discs, believing that when the last disc was correctly placed, the world would come to an end.
However, even if the monks moved one disc every second, it would take about $2^{64} \approx 1.84×10^{19}$ —approximately 585 billion years—far exceeding current estimates of the age of the universe. Thus, if the legend is true, we probably do not need to worry about the world ending.
However, even if the monks moved one disc per second, it would take approximately $2^{64} \approx 1.84×10^{19}$ seconds, which is about $5850$ billion years, far exceeding current estimates of the age of the universe. Therefore, if this legend is true, we should not need to worry about the end of the world.
en/docs/chapter_divide_and_conquer/index.md
+7 -7
View File
@@ -3,20 +3,20 @@ comments: true
icon: material/set-split
---
# Chapter 12. &nbsp; Divide and conquer
# Chapter 12. &nbsp; Divide and Conquer
![Divide and Conquer](../assets/covers/chapter_divide_and_conquer.jpg){ class="cover-image" }
![Divide and conquer](../assets/covers/chapter_divide_and_conquer.jpg){ class="cover-image" }
!!! abstract
Difficult problems are decomposed layer by layer, with each decomposition making them simpler.
Divide and conquer unveils a profound truth: begin with simplicity, and complexity dissolves.
Divide and conquer reveals an important truth: start with simplicity, and nothing remains complex.
## Chapter contents
- [12.1 &nbsp; Divide and conquer algorithms](divide_and_conquer.md)
- [12.2 &nbsp; Divide and conquer search strategy](binary_search_recur.md)
- [12.3 &nbsp; Building binary tree problem](build_binary_tree_problem.md)
- [12.4 &nbsp; Tower of Hanoi Problem](hanota_problem.md)
- [12.1 &nbsp; Divide and Conquer Algorithms](divide_and_conquer.md)
- [12.2 &nbsp; Divide and Conquer Search Strategy](binary_search_recur.md)
- [12.3 &nbsp; Building a Binary Tree Problem](build_binary_tree_problem.md)
- [12.4 &nbsp; Hanoi Tower Problem](hanota_problem.md)
- [12.5 &nbsp; Summary](summary.md)
en/docs/chapter_divide_and_conquer/summary.md
+11 -9
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@@ -4,12 +4,14 @@ comments: true
# 12.5 &nbsp; Summary
- Divide and conquer is a common algorithm design strategy that consists of two stages—divide (partition) and conquer (merge)—and is generally implemented using recursion.
- To determine whether a problem is suited for a divide and conquer approach, we check if the problem can be decomposed, whether the subproblems are independent, and whether the subproblems can be merged.
- Merge sort is a typical example of the divide and conquer strategy. It recursively splits an array into two equal-length subarrays until only one element remains, and then merges these subarrays layer by layer to complete the sorting.
- Introducing the divide and conquer strategy often improves algorithm efficiency. On one hand, it reduces the number of operations; on the other hand, it facilitates parallel optimization of the system after division.
- Divide and conquer can be applied to numerous algorithmic problems and is widely used in data structures and algorithm design, appearing in many scenarios.
- Compared to brute force search, adaptive search is more efficient. Search algorithms with a time complexity of $O(\log n)$ are typically based on the divide and conquer strategy.
- Binary search is another classic application of the divide-and-conquer strategy. It does not involve merging subproblem solutions and can be implemented via a recursive divide-and-conquer approach.
- In the problem of constructing binary trees, building the tree (the original problem) can be divided into building the left subtree and right subtree (the subproblems). This can be achieved by partitioning the index ranges of the preorder and inorder traversals.
- In the Tower of Hanoi problem, a problem of size $n$ can be broken down into two subproblems of size $n-1$ and one subproblem of size $1$. By solving these three subproblems in sequence, the original problem is resolved.
### 1. &nbsp; Key Review
- Divide and conquer is a common algorithm design strategy, consisting of two phases: divide (partition) and conquer (merge), typically implemented based on recursion.
- The criteria for determining whether a problem is a divide and conquer problem include: whether the problem can be decomposed, whether subproblems are independent, and whether subproblems can be merged.
- Merge sort is a typical application of the divide and conquer strategy. It recursively divides an array into two equal-length subarrays until only one element remains, then merges them layer by layer to complete the sorting.
- Introducing the divide and conquer strategy can often improve algorithm efficiency. On one hand, the divide and conquer strategy reduces the number of operations; on the other hand, it facilitates parallel optimization of the system after division.
- Divide and conquer can both solve many algorithmic problems and is widely applied in data structure and algorithm design, appearing everywhere.
- Compared to brute-force search, adaptive search is more efficient. Search algorithms with time complexity of $O(\log n)$ are typically implemented based on the divide and conquer strategy.
- Binary search is another typical application of divide and conquer. It does not include the step of merging solutions of subproblems. We can implement binary search through recursive divide and conquer.
- In the problem of building a binary tree, building the tree (original problem) can be divided into building the left subtree and right subtree (subproblems), which can be achieved by dividing the index intervals of the preorder and inorder traversals.
- In the hanota problem, a problem of size $n$ can be divided into two subproblems of size $n-1$ and one subproblem of size $1$. After solving these three subproblems in order, the original problem is solved.
en/docs/chapter_dynamic_programming/dp_problem_features.md
+576 -109
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@@ -2,55 +2,55 @@
comments: true
---
# 14.2 &nbsp; Characteristics of dynamic programming problems
# 14.2 &nbsp; Characteristics of Dynamic Programming Problems
In the previous section, we learned how dynamic programming solves the original problem by decomposing it into subproblems. In fact, subproblem decomposition is a general algorithmic approach, with different emphases in divide and conquer, dynamic programming, and backtracking.
- Divide and conquer algorithms recursively divide the original problem into multiple independent subproblems until the smallest subproblems are reached, and combine the solutions of the subproblems during backtracking to ultimately obtain the solution to the original problem.
- Dynamic programming also decomposes the problem recursively, but the main difference from divide and conquer algorithms is that the subproblems in dynamic programming are interdependent, and many overlapping subproblems will appear during the decomposition process.
- Backtracking algorithms exhaust all possible solutions through trial and error and avoid unnecessary search branches by pruning. The solution to the original problem consists of a series of decision steps, and we can consider each sub-sequence before each decision step as a subproblem.
- Divide and conquer algorithms recursively divide the original problem into multiple independent subproblems until the smallest subproblems are reached, and merge the solutions to the subproblems during backtracking to ultimately obtain the solution to the original problem.
- Dynamic programming also recursively decomposes problems, but the main difference from divide and conquer algorithms is that subproblems in dynamic programming are interdependent, and many overlapping subproblems appear during the decomposition process.
- Backtracking algorithms enumerate all possible solutions through trial and error, and avoid unnecessary search branches through pruning. The solution to the original problem consists of a series of decision steps, and we can regard the subsequence before each decision step as a subproblem.
In fact, dynamic programming is commonly used to solve optimization problems, which not only include overlapping subproblems but also have two other major characteristics: optimal substructure and statelessness.
In fact, dynamic programming is commonly used to solve optimization problems, which not only contain overlapping subproblems but also have two other major characteristics: optimal substructure and no aftereffects.
## 14.2.1 &nbsp; Optimal substructure
## 14.2.1 &nbsp; Optimal Substructure
We make a slight modification to the stair climbing problem to make it more suitable to demonstrate the concept of optimal substructure.
We make a slight modification to the stair climbing problem to make it more suitable for demonstrating the concept of optimal substructure.
!!! question "Minimum cost of climbing stairs"
!!! question "Climbing stairs with minimum cost"
Given a staircase, you can step up 1 or 2 steps at a time, and each step on the staircase has a non-negative integer representing the cost you need to pay at that step. Given a non-negative integer array $cost$, where $cost[i]$ represents the cost you need to pay at the $i$-th step, $cost[0]$ is the ground (starting point). What is the minimum cost required to reach the top?
Given a staircase, where you can climb $1$ or $2$ steps at a time, and each step has a non-negative integer representing the cost you need to pay at that step. Given a non-negative integer array $cost$, where $cost[i]$ represents the cost at the $i$-th step, and $cost[0]$ is the ground (starting point). What is the minimum cost required to reach the top?
As shown in Figure 14-6, if the costs of the 1st, 2nd, and 3rd steps are $1$, $10$, and $1$ respectively, then the minimum cost to climb to the 3rd step from the ground is $2$.
As shown in Figure 14-6, if the costs of the $1$st, $2$nd, and $3$rd steps are $1$, $10$, and $1$ respectively, then climbing from the ground to the $3$rd step requires a minimum cost of $2$.
![Minimum cost to climb to the 3rd step](dp_problem_features.assets/min_cost_cs_example.png){ class="animation-figure" }
<p align="center"> Figure 14-6 &nbsp; Minimum cost to climb to the 3rd step </p>
Let $dp[i]$ be the cumulative cost of climbing to the $i$-th step. Since the $i$-th step can only come from the $i-1$ or $i-2$ step, $dp[i]$ can only be either $dp[i-1] + cost[i]$ or $dp[i-2] + cost[i]$. To minimize the cost, we should choose the smaller of the two:
Let $dp[i]$ be the accumulated cost of climbing to the $i$-th step. Since the $i$-th step can only come from the $i-1$-th or $i-2$-th step, $dp[i]$ can only equal $dp[i-1] + cost[i]$ or $dp[i-2] + cost[i]$. To minimize the cost, we should choose the smaller of the two:
$$
dp[i] = \min(dp[i-1], dp[i-2]) + cost[i]
$$
This leads us to the meaning of optimal substructure: **The optimal solution to the original problem is constructed from the optimal solutions of subproblems**.
This leads us to the meaning of optimal substructure: **the optimal solution to the original problem is constructed from the optimal solutions to the subproblems**.
This problem obviously has optimal substructure: we select the better one from the optimal solutions of the two subproblems, $dp[i-1]$ and $dp[i-2]$, and use it to construct the optimal solution for the original problem $dp[i]$.
This problem clearly has optimal substructure: we select the better one from the optimal solutions to the two subproblems $dp[i-1]$ and $dp[i-2]$, and use it to construct the optimal solution to the original problem $dp[i]$.
So, does the stair climbing problem from the previous section have optimal substructure? Its goal is to solve for the number of solutions, which seems to be a counting problem, but if we ask in another way: "Solve for the maximum number of solutions". We surprisingly find that **although the problem has changed, the optimal substructure has emerged**: the maximum number of solutions at the $n$-th step equals the sum of the maximum number of solutions at the $n-1$ and $n-2$ steps. Thus, the interpretation of optimal substructure is quite flexible and will have different meanings in different problems.
So, does the stair climbing problem from the previous section have optimal substructure? Its goal is to find the number of ways, which seems to be a counting problem, but if we change the question: "Find the maximum number of ways". We surprisingly discover that **although the problem before and after modification are equivalent, the optimal substructure has emerged**: the maximum number of ways for the $n$-th step equals the sum of the maximum number of ways for the $n-1$-th and $n-2$-th steps. Therefore, the interpretation of optimal substructure is quite flexible and will have different meanings in different problems.
According to the state transition equation, and the initial states $dp[1] = cost[1]$ and $dp[2] = cost[2]$, we can obtain the dynamic programming code:
According to the state transition equation and the initial states $dp[1] = cost[1]$ and $dp[2] = cost[2]$, we can obtain the dynamic programming code:
=== "Python"
```python title="min_cost_climbing_stairs_dp.py"
def min_cost_climbing_stairs_dp(cost: list[int]) -> int:
"""Climbing stairs with minimum cost: Dynamic programming"""
"""Minimum cost climbing stairs: Dynamic programming"""
n = len(cost) - 1
if n == 1 or n == 2:
return cost[n]
# Initialize dp table, used to store subproblem solutions
# Initialize dp table, used to store solutions to subproblems
dp = [0] * (n + 1)
# Initial state: preset the smallest subproblem solution
# Initial state: preset the solution to the smallest subproblem
dp[1], dp[2] = cost[1], cost[2]
# State transition: gradually solve larger subproblems from smaller ones
for i in range(3, n + 1):
@@ -61,14 +61,14 @@ According to the state transition equation, and the initial states $dp[1] = cost
=== "C++"
```cpp title="min_cost_climbing_stairs_dp.cpp"
/* Climbing stairs with minimum cost: Dynamic programming */
/* Minimum cost climbing stairs: Dynamic programming */
int minCostClimbingStairsDP(vector<int> &cost) {
int n = cost.size() - 1;
if (n == 1 || n == 2)
return cost[n];
// Initialize dp table, used to store subproblem solutions
// Initialize dp table, used to store solutions to subproblems
vector<int> dp(n + 1);
// Initial state: preset the smallest subproblem solution
// Initial state: preset the solution to the smallest subproblem
dp[1] = cost[1];
dp[2] = cost[2];
// State transition: gradually solve larger subproblems from smaller ones
@@ -82,14 +82,14 @@ According to the state transition equation, and the initial states $dp[1] = cost
=== "Java"
```java title="min_cost_climbing_stairs_dp.java"
/* Climbing stairs with minimum cost: Dynamic programming */
/* Minimum cost climbing stairs: Dynamic programming */
int minCostClimbingStairsDP(int[] cost) {
int n = cost.length - 1;
if (n == 1 || n == 2)
return cost[n];
// Initialize dp table, used to store subproblem solutions
// Initialize dp table, used to store solutions to subproblems
int[] dp = new int[n + 1];
// Initial state: preset the smallest subproblem solution
// Initial state: preset the solution to the smallest subproblem
dp[1] = cost[1];
dp[2] = cost[2];
// State transition: gradually solve larger subproblems from smaller ones
@@ -103,82 +103,234 @@ According to the state transition equation, and the initial states $dp[1] = cost
=== "C#"
```csharp title="min_cost_climbing_stairs_dp.cs"
[class]{min_cost_climbing_stairs_dp}-[func]{MinCostClimbingStairsDP}
/* Minimum cost climbing stairs: Dynamic programming */
int MinCostClimbingStairsDP(int[] cost) {
int n = cost.Length - 1;
if (n == 1 || n == 2)
return cost[n];
// Initialize dp table, used to store solutions to subproblems
int[] dp = new int[n + 1];
// Initial state: preset the solution to the smallest subproblem
dp[1] = cost[1];
dp[2] = cost[2];
// State transition: gradually solve larger subproblems from smaller ones
for (int i = 3; i <= n; i++) {
dp[i] = Math.Min(dp[i - 1], dp[i - 2]) + cost[i];
}
return dp[n];
}
```
=== "Go"
```go title="min_cost_climbing_stairs_dp.go"
[class]{}-[func]{minCostClimbingStairsDP}
/* Minimum cost climbing stairs: Dynamic programming */
func minCostClimbingStairsDP(cost []int) int {
n := len(cost) - 1
if n == 1 || n == 2 {
return cost[n]
}
min := func(a, b int) int {
if a < b {
return a
}
return b
}
// Initialize dp table, used to store solutions to subproblems
dp := make([]int, n+1)
// Initial state: preset the solution to the smallest subproblem
dp[1] = cost[1]
dp[2] = cost[2]
// State transition: gradually solve larger subproblems from smaller ones
for i := 3; i <= n; i++ {
dp[i] = min(dp[i-1], dp[i-2]) + cost[i]
}
return dp[n]
}
```
=== "Swift"
```swift title="min_cost_climbing_stairs_dp.swift"
[class]{}-[func]{minCostClimbingStairsDP}
/* Minimum cost climbing stairs: Dynamic programming */
func minCostClimbingStairsDP(cost: [Int]) -> Int {
let n = cost.count - 1
if n == 1 || n == 2 {
return cost[n]
}
// Initialize dp table, used to store solutions to subproblems
var dp = Array(repeating: 0, count: n + 1)
// Initial state: preset the solution to the smallest subproblem
dp[1] = cost[1]
dp[2] = cost[2]
// State transition: gradually solve larger subproblems from smaller ones
for i in 3 ... n {
dp[i] = min(dp[i - 1], dp[i - 2]) + cost[i]
}
return dp[n]
}
```
=== "JS"
```javascript title="min_cost_climbing_stairs_dp.js"
[class]{}-[func]{minCostClimbingStairsDP}
/* Minimum cost climbing stairs: Dynamic programming */
function minCostClimbingStairsDP(cost) {
const n = cost.length - 1;
if (n === 1 || n === 2) {
return cost[n];
}
// Initialize dp table, used to store solutions to subproblems
const dp = new Array(n + 1);
// Initial state: preset the solution to the smallest subproblem
dp[1] = cost[1];
dp[2] = cost[2];
// State transition: gradually solve larger subproblems from smaller ones
for (let i = 3; i <= n; i++) {
dp[i] = Math.min(dp[i - 1], dp[i - 2]) + cost[i];
}
return dp[n];
}
```
=== "TS"
```typescript title="min_cost_climbing_stairs_dp.ts"
[class]{}-[func]{minCostClimbingStairsDP}
/* Minimum cost climbing stairs: Dynamic programming */
function minCostClimbingStairsDP(cost: Array<number>): number {
const n = cost.length - 1;
if (n === 1 || n === 2) {
return cost[n];
}
// Initialize dp table, used to store solutions to subproblems
const dp = new Array(n + 1);
// Initial state: preset the solution to the smallest subproblem
dp[1] = cost[1];
dp[2] = cost[2];
// State transition: gradually solve larger subproblems from smaller ones
for (let i = 3; i <= n; i++) {
dp[i] = Math.min(dp[i - 1], dp[i - 2]) + cost[i];
}
return dp[n];
}
```
=== "Dart"
```dart title="min_cost_climbing_stairs_dp.dart"
[class]{}-[func]{minCostClimbingStairsDP}
/* Minimum cost climbing stairs: Dynamic programming */
int minCostClimbingStairsDP(List<int> cost) {
int n = cost.length - 1;
if (n == 1 || n == 2) return cost[n];
// Initialize dp table, used to store solutions to subproblems
List<int> dp = List.filled(n + 1, 0);
// Initial state: preset the solution to the smallest subproblem
dp[1] = cost[1];
dp[2] = cost[2];
// State transition: gradually solve larger subproblems from smaller ones
for (int i = 3; i <= n; i++) {
dp[i] = min(dp[i - 1], dp[i - 2]) + cost[i];
}
return dp[n];
}
```
=== "Rust"
```rust title="min_cost_climbing_stairs_dp.rs"
[class]{}-[func]{min_cost_climbing_stairs_dp}
/* Minimum cost climbing stairs: Dynamic programming */
fn min_cost_climbing_stairs_dp(cost: &[i32]) -> i32 {
let n = cost.len() - 1;
if n == 1 || n == 2 {
return cost[n];
}
// Initialize dp table, used to store solutions to subproblems
let mut dp = vec![-1; n + 1];
// Initial state: preset the solution to the smallest subproblem
dp[1] = cost[1];
dp[2] = cost[2];
// State transition: gradually solve larger subproblems from smaller ones
for i in 3..=n {
dp[i] = cmp::min(dp[i - 1], dp[i - 2]) + cost[i];
}
dp[n]
}
```
=== "C"
```c title="min_cost_climbing_stairs_dp.c"
[class]{}-[func]{minCostClimbingStairsDP}
/* Minimum cost climbing stairs: Dynamic programming */
int minCostClimbingStairsDP(int cost[], int costSize) {
int n = costSize - 1;
if (n == 1 || n == 2)
return cost[n];
// Initialize dp table, used to store solutions to subproblems
int *dp = calloc(n + 1, sizeof(int));
// Initial state: preset the solution to the smallest subproblem
dp[1] = cost[1];
dp[2] = cost[2];
// State transition: gradually solve larger subproblems from smaller ones
for (int i = 3; i <= n; i++) {
dp[i] = myMin(dp[i - 1], dp[i - 2]) + cost[i];
}
int res = dp[n];
// Free memory
free(dp);
return res;
}
```
=== "Kotlin"
```kotlin title="min_cost_climbing_stairs_dp.kt"
[class]{}-[func]{minCostClimbingStairsDP}
/* Minimum cost climbing stairs: Dynamic programming */
fun minCostClimbingStairsDP(cost: IntArray): Int {
val n = cost.size - 1
if (n == 1 || n == 2) return cost[n]
// Initialize dp table, used to store solutions to subproblems
val dp = IntArray(n + 1)
// Initial state: preset the solution to the smallest subproblem
dp[1] = cost[1]
dp[2] = cost[2]
// State transition: gradually solve larger subproblems from smaller ones
for (i in 3..n) {
dp[i] = min(dp[i - 1], dp[i - 2]) + cost[i]
}
return dp[n]
}
```
=== "Ruby"
```ruby title="min_cost_climbing_stairs_dp.rb"
[class]{}-[func]{min_cost_climbing_stairs_dp}
```
=== "Zig"
```zig title="min_cost_climbing_stairs_dp.zig"
[class]{}-[func]{minCostClimbingStairsDP}
### Minimum cost climbing stairs: DP ###
def min_cost_climbing_stairs_dp(cost)
n = cost.length - 1
return cost[n] if n == 1 || n == 2
# Initialize dp table, used to store solutions to subproblems
dp = Array.new(n + 1, 0)
# Initial state: preset the solution to the smallest subproblem
dp[1], dp[2] = cost[1], cost[2]
# State transition: gradually solve larger subproblems from smaller ones
(3...(n + 1)).each { |i| dp[i] = [dp[i - 1], dp[i - 2]].min + cost[i] }
dp[n]
end
```
Figure 14-7 shows the dynamic programming process for the above code.
![Dynamic programming process for minimum cost of climbing stairs](dp_problem_features.assets/min_cost_cs_dp.png){ class="animation-figure" }
![Dynamic programming process for climbing stairs with minimum cost](dp_problem_features.assets/min_cost_cs_dp.png){ class="animation-figure" }
<p align="center"> Figure 14-7 &nbsp; Dynamic programming process for minimum cost of climbing stairs </p>
<p align="center"> Figure 14-7 &nbsp; Dynamic programming process for climbing stairs with minimum cost </p>
This problem can also be space-optimized, compressing one dimension to zero, reducing the space complexity from $O(n)$ to $O(1)$:
This problem can also be space-optimized, compressing from one dimension to zero, reducing the space complexity from $O(n)$ to $O(1)$:
=== "Python"
```python title="min_cost_climbing_stairs_dp.py"
def min_cost_climbing_stairs_dp_comp(cost: list[int]) -> int:
"""Climbing stairs with minimum cost: Space-optimized dynamic programming"""
"""Minimum cost climbing stairs: Space-optimized dynamic programming"""
n = len(cost) - 1
if n == 1 or n == 2:
return cost[n]
@@ -191,7 +343,7 @@ This problem can also be space-optimized, compressing one dimension to zero, red
=== "C++"
```cpp title="min_cost_climbing_stairs_dp.cpp"
/* Climbing stairs with minimum cost: Space-optimized dynamic programming */
/* Minimum cost climbing stairs: Space-optimized dynamic programming */
int minCostClimbingStairsDPComp(vector<int> &cost) {
int n = cost.size() - 1;
if (n == 1 || n == 2)
@@ -209,7 +361,7 @@ This problem can also be space-optimized, compressing one dimension to zero, red
=== "Java"
```java title="min_cost_climbing_stairs_dp.java"
/* Climbing stairs with minimum cost: Space-optimized dynamic programming */
/* Minimum cost climbing stairs: Space-optimized dynamic programming */
int minCostClimbingStairsDPComp(int[] cost) {
int n = cost.length - 1;
if (n == 1 || n == 2)
@@ -227,97 +379,231 @@ This problem can also be space-optimized, compressing one dimension to zero, red
=== "C#"
```csharp title="min_cost_climbing_stairs_dp.cs"
[class]{min_cost_climbing_stairs_dp}-[func]{MinCostClimbingStairsDPComp}
/* Minimum cost climbing stairs: Space-optimized dynamic programming */
int MinCostClimbingStairsDPComp(int[] cost) {
int n = cost.Length - 1;
if (n == 1 || n == 2)
return cost[n];
int a = cost[1], b = cost[2];
for (int i = 3; i <= n; i++) {
int tmp = b;
b = Math.Min(a, tmp) + cost[i];
a = tmp;
}
return b;
}
```
=== "Go"
```go title="min_cost_climbing_stairs_dp.go"
[class]{}-[func]{minCostClimbingStairsDPComp}
/* Minimum cost climbing stairs: Space-optimized dynamic programming */
func minCostClimbingStairsDPComp(cost []int) int {
n := len(cost) - 1
if n == 1 || n == 2 {
return cost[n]
}
min := func(a, b int) int {
if a < b {
return a
}
return b
}
// Initial state: preset the solution to the smallest subproblem
a, b := cost[1], cost[2]
// State transition: gradually solve larger subproblems from smaller ones
for i := 3; i <= n; i++ {
tmp := b
b = min(a, tmp) + cost[i]
a = tmp
}
return b
}
```
=== "Swift"
```swift title="min_cost_climbing_stairs_dp.swift"
[class]{}-[func]{minCostClimbingStairsDPComp}
/* Minimum cost climbing stairs: Space-optimized dynamic programming */
func minCostClimbingStairsDPComp(cost: [Int]) -> Int {
let n = cost.count - 1
if n == 1 || n == 2 {
return cost[n]
}
var (a, b) = (cost[1], cost[2])
for i in 3 ... n {
(a, b) = (b, min(a, b) + cost[i])
}
return b
}
```
=== "JS"
```javascript title="min_cost_climbing_stairs_dp.js"
[class]{}-[func]{minCostClimbingStairsDPComp}
/* Minimum cost climbing stairs: Space-optimized dynamic programming */
function minCostClimbingStairsDPComp(cost) {
const n = cost.length - 1;
if (n === 1 || n === 2) {
return cost[n];
}
let a = cost[1],
b = cost[2];
for (let i = 3; i <= n; i++) {
const tmp = b;
b = Math.min(a, tmp) + cost[i];
a = tmp;
}
return b;
}
```
=== "TS"
```typescript title="min_cost_climbing_stairs_dp.ts"
[class]{}-[func]{minCostClimbingStairsDPComp}
/* Minimum cost climbing stairs: Space-optimized dynamic programming */
function minCostClimbingStairsDPComp(cost: Array<number>): number {
const n = cost.length - 1;
if (n === 1 || n === 2) {
return cost[n];
}
let a = cost[1],
b = cost[2];
for (let i = 3; i <= n; i++) {
const tmp = b;
b = Math.min(a, tmp) + cost[i];
a = tmp;
}
return b;
}
```
=== "Dart"
```dart title="min_cost_climbing_stairs_dp.dart"
[class]{}-[func]{minCostClimbingStairsDPComp}
/* Minimum cost climbing stairs: Space-optimized dynamic programming */
int minCostClimbingStairsDPComp(List<int> cost) {
int n = cost.length - 1;
if (n == 1 || n == 2) return cost[n];
int a = cost[1], b = cost[2];
for (int i = 3; i <= n; i++) {
int tmp = b;
b = min(a, tmp) + cost[i];
a = tmp;
}
return b;
}
```
=== "Rust"
```rust title="min_cost_climbing_stairs_dp.rs"
[class]{}-[func]{min_cost_climbing_stairs_dp_comp}
/* Minimum cost climbing stairs: Space-optimized dynamic programming */
fn min_cost_climbing_stairs_dp_comp(cost: &[i32]) -> i32 {
let n = cost.len() - 1;
if n == 1 || n == 2 {
return cost[n];
};
let (mut a, mut b) = (cost[1], cost[2]);
for i in 3..=n {
let tmp = b;
b = cmp::min(a, tmp) + cost[i];
a = tmp;
}
b
}
```
=== "C"
```c title="min_cost_climbing_stairs_dp.c"
[class]{}-[func]{minCostClimbingStairsDPComp}
/* Minimum cost climbing stairs: Space-optimized dynamic programming */
int minCostClimbingStairsDPComp(int cost[], int costSize) {
int n = costSize - 1;
if (n == 1 || n == 2)
return cost[n];
int a = cost[1], b = cost[2];
for (int i = 3; i <= n; i++) {
int tmp = b;
b = myMin(a, tmp) + cost[i];
a = tmp;
}
return b;
}
```
=== "Kotlin"
```kotlin title="min_cost_climbing_stairs_dp.kt"
[class]{}-[func]{minCostClimbingStairsDPComp}
/* Minimum cost climbing stairs: Space-optimized dynamic programming */
fun minCostClimbingStairsDPComp(cost: IntArray): Int {
val n = cost.size - 1
if (n == 1 || n == 2) return cost[n]
var a = cost[1]
var b = cost[2]
for (i in 3..n) {
val tmp = b
b = min(a, tmp) + cost[i]
a = tmp
}
return b
}
```
=== "Ruby"
```ruby title="min_cost_climbing_stairs_dp.rb"
[class]{}-[func]{min_cost_climbing_stairs_dp_comp}
### Minimum cost climbing stairs: DP ###
def min_cost_climbing_stairs_dp(cost)
n = cost.length - 1
return cost[n] if n == 1 || n == 2
# Initialize dp table, used to store solutions to subproblems
dp = Array.new(n + 1, 0)
# Initial state: preset the solution to the smallest subproblem
dp[1], dp[2] = cost[1], cost[2]
# State transition: gradually solve larger subproblems from smaller ones
(3...(n + 1)).each { |i| dp[i] = [dp[i - 1], dp[i - 2]].min + cost[i] }
dp[n]
end
# Minimum cost climbing stairs: Space-optimized dynamic programming
def min_cost_climbing_stairs_dp_comp(cost)
n = cost.length - 1
return cost[n] if n == 1 || n == 2
a, b = cost[1], cost[2]
(3...(n + 1)).each { |i| a, b = b, [a, b].min + cost[i] }
b
end
```
=== "Zig"
## 14.2.2 &nbsp; No Aftereffects
```zig title="min_cost_climbing_stairs_dp.zig"
[class]{}-[func]{minCostClimbingStairsDPComp}
```
No aftereffects is one of the important characteristics that enable dynamic programming to solve problems effectively. Its definition is: **given a certain state, its future development is only related to the current state and has nothing to do with all past states**.
## 14.2.2 &nbsp; Statelessness
Statelessness is one of the important characteristics that make dynamic programming effective in solving problems. Its definition is: **Given a certain state, its future development is only related to the current state and unrelated to all past states experienced**.
Taking the stair climbing problem as an example, given state $i$, it will develop into states $i+1$ and $i+2$, corresponding to jumping 1 step and 2 steps respectively. When making these two choices, we do not need to consider the states before state $i$, as they do not affect the future of state $i$.
Taking the stair climbing problem as an example, given state $i$, it will develop into states $i+1$ and $i+2$, corresponding to jumping $1$ step and jumping $2$ steps, respectively. When making these two choices, we do not need to consider the states before state $i$, as they have no effect on the future of state $i$.
However, if we add a constraint to the stair climbing problem, the situation changes.
!!! question "Stair climbing with constraints"
!!! question "Climbing stairs with constraint"
Given a staircase with $n$ steps, you can go up 1 or 2 steps each time, **but you cannot jump 1 step twice in a row**. How many ways are there to climb to the top?
Given a staircase with $n$ steps, where you can climb $1$ or $2$ steps at a time, **but you cannot jump $1$ step in two consecutive rounds**. How many ways are there to climb to the top?
As shown in Figure 14-8, there are only 2 feasible options for climbing to the 3rd step, among which the option of jumping 1 step three times in a row does not meet the constraint condition and is therefore discarded.
As shown in Figure 14-8, there are only $2$ feasible ways to climb to the $3$rd step. The way of jumping $1$ step three consecutive times does not satisfy the constraint and is therefore discarded.
![Number of feasible options for climbing to the 3rd step with constraints](dp_problem_features.assets/climbing_stairs_constraint_example.png){ class="animation-figure" }
![Number of ways to climb to the 3rd step with constraint](dp_problem_features.assets/climbing_stairs_constraint_example.png){ class="animation-figure" }
<p align="center"> Figure 14-8 &nbsp; Number of feasible options for climbing to the 3rd step with constraints </p>
<p align="center"> Figure 14-8 &nbsp; Number of ways to climb to the 3rd step with constraint </p>
In this problem, if the last round was a jump of 1 step, then the next round must be a jump of 2 steps. This means that **the next step choice cannot be independently determined by the current state (current stair step), but also depends on the previous state (last round's stair step)**.
In this problem, if the previous round was a jump of $1$ step, then the next round must jump $2$ steps. This means that **the next choice cannot be determined solely by the current state (current stair step number), but also depends on the previous state (the stair step number from the previous round)**.
It is not difficult to find that this problem no longer satisfies statelessness, and the state transition equation $dp[i] = dp[i-1] + dp[i-2]$ also fails, because $dp[i-1]$ represents this round's jump of 1 step, but it includes many "last round was a jump of 1 step" options, which, to meet the constraint, cannot be directly included in $dp[i]$.
It is not difficult to see that this problem no longer satisfies no aftereffects, and the state transition equation $dp[i] = dp[i-1] + dp[i-2]$ also fails, because $dp[i-1]$ represents jumping $1$ step in this round, but it includes many solutions where "the previous round was a jump of $1$ step", which cannot be directly counted in $dp[i]$ to satisfy the constraint.
For this, we need to expand the state definition: **State $[i, j]$ represents being on the $i$-th step and the last round was a jump of $j$ steps**, where $j \in \{1, 2\}$. This state definition effectively distinguishes whether the last round was a jump of 1 step or 2 steps, and we can judge accordingly where the current state came from.
For this reason, we need to expand the state definition: **state $[i, j]$ represents being on the $i$-th step with the previous round having jumped $j$ steps**, where $j \in \{1, 2\}$. This state definition effectively distinguishes whether the previous round was a jump of $1$ step or $2$ steps, allowing us to determine where the current state came from.
- When the last round was a jump of 1 step, the round before last could only choose to jump 2 steps, that is, $dp[i, 1]$ can only be transferred from $dp[i-1, 2]$.
- When the last round was a jump of 2 steps, the round before last could choose to jump 1 step or 2 steps, that is, $dp[i, 2]$ can be transferred from $dp[i-2, 1]$ or $dp[i-2, 2]$.
- When the previous round jumped $1$ step, the round before that could only choose to jump $2$ steps, i.e., $dp[i, 1]$ can only be transferred from $dp[i-1, 2]$.
- When the previous round jumped $2$ steps, the round before that could choose to jump $1$ step or $2$ steps, i.e., $dp[i, 2]$ can be transferred from $dp[i-2, 1]$ or $dp[i-2, 2]$.
As shown in Figure 14-9, $dp[i, j]$ represents the number of solutions for state $[i, j]$. At this point, the state transition equation is:
As shown in Figure 14-9, under this definition, $dp[i, j]$ represents the number of ways for state $[i, j]$. The state transition equation is then:
$$
\begin{cases}
@@ -326,22 +612,22 @@ dp[i, 2] = dp[i-2, 1] + dp[i-2, 2]
\end{cases}
$$
![Recursive relationship considering constraints](dp_problem_features.assets/climbing_stairs_constraint_state_transfer.png){ class="animation-figure" }
![Recurrence relation considering constraints](dp_problem_features.assets/climbing_stairs_constraint_state_transfer.png){ class="animation-figure" }
<p align="center"> Figure 14-9 &nbsp; Recursive relationship considering constraints </p>
<p align="center"> Figure 14-9 &nbsp; Recurrence relation considering constraints </p>
In the end, returning $dp[n, 1] + dp[n, 2]$ will do, the sum of the two representing the total number of solutions for climbing to the $n$-th step:
Finally, return $dp[n, 1] + dp[n, 2]$, where the sum of the two represents the total number of ways to climb to the $n$-th step:
=== "Python"
```python title="climbing_stairs_constraint_dp.py"
def climbing_stairs_constraint_dp(n: int) -> int:
"""Constrained climbing stairs: Dynamic programming"""
"""Climbing stairs with constraint: Dynamic programming"""
if n == 1 or n == 2:
return 1
# Initialize dp table, used to store subproblem solutions
# Initialize dp table, used to store solutions to subproblems
dp = [[0] * 3 for _ in range(n + 1)]
# Initial state: preset the smallest subproblem solution
# Initial state: preset the solution to the smallest subproblem
dp[1][1], dp[1][2] = 1, 0
dp[2][1], dp[2][2] = 0, 1
# State transition: gradually solve larger subproblems from smaller ones
@@ -354,14 +640,14 @@ In the end, returning $dp[n, 1] + dp[n, 2]$ will do, the sum of the two represen
=== "C++"
```cpp title="climbing_stairs_constraint_dp.cpp"
/* Constrained climbing stairs: Dynamic programming */
/* Climbing stairs with constraint: Dynamic programming */
int climbingStairsConstraintDP(int n) {
if (n == 1 || n == 2) {
return 1;
}
// Initialize dp table, used to store subproblem solutions
// Initialize dp table, used to store solutions to subproblems
vector<vector<int>> dp(n + 1, vector<int>(3, 0));
// Initial state: preset the smallest subproblem solution
// Initial state: preset the solution to the smallest subproblem
dp[1][1] = 1;
dp[1][2] = 0;
dp[2][1] = 0;
@@ -378,14 +664,14 @@ In the end, returning $dp[n, 1] + dp[n, 2]$ will do, the sum of the two represen
=== "Java"
```java title="climbing_stairs_constraint_dp.java"
/* Constrained climbing stairs: Dynamic programming */
/* Climbing stairs with constraint: Dynamic programming */
int climbingStairsConstraintDP(int n) {
if (n == 1 || n == 2) {
return 1;
}
// Initialize dp table, used to store subproblem solutions
// Initialize dp table, used to store solutions to subproblems
int[][] dp = new int[n + 1][3];
// Initial state: preset the smallest subproblem solution
// Initial state: preset the solution to the smallest subproblem
dp[1][1] = 1;
dp[1][2] = 0;
dp[2][1] = 0;
@@ -402,75 +688,256 @@ In the end, returning $dp[n, 1] + dp[n, 2]$ will do, the sum of the two represen
=== "C#"
```csharp title="climbing_stairs_constraint_dp.cs"
[class]{climbing_stairs_constraint_dp}-[func]{ClimbingStairsConstraintDP}
/* Climbing stairs with constraint: Dynamic programming */
int ClimbingStairsConstraintDP(int n) {
if (n == 1 || n == 2) {
return 1;
}
// Initialize dp table, used to store solutions to subproblems
int[,] dp = new int[n + 1, 3];
// Initial state: preset the solution to the smallest subproblem
dp[1, 1] = 1;
dp[1, 2] = 0;
dp[2, 1] = 0;
dp[2, 2] = 1;
// State transition: gradually solve larger subproblems from smaller ones
for (int i = 3; i <= n; i++) {
dp[i, 1] = dp[i - 1, 2];
dp[i, 2] = dp[i - 2, 1] + dp[i - 2, 2];
}
return dp[n, 1] + dp[n, 2];
}
```
=== "Go"
```go title="climbing_stairs_constraint_dp.go"
[class]{}-[func]{climbingStairsConstraintDP}
/* Climbing stairs with constraint: Dynamic programming */
func climbingStairsConstraintDP(n int) int {
if n == 1 || n == 2 {
return 1
}
// Initialize dp table, used to store solutions to subproblems
dp := make([][3]int, n+1)
// Initial state: preset the solution to the smallest subproblem
dp[1][1] = 1
dp[1][2] = 0
dp[2][1] = 0
dp[2][2] = 1
// State transition: gradually solve larger subproblems from smaller ones
for i := 3; i <= n; i++ {
dp[i][1] = dp[i-1][2]
dp[i][2] = dp[i-2][1] + dp[i-2][2]
}
return dp[n][1] + dp[n][2]
}
```
=== "Swift"
```swift title="climbing_stairs_constraint_dp.swift"
[class]{}-[func]{climbingStairsConstraintDP}
/* Climbing stairs with constraint: Dynamic programming */
func climbingStairsConstraintDP(n: Int) -> Int {
if n == 1 || n == 2 {
return 1
}
// Initialize dp table, used to store solutions to subproblems
var dp = Array(repeating: Array(repeating: 0, count: 3), count: n + 1)
// Initial state: preset the solution to the smallest subproblem
dp[1][1] = 1
dp[1][2] = 0
dp[2][1] = 0
dp[2][2] = 1
// State transition: gradually solve larger subproblems from smaller ones
for i in 3 ... n {
dp[i][1] = dp[i - 1][2]
dp[i][2] = dp[i - 2][1] + dp[i - 2][2]
}
return dp[n][1] + dp[n][2]
}
```
=== "JS"
```javascript title="climbing_stairs_constraint_dp.js"
[class]{}-[func]{climbingStairsConstraintDP}
/* Climbing stairs with constraint: Dynamic programming */
function climbingStairsConstraintDP(n) {
if (n === 1 || n === 2) {
return 1;
}
// Initialize dp table, used to store solutions to subproblems
const dp = Array.from(new Array(n + 1), () => new Array(3));
// Initial state: preset the solution to the smallest subproblem
dp[1][1] = 1;
dp[1][2] = 0;
dp[2][1] = 0;
dp[2][2] = 1;
// State transition: gradually solve larger subproblems from smaller ones
for (let i = 3; i <= n; i++) {
dp[i][1] = dp[i - 1][2];
dp[i][2] = dp[i - 2][1] + dp[i - 2][2];
}
return dp[n][1] + dp[n][2];
}
```
=== "TS"
```typescript title="climbing_stairs_constraint_dp.ts"
[class]{}-[func]{climbingStairsConstraintDP}
/* Climbing stairs with constraint: Dynamic programming */
function climbingStairsConstraintDP(n: number): number {
if (n === 1 || n === 2) {
return 1;
}
// Initialize dp table, used to store solutions to subproblems
const dp = Array.from({ length: n + 1 }, () => new Array(3));
// Initial state: preset the solution to the smallest subproblem
dp[1][1] = 1;
dp[1][2] = 0;
dp[2][1] = 0;
dp[2][2] = 1;
// State transition: gradually solve larger subproblems from smaller ones
for (let i = 3; i <= n; i++) {
dp[i][1] = dp[i - 1][2];
dp[i][2] = dp[i - 2][1] + dp[i - 2][2];
}
return dp[n][1] + dp[n][2];
}
```
=== "Dart"
```dart title="climbing_stairs_constraint_dp.dart"
[class]{}-[func]{climbingStairsConstraintDP}
/* Climbing stairs with constraint: Dynamic programming */
int climbingStairsConstraintDP(int n) {
if (n == 1 || n == 2) {
return 1;
}
// Initialize dp table, used to store solutions to subproblems
List<List<int>> dp = List.generate(n + 1, (index) => List.filled(3, 0));
// Initial state: preset the solution to the smallest subproblem
dp[1][1] = 1;
dp[1][2] = 0;
dp[2][1] = 0;
dp[2][2] = 1;
// State transition: gradually solve larger subproblems from smaller ones
for (int i = 3; i <= n; i++) {
dp[i][1] = dp[i - 1][2];
dp[i][2] = dp[i - 2][1] + dp[i - 2][2];
}
return dp[n][1] + dp[n][2];
}
```
=== "Rust"
```rust title="climbing_stairs_constraint_dp.rs"
[class]{}-[func]{climbing_stairs_constraint_dp}
/* Climbing stairs with constraint: Dynamic programming */
fn climbing_stairs_constraint_dp(n: usize) -> i32 {
if n == 1 || n == 2 {
return 1;
};
// Initialize dp table, used to store solutions to subproblems
let mut dp = vec![vec![-1; 3]; n + 1];
// Initial state: preset the solution to the smallest subproblem
dp[1][1] = 1;
dp[1][2] = 0;
dp[2][1] = 0;
dp[2][2] = 1;
// State transition: gradually solve larger subproblems from smaller ones
for i in 3..=n {
dp[i][1] = dp[i - 1][2];
dp[i][2] = dp[i - 2][1] + dp[i - 2][2];
}
dp[n][1] + dp[n][2]
}
```
=== "C"
```c title="climbing_stairs_constraint_dp.c"
[class]{}-[func]{climbingStairsConstraintDP}
/* Climbing stairs with constraint: Dynamic programming */
int climbingStairsConstraintDP(int n) {
if (n == 1 || n == 2) {
return 1;
}
// Initialize dp table, used to store solutions to subproblems
int **dp = malloc((n + 1) * sizeof(int *));
for (int i = 0; i <= n; i++) {
dp[i] = calloc(3, sizeof(int));
}
// Initial state: preset the solution to the smallest subproblem
dp[1][1] = 1;
dp[1][2] = 0;
dp[2][1] = 0;
dp[2][2] = 1;
// State transition: gradually solve larger subproblems from smaller ones
for (int i = 3; i <= n; i++) {
dp[i][1] = dp[i - 1][2];
dp[i][2] = dp[i - 2][1] + dp[i - 2][2];
}
int res = dp[n][1] + dp[n][2];
// Free memory
for (int i = 0; i <= n; i++) {
free(dp[i]);
}
free(dp);
return res;
}
```
=== "Kotlin"
```kotlin title="climbing_stairs_constraint_dp.kt"
[class]{}-[func]{climbingStairsConstraintDP}
/* Climbing stairs with constraint: Dynamic programming */
fun climbingStairsConstraintDP(n: Int): Int {
if (n == 1 || n == 2) {
return 1
}
// Initialize dp table, used to store solutions to subproblems
val dp = Array(n + 1) { IntArray(3) }
// Initial state: preset the solution to the smallest subproblem
dp[1][1] = 1
dp[1][2] = 0
dp[2][1] = 0
dp[2][2] = 1
// State transition: gradually solve larger subproblems from smaller ones
for (i in 3..n) {
dp[i][1] = dp[i - 1][2]
dp[i][2] = dp[i - 2][1] + dp[i - 2][2]
}
return dp[n][1] + dp[n][2]
}
```
=== "Ruby"
```ruby title="climbing_stairs_constraint_dp.rb"
[class]{}-[func]{climbing_stairs_constraint_dp}
### Climbing stairs with constraint: DP ###
def climbing_stairs_constraint_dp(n)
return 1 if n == 1 || n == 2
# Initialize dp table, used to store solutions to subproblems
dp = Array.new(n + 1) { Array.new(3, 0) }
# Initial state: preset the solution to the smallest subproblem
dp[1][1], dp[1][2] = 1, 0
dp[2][1], dp[2][2] = 0, 1
# State transition: gradually solve larger subproblems from smaller ones
for i in 3...(n + 1)
dp[i][1] = dp[i - 1][2]
dp[i][2] = dp[i - 2][1] + dp[i - 2][2]
end
dp[n][1] + dp[n][2]
end
```
=== "Zig"
In the above case, since we only need to consider one more preceding state, we can still make the problem satisfy no aftereffects by expanding the state definition. However, some problems have very severe "aftereffects".
```zig title="climbing_stairs_constraint_dp.zig"
[class]{}-[func]{climbingStairsConstraintDP}
```
!!! question "Climbing stairs with obstacle generation"
In the above cases, since we only need to consider the previous state, we can still meet the statelessness by expanding the state definition. However, some problems have very serious "state effects".
Given a staircase with $n$ steps, where you can climb $1$ or $2$ steps at a time. **It is stipulated that when climbing to the $i$-th step, the system will automatically place an obstacle on the $2i$-th step, and thereafter no round is allowed to jump to the $2i$-th step**. For example, if the first two rounds jump to the $2$nd and $3$rd steps, then afterwards you cannot jump to the $4$th and $6$th steps. How many ways are there to climb to the top?
!!! question "Stair climbing with obstacle generation"
In this problem, the next jump depends on all past states, because each jump places obstacles on higher steps, affecting future jumps. For such problems, dynamic programming is often difficult to solve.
Given a staircase with $n$ steps, you can go up 1 or 2 steps each time. **It is stipulated that when climbing to the $i$-th step, the system automatically places an obstacle on the $2i$-th step, and thereafter all rounds are not allowed to jump to the $2i$-th step**. For example, if the first two rounds jump to the 2nd and 3rd steps, then later you cannot jump to the 4th and 6th steps. How many ways are there to climb to the top?
In this problem, the next jump depends on all past states, as each jump places obstacles on higher steps, affecting future jumps. For such problems, dynamic programming often struggles to solve.
In fact, many complex combinatorial optimization problems (such as the traveling salesman problem) do not satisfy statelessness. For these kinds of problems, we usually choose to use other methods, such as heuristic search, genetic algorithms, reinforcement learning, etc., to obtain usable local optimal solutions within a limited time.
In fact, many complex combinatorial optimization problems (such as the traveling salesman problem) do not satisfy no aftereffects. For such problems, we usually choose to use other methods, such as heuristic search, genetic algorithms, reinforcement learning, etc., to obtain usable local optimal solutions within a limited time.
en/docs/chapter_dynamic_programming/dp_solution_pipeline.md
+920 -164
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en/docs/chapter_dynamic_programming/edit_distance_problem.md
+608 -89
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@@ -2,80 +2,80 @@
comments: true
---
# 14.6 &nbsp; Edit distance problem
# 14.6 &nbsp; Edit Distance Problem
Edit distance, also known as Levenshtein distance, refers to the minimum number of modifications required to transform one string into another, commonly used in information retrieval and natural language processing to measure the similarity between two sequences.
Edit distance, also known as Levenshtein distance, refers to the minimum number of edits required to transform one string into another, commonly used in information retrieval and natural language processing to measure the similarity between two sequences.
!!! question
Given two strings $s$ and $t$, return the minimum number of edits required to transform $s$ into $t$.
You can perform three types of edits on a string: insert a character, delete a character, or replace a character with any other character.
You can perform three types of edit operations on a string: insert a character, delete a character, or replace a character with any other character.
As shown in Figure 14-27, transforming `kitten` into `sitting` requires 3 edits, including 2 replacements and 1 insertion; transforming `hello` into `algo` requires 3 steps, including 2 replacements and 1 deletion.
![Example data of edit distance](edit_distance_problem.assets/edit_distance_example.png){ class="animation-figure" }
![Example data for edit distance](edit_distance_problem.assets/edit_distance_example.png){ class="animation-figure" }
<p align="center"> Figure 14-27 &nbsp; Example data of edit distance </p>
<p align="center"> Figure 14-27 &nbsp; Example data for edit distance </p>
**The edit distance problem can naturally be explained with a decision tree model**. Strings correspond to tree nodes, and a round of decision (an edit operation) corresponds to an edge of the tree.
**The edit distance problem can be naturally explained using the decision tree model**. Strings correspond to tree nodes, and a round of decision (one edit operation) corresponds to an edge of the tree.
As shown in Figure 14-28, with unrestricted operations, each node can derive many edges, each corresponding to one operation, meaning there are many possible paths to transform `hello` into `algo`.
As shown in Figure 14-28, without restricting operations, each node can branch into many edges, with each edge corresponding to one operation, meaning there are many possible paths to transform `hello` into `algo`.
From the perspective of the decision tree, the goal of this problem is to find the shortest path between the node `hello` and the node `algo`.
From the perspective of the decision tree, the goal of this problem is to find the shortest path between node `hello` and node `algo`.
![Edit distance problem represented based on decision tree model](edit_distance_problem.assets/edit_distance_decision_tree.png){ class="animation-figure" }
![Representing edit distance problem based on decision tree model](edit_distance_problem.assets/edit_distance_decision_tree.png){ class="animation-figure" }
<p align="center"> Figure 14-28 &nbsp; Edit distance problem represented based on decision tree model </p>
<p align="center"> Figure 14-28 &nbsp; Representing edit distance problem based on decision tree model </p>
### 1. &nbsp; Dynamic programming approach
### 1. &nbsp; Dynamic Programming Approach
**Step one: Think about each round of decision, define the state, thus obtaining the $dp$ table**
**Step 1: Think about the decisions in each round, define the state, and thus obtain the $dp$ table**
Each round of decision involves performing one edit operation on string $s$.
We aim to gradually reduce the problem size during the edit process, which enables us to construct subproblems. Let the lengths of strings $s$ and $t$ be $n$ and $m$, respectively. We first consider the tail characters of both strings $s[n-1]$ and $t[m-1]$.
We want the problem scale to gradually decrease during the editing process, which allows us to construct subproblems. Let the lengths of strings $s$ and $t$ be $n$ and $m$ respectively. We first consider the tail characters of the two strings, $s[n-1]$ and $t[m-1]$.
- If $s[n-1]$ and $t[m-1]$ are the same, we can skip them and directly consider $s[n-2]$ and $t[m-2]$.
- If $s[n-1]$ and $t[m-1]$ are different, we need to perform one edit on $s$ (insert, delete, replace) so that the tail characters of the two strings match, allowing us to skip them and consider a smaller-scale problem.
- If $s[n-1]$ and $t[m-1]$ are different, we need to perform one edit on $s$ (insert, delete, or replace) to make the tail characters of the two strings the same, allowing us to skip them and consider a smaller-scale problem.
Thus, each round of decision (edit operation) in string $s$ changes the remaining characters in $s$ and $t$ to be matched. Therefore, the state is the $i$-th and $j$-th characters currently considered in $s$ and $t$, denoted as $[i, j]$.
In other words, each round of decision (edit operation) we make on string $s$ will change the remaining characters to be matched in $s$ and $t$. Therefore, the state is the $i$-th and $j$-th characters currently being considered in $s$ and $t$, denoted as $[i, j]$.
State $[i, j]$ corresponds to the subproblem: **The minimum number of edits required to change the first $i$ characters of $s$ into the first $j$ characters of $t$**.
State $[i, j]$ corresponds to the subproblem: **the minimum number of edits required to change the first $i$ characters of $s$ into the first $j$ characters of $t$**.
From this, we obtain a two-dimensional $dp$ table of size $(i+1) \times (j+1)$.
**Step two: Identify the optimal substructure and then derive the state transition equation**
**Step 2: Identify the optimal substructure, and then derive the state transition equation**
Consider the subproblem $dp[i, j]$, whose corresponding tail characters of the two strings are $s[i-1]$ and $t[j-1]$, which can be divided into three scenarios as shown in Figure 14-29.
Consider subproblem $dp[i, j]$, where the tail characters of the corresponding two strings are $s[i-1]$ and $t[j-1]$, which can be divided into the three cases shown in Figure 14-29 based on different edit operations.
1. Add $t[j-1]$ after $s[i-1]$, then the remaining subproblem is $dp[i, j-1]$.
1. Insert $t[j-1]$ after $s[i-1]$, then the remaining subproblem is $dp[i, j-1]$.
2. Delete $s[i-1]$, then the remaining subproblem is $dp[i-1, j]$.
3. Replace $s[i-1]$ with $t[j-1]$, then the remaining subproblem is $dp[i-1, j-1]$.
![State transition of edit distance](edit_distance_problem.assets/edit_distance_state_transfer.png){ class="animation-figure" }
![State transition for edit distance](edit_distance_problem.assets/edit_distance_state_transfer.png){ class="animation-figure" }
<p align="center"> Figure 14-29 &nbsp; State transition of edit distance </p>
<p align="center"> Figure 14-29 &nbsp; State transition for edit distance </p>
Based on the analysis above, we can determine the optimal substructure: The minimum number of edits for $dp[i, j]$ is the minimum among $dp[i, j-1]$, $dp[i-1, j]$, and $dp[i-1, j-1]$, plus the edit step $1$. The corresponding state transition equation is:
Based on the above analysis, the optimal substructure can be obtained: the minimum number of edits for $dp[i, j]$ equals the minimum among the minimum edit steps of $dp[i, j-1]$, $dp[i-1, j]$, and $dp[i-1, j-1]$, plus the edit step $1$ for this time. The corresponding state transition equation is:
$$
dp[i, j] = \min(dp[i, j-1], dp[i-1, j], dp[i-1, j-1]) + 1
$$
Please note, **when $s[i-1]$ and $t[j-1]$ are the same, no edit is required for the current character**, in which case the state transition equation is:
Please note that **when $s[i-1]$ and $t[j-1]$ are the same, no edit is required for the current character**, in which case the state transition equation is:
$$
dp[i, j] = dp[i-1, j-1]
$$
**Step three: Determine the boundary conditions and the order of state transitions**
**Step 3: Determine boundary conditions and state transition order**
When both strings are empty, the number of edits is $0$, i.e., $dp[0, 0] = 0$. When $s$ is empty but $t$ is not, the minimum number of edits equals the length of $t$, that is, the first row $dp[0, j] = j$. When $s$ is not empty but $t$ is, the minimum number of edits equals the length of $s$, that is, the first column $dp[i, 0] = i$.
When both strings are empty, the number of edit steps is $0$, i.e., $dp[0, 0] = 0$. When $s$ is empty but $t$ is not, the minimum number of edit steps equals the length of $t$, i.e., the first row $dp[0, j] = j$. When $s$ is not empty but $t$ is empty, the minimum number of edit steps equals the length of $s$, i.e., the first column $dp[i, 0] = i$.
Observing the state transition equation, solving $dp[i, j]$ depends on the solutions to the left, above, and upper left, so a double loop can be used to traverse the entire $dp$ table in the correct order.
Observing the state transition equation, the solution $dp[i, j]$ depends on solutions to the left, above, and upper-left, so the entire $dp$ table can be traversed in order through two nested loops.
### 2. &nbsp; Code implementation
### 2. &nbsp; Code Implementation
=== "Python"
@@ -89,14 +89,14 @@ Observing the state transition equation, solving $dp[i, j]$ depends on the solut
dp[i][0] = i
for j in range(1, m + 1):
dp[0][j] = j
# State transition: the rest of the rows and columns
# State transition: rest of the rows and columns
for i in range(1, n + 1):
for j in range(1, m + 1):
if s[i - 1] == t[j - 1]:
# If the two characters are equal, skip these two characters
# If two characters are equal, skip both characters
dp[i][j] = dp[i - 1][j - 1]
else:
# The minimum number of edits = the minimum number of edits from three operations (insert, remove, replace) + 1
# Minimum edit steps = minimum edit steps of insert, delete, replace + 1
dp[i][j] = min(dp[i][j - 1], dp[i - 1][j], dp[i - 1][j - 1]) + 1
return dp[n][m]
```
@@ -115,14 +115,14 @@ Observing the state transition equation, solving $dp[i, j]$ depends on the solut
for (int j = 1; j <= m; j++) {
dp[0][j] = j;
}
// State transition: the rest of the rows and columns
// State transition: rest of the rows and columns
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++) {
if (s[i - 1] == t[j - 1]) {
// If the two characters are equal, skip these two characters
// If two characters are equal, skip both characters
dp[i][j] = dp[i - 1][j - 1];
} else {
// The minimum number of edits = the minimum number of edits from three operations (insert, remove, replace) + 1
// Minimum edit steps = minimum edit steps of insert, delete, replace + 1
dp[i][j] = min(min(dp[i][j - 1], dp[i - 1][j]), dp[i - 1][j - 1]) + 1;
}
}
@@ -145,14 +145,14 @@ Observing the state transition equation, solving $dp[i, j]$ depends on the solut
for (int j = 1; j <= m; j++) {
dp[0][j] = j;
}
// State transition: the rest of the rows and columns
// State transition: rest of the rows and columns
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++) {
if (s.charAt(i - 1) == t.charAt(j - 1)) {
// If the two characters are equal, skip these two characters
// If two characters are equal, skip both characters
dp[i][j] = dp[i - 1][j - 1];
} else {
// The minimum number of edits = the minimum number of edits from three operations (insert, remove, replace) + 1
// Minimum edit steps = minimum edit steps of insert, delete, replace + 1
dp[i][j] = Math.min(Math.min(dp[i][j - 1], dp[i - 1][j]), dp[i - 1][j - 1]) + 1;
}
}
@@ -164,73 +164,323 @@ Observing the state transition equation, solving $dp[i, j]$ depends on the solut
=== "C#"
```csharp title="edit_distance.cs"
[class]{edit_distance}-[func]{EditDistanceDP}
/* Edit distance: Dynamic programming */
int EditDistanceDP(string s, string t) {
int n = s.Length, m = t.Length;
int[,] dp = new int[n + 1, m + 1];
// State transition: first row and first column
for (int i = 1; i <= n; i++) {
dp[i, 0] = i;
}
for (int j = 1; j <= m; j++) {
dp[0, j] = j;
}
// State transition: rest of the rows and columns
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++) {
if (s[i - 1] == t[j - 1]) {
// If two characters are equal, skip both characters
dp[i, j] = dp[i - 1, j - 1];
} else {
// Minimum edit steps = minimum edit steps of insert, delete, replace + 1
dp[i, j] = Math.Min(Math.Min(dp[i, j - 1], dp[i - 1, j]), dp[i - 1, j - 1]) + 1;
}
}
}
return dp[n, m];
}
```
=== "Go"
```go title="edit_distance.go"
[class]{}-[func]{editDistanceDP}
/* Edit distance: Dynamic programming */
func editDistanceDP(s string, t string) int {
n := len(s)
m := len(t)
dp := make([][]int, n+1)
for i := 0; i <= n; i++ {
dp[i] = make([]int, m+1)
}
// State transition: first row and first column
for i := 1; i <= n; i++ {
dp[i][0] = i
}
for j := 1; j <= m; j++ {
dp[0][j] = j
}
// State transition: rest of the rows and columns
for i := 1; i <= n; i++ {
for j := 1; j <= m; j++ {
if s[i-1] == t[j-1] {
// If two characters are equal, skip both characters
dp[i][j] = dp[i-1][j-1]
} else {
// Minimum edit steps = minimum edit steps of insert, delete, replace + 1
dp[i][j] = MinInt(MinInt(dp[i][j-1], dp[i-1][j]), dp[i-1][j-1]) + 1
}
}
}
return dp[n][m]
}
```
=== "Swift"
```swift title="edit_distance.swift"
[class]{}-[func]{editDistanceDP}
/* Edit distance: Dynamic programming */
func editDistanceDP(s: String, t: String) -> Int {
let n = s.utf8CString.count
let m = t.utf8CString.count
var dp = Array(repeating: Array(repeating: 0, count: m + 1), count: n + 1)
// State transition: first row and first column
for i in 1 ... n {
dp[i][0] = i
}
for j in 1 ... m {
dp[0][j] = j
}
// State transition: rest of the rows and columns
for i in 1 ... n {
for j in 1 ... m {
if s.utf8CString[i - 1] == t.utf8CString[j - 1] {
// If two characters are equal, skip both characters
dp[i][j] = dp[i - 1][j - 1]
} else {
// Minimum edit steps = minimum edit steps of insert, delete, replace + 1
dp[i][j] = min(min(dp[i][j - 1], dp[i - 1][j]), dp[i - 1][j - 1]) + 1
}
}
}
return dp[n][m]
}
```
=== "JS"
```javascript title="edit_distance.js"
[class]{}-[func]{editDistanceDP}
/* Edit distance: Dynamic programming */
function editDistanceDP(s, t) {
const n = s.length,
m = t.length;
const dp = Array.from({ length: n + 1 }, () => new Array(m + 1).fill(0));
// State transition: first row and first column
for (let i = 1; i <= n; i++) {
dp[i][0] = i;
}
for (let j = 1; j <= m; j++) {
dp[0][j] = j;
}
// State transition: rest of the rows and columns
for (let i = 1; i <= n; i++) {
for (let j = 1; j <= m; j++) {
if (s.charAt(i - 1) === t.charAt(j - 1)) {
// If two characters are equal, skip both characters
dp[i][j] = dp[i - 1][j - 1];
} else {
// Minimum edit steps = minimum edit steps of insert, delete, replace + 1
dp[i][j] =
Math.min(dp[i][j - 1], dp[i - 1][j], dp[i - 1][j - 1]) + 1;
}
}
}
return dp[n][m];
}
```
=== "TS"
```typescript title="edit_distance.ts"
[class]{}-[func]{editDistanceDP}
/* Edit distance: Dynamic programming */
function editDistanceDP(s: string, t: string): number {
const n = s.length,
m = t.length;
const dp = Array.from({ length: n + 1 }, () =>
Array.from({ length: m + 1 }, () => 0)
);
// State transition: first row and first column
for (let i = 1; i <= n; i++) {
dp[i][0] = i;
}
for (let j = 1; j <= m; j++) {
dp[0][j] = j;
}
// State transition: rest of the rows and columns
for (let i = 1; i <= n; i++) {
for (let j = 1; j <= m; j++) {
if (s.charAt(i - 1) === t.charAt(j - 1)) {
// If two characters are equal, skip both characters
dp[i][j] = dp[i - 1][j - 1];
} else {
// Minimum edit steps = minimum edit steps of insert, delete, replace + 1
dp[i][j] =
Math.min(dp[i][j - 1], dp[i - 1][j], dp[i - 1][j - 1]) + 1;
}
}
}
return dp[n][m];
}
```
=== "Dart"
```dart title="edit_distance.dart"
[class]{}-[func]{editDistanceDP}
/* Edit distance: Dynamic programming */
int editDistanceDP(String s, String t) {
int n = s.length, m = t.length;
List<List<int>> dp = List.generate(n + 1, (_) => List.filled(m + 1, 0));
// State transition: first row and first column
for (int i = 1; i <= n; i++) {
dp[i][0] = i;
}
for (int j = 1; j <= m; j++) {
dp[0][j] = j;
}
// State transition: rest of the rows and columns
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++) {
if (s[i - 1] == t[j - 1]) {
// If two characters are equal, skip both characters
dp[i][j] = dp[i - 1][j - 1];
} else {
// Minimum edit steps = minimum edit steps of insert, delete, replace + 1
dp[i][j] = min(min(dp[i][j - 1], dp[i - 1][j]), dp[i - 1][j - 1]) + 1;
}
}
}
return dp[n][m];
}
```
=== "Rust"
```rust title="edit_distance.rs"
[class]{}-[func]{edit_distance_dp}
/* Edit distance: Dynamic programming */
fn edit_distance_dp(s: &str, t: &str) -> i32 {
let (n, m) = (s.len(), t.len());
let mut dp = vec![vec![0; m + 1]; n + 1];
// State transition: first row and first column
for i in 1..=n {
dp[i][0] = i as i32;
}
for j in 1..m {
dp[0][j] = j as i32;
}
// State transition: rest of the rows and columns
for i in 1..=n {
for j in 1..=m {
if s.chars().nth(i - 1) == t.chars().nth(j - 1) {
// If two characters are equal, skip both characters
dp[i][j] = dp[i - 1][j - 1];
} else {
// Minimum edit steps = minimum edit steps of insert, delete, replace + 1
dp[i][j] =
std::cmp::min(std::cmp::min(dp[i][j - 1], dp[i - 1][j]), dp[i - 1][j - 1]) + 1;
}
}
}
dp[n][m]
}
```
=== "C"
```c title="edit_distance.c"
[class]{}-[func]{editDistanceDP}
/* Edit distance: Dynamic programming */
int editDistanceDP(char *s, char *t, int n, int m) {
int **dp = malloc((n + 1) * sizeof(int *));
for (int i = 0; i <= n; i++) {
dp[i] = calloc(m + 1, sizeof(int));
}
// State transition: first row and first column
for (int i = 1; i <= n; i++) {
dp[i][0] = i;
}
for (int j = 1; j <= m; j++) {
dp[0][j] = j;
}
// State transition: rest of the rows and columns
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++) {
if (s[i - 1] == t[j - 1]) {
// If two characters are equal, skip both characters
dp[i][j] = dp[i - 1][j - 1];
} else {
// Minimum edit steps = minimum edit steps of insert, delete, replace + 1
dp[i][j] = myMin(myMin(dp[i][j - 1], dp[i - 1][j]), dp[i - 1][j - 1]) + 1;
}
}
}
int res = dp[n][m];
// Free memory
for (int i = 0; i <= n; i++) {
free(dp[i]);
}
return res;
}
```
=== "Kotlin"
```kotlin title="edit_distance.kt"
[class]{}-[func]{editDistanceDP}
/* Edit distance: Dynamic programming */
fun editDistanceDP(s: String, t: String): Int {
val n = s.length
val m = t.length
val dp = Array(n + 1) { IntArray(m + 1) }
// State transition: first row and first column
for (i in 1..n) {
dp[i][0] = i
}
for (j in 1..m) {
dp[0][j] = j
}
// State transition: rest of the rows and columns
for (i in 1..n) {
for (j in 1..m) {
if (s[i - 1] == t[j - 1]) {
// If two characters are equal, skip both characters
dp[i][j] = dp[i - 1][j - 1]
} else {
// Minimum edit steps = minimum edit steps of insert, delete, replace + 1
dp[i][j] = min(min(dp[i][j - 1], dp[i - 1][j]), dp[i - 1][j - 1]) + 1
}
}
}
return dp[n][m]
}
```
=== "Ruby"
```ruby title="edit_distance.rb"
[class]{}-[func]{edit_distance_dp}
### Edit distance: dynamic programming ###
def edit_distance_dp(s, t)
n, m = s.length, t.length
dp = Array.new(n + 1) { Array.new(m + 1, 0) }
# State transition: first row and first column
(1...(n + 1)).each { |i| dp[i][0] = i }
(1...(m + 1)).each { |j| dp[0][j] = j }
# State transition: rest of the rows and columns
for i in 1...(n + 1)
for j in 1...(m +1)
if s[i - 1] == t[j - 1]
# If two characters are equal, skip both characters
dp[i][j] = dp[i - 1][j - 1]
else
# Minimum edit steps = minimum edit steps of insert, delete, replace + 1
dp[i][j] = [dp[i][j - 1], dp[i - 1][j], dp[i - 1][j - 1]].min + 1
end
end
end
dp[n][m]
end
```
=== "Zig"
```zig title="edit_distance.zig"
[class]{}-[func]{editDistanceDP}
```
As shown in Figure 14-30, the process of state transition in the edit distance problem is very similar to that in the knapsack problem, which can be seen as filling a two-dimensional grid.
As shown in Figure 14-30, the state transition process for the edit distance problem is very similar to the knapsack problem and can both be viewed as the process of filling a two-dimensional grid.
=== "<1>"
![Dynamic programming process of edit distance](edit_distance_problem.assets/edit_distance_dp_step1.png){ class="animation-figure" }
![Dynamic programming process for edit distance](edit_distance_problem.assets/edit_distance_dp_step1.png){ class="animation-figure" }
=== "<2>"
![edit_distance_dp_step2](edit_distance_problem.assets/edit_distance_dp_step2.png){ class="animation-figure" }
@@ -274,13 +524,13 @@ As shown in Figure 14-30, the process of state transition in the edit distance p
=== "<15>"
![edit_distance_dp_step15](edit_distance_problem.assets/edit_distance_dp_step15.png){ class="animation-figure" }
<p align="center"> Figure 14-30 &nbsp; Dynamic programming process of edit distance </p>
<p align="center"> Figure 14-30 &nbsp; Dynamic programming process for edit distance </p>
### 3. &nbsp; Space optimization
### 3. &nbsp; Space Optimization
Since $dp[i, j]$ is derived from the solutions above $dp[i-1, j]$, to the left $dp[i, j-1]$, and to the upper left $dp[i-1, j-1]$, and direct traversal will lose the upper left solution $dp[i-1, j-1]$, and reverse traversal cannot build $dp[i, j-1]$ in advance, therefore, both traversal orders are not feasible.
Since $dp[i, j]$ is transferred from the solutions above $dp[i-1, j]$, to the left $dp[i, j-1]$, and to the upper-left $dp[i-1, j-1]$, forward traversal will lose the upper-left solution $dp[i-1, j-1]$, and reverse traversal cannot build $dp[i, j-1]$ in advance, so neither traversal order is feasible.
For this reason, we can use a variable `leftup` to temporarily store the solution from the upper left $dp[i-1, j-1]$, thus only needing to consider the solutions to the left and above. This situation is similar to the unbounded knapsack problem, allowing for direct traversal. The code is as follows:
For this reason, we can use a variable `leftup` to temporarily store the upper-left solution $dp[i-1, j-1]$, so we only need to consider the solutions to the left and above. This situation is the same as the unbounded knapsack problem, allowing for forward traversal. The code is as follows:
=== "Python"
@@ -292,21 +542,21 @@ For this reason, we can use a variable `leftup` to temporarily store the solutio
# State transition: first row
for j in range(1, m + 1):
dp[j] = j
# State transition: the rest of the rows
# State transition: rest of the rows
for i in range(1, n + 1):
# State transition: first column
leftup = dp[0] # Temporarily store dp[i-1, j-1]
dp[0] += 1
# State transition: the rest of the columns
# State transition: rest of the columns
for j in range(1, m + 1):
temp = dp[j]
if s[i - 1] == t[j - 1]:
# If the two characters are equal, skip these two characters
# If two characters are equal, skip both characters
dp[j] = leftup
else:
# The minimum number of edits = the minimum number of edits from three operations (insert, remove, replace) + 1
# Minimum edit steps = minimum edit steps of insert, delete, replace + 1
dp[j] = min(dp[j - 1], dp[j], leftup) + 1
leftup = temp # Update for the next round of dp[i-1, j-1]
leftup = temp # Update for next round's dp[i-1, j-1]
return dp[m]
```
@@ -321,22 +571,22 @@ For this reason, we can use a variable `leftup` to temporarily store the solutio
for (int j = 1; j <= m; j++) {
dp[j] = j;
}
// State transition: the rest of the rows
// State transition: rest of the rows
for (int i = 1; i <= n; i++) {
// State transition: first column
int leftup = dp[0]; // Temporarily store dp[i-1, j-1]
dp[0] = i;
// State transition: the rest of the columns
// State transition: rest of the columns
for (int j = 1; j <= m; j++) {
int temp = dp[j];
if (s[i - 1] == t[j - 1]) {
// If the two characters are equal, skip these two characters
// If two characters are equal, skip both characters
dp[j] = leftup;
} else {
// The minimum number of edits = the minimum number of edits from three operations (insert, remove, replace) + 1
// Minimum edit steps = minimum edit steps of insert, delete, replace + 1
dp[j] = min(min(dp[j - 1], dp[j]), leftup) + 1;
}
leftup = temp; // Update for the next round of dp[i-1, j-1]
leftup = temp; // Update for next round's dp[i-1, j-1]
}
}
return dp[m];
@@ -354,22 +604,22 @@ For this reason, we can use a variable `leftup` to temporarily store the solutio
for (int j = 1; j <= m; j++) {
dp[j] = j;
}
// State transition: the rest of the rows
// State transition: rest of the rows
for (int i = 1; i <= n; i++) {
// State transition: first column
int leftup = dp[0]; // Temporarily store dp[i-1, j-1]
dp[0] = i;
// State transition: the rest of the columns
// State transition: rest of the columns
for (int j = 1; j <= m; j++) {
int temp = dp[j];
if (s.charAt(i - 1) == t.charAt(j - 1)) {
// If the two characters are equal, skip these two characters
// If two characters are equal, skip both characters
dp[j] = leftup;
} else {
// The minimum number of edits = the minimum number of edits from three operations (insert, remove, replace) + 1
// Minimum edit steps = minimum edit steps of insert, delete, replace + 1
dp[j] = Math.min(Math.min(dp[j - 1], dp[j]), leftup) + 1;
}
leftup = temp; // Update for the next round of dp[i-1, j-1]
leftup = temp; // Update for next round's dp[i-1, j-1]
}
}
return dp[m];
@@ -379,65 +629,334 @@ For this reason, we can use a variable `leftup` to temporarily store the solutio
=== "C#"
```csharp title="edit_distance.cs"
[class]{edit_distance}-[func]{EditDistanceDPComp}
/* Edit distance: Space-optimized dynamic programming */
int EditDistanceDPComp(string s, string t) {
int n = s.Length, m = t.Length;
int[] dp = new int[m + 1];
// State transition: first row
for (int j = 1; j <= m; j++) {
dp[j] = j;
}
// State transition: rest of the rows
for (int i = 1; i <= n; i++) {
// State transition: first column
int leftup = dp[0]; // Temporarily store dp[i-1, j-1]
dp[0] = i;
// State transition: rest of the columns
for (int j = 1; j <= m; j++) {
int temp = dp[j];
if (s[i - 1] == t[j - 1]) {
// If two characters are equal, skip both characters
dp[j] = leftup;
} else {
// Minimum edit steps = minimum edit steps of insert, delete, replace + 1
dp[j] = Math.Min(Math.Min(dp[j - 1], dp[j]), leftup) + 1;
}
leftup = temp; // Update for next round's dp[i-1, j-1]
}
}
return dp[m];
}
```
=== "Go"
```go title="edit_distance.go"
[class]{}-[func]{editDistanceDPComp}
/* Edit distance: Space-optimized dynamic programming */
func editDistanceDPComp(s string, t string) int {
n := len(s)
m := len(t)
dp := make([]int, m+1)
// State transition: first row
for j := 1; j <= m; j++ {
dp[j] = j
}
// State transition: rest of the rows
for i := 1; i <= n; i++ {
// State transition: first column
leftUp := dp[0] // Temporarily store dp[i-1, j-1]
dp[0] = i
// State transition: rest of the columns
for j := 1; j <= m; j++ {
temp := dp[j]
if s[i-1] == t[j-1] {
// If two characters are equal, skip both characters
dp[j] = leftUp
} else {
// Minimum edit steps = minimum edit steps of insert, delete, replace + 1
dp[j] = MinInt(MinInt(dp[j-1], dp[j]), leftUp) + 1
}
leftUp = temp // Update for next round's dp[i-1, j-1]
}
}
return dp[m]
}
```
=== "Swift"
```swift title="edit_distance.swift"
[class]{}-[func]{editDistanceDPComp}
/* Edit distance: Space-optimized dynamic programming */
func editDistanceDPComp(s: String, t: String) -> Int {
let n = s.utf8CString.count
let m = t.utf8CString.count
var dp = Array(repeating: 0, count: m + 1)
// State transition: first row
for j in 1 ... m {
dp[j] = j
}
// State transition: rest of the rows
for i in 1 ... n {
// State transition: first column
var leftup = dp[0] // Temporarily store dp[i-1, j-1]
dp[0] = i
// State transition: rest of the columns
for j in 1 ... m {
let temp = dp[j]
if s.utf8CString[i - 1] == t.utf8CString[j - 1] {
// If two characters are equal, skip both characters
dp[j] = leftup
} else {
// Minimum edit steps = minimum edit steps of insert, delete, replace + 1
dp[j] = min(min(dp[j - 1], dp[j]), leftup) + 1
}
leftup = temp // Update for next round's dp[i-1, j-1]
}
}
return dp[m]
}
```
=== "JS"
```javascript title="edit_distance.js"
[class]{}-[func]{editDistanceDPComp}
/* Edit distance: Space-optimized dynamic programming */
function editDistanceDPComp(s, t) {
const n = s.length,
m = t.length;
const dp = new Array(m + 1).fill(0);
// State transition: first row
for (let j = 1; j <= m; j++) {
dp[j] = j;
}
// State transition: rest of the rows
for (let i = 1; i <= n; i++) {
// State transition: first column
let leftup = dp[0]; // Temporarily store dp[i-1, j-1]
dp[0] = i;
// State transition: rest of the columns
for (let j = 1; j <= m; j++) {
const temp = dp[j];
if (s.charAt(i - 1) === t.charAt(j - 1)) {
// If two characters are equal, skip both characters
dp[j] = leftup;
} else {
// Minimum edit steps = minimum edit steps of insert, delete, replace + 1
dp[j] = Math.min(dp[j - 1], dp[j], leftup) + 1;
}
leftup = temp; // Update for next round's dp[i-1, j-1]
}
}
return dp[m];
}
```
=== "TS"
```typescript title="edit_distance.ts"
[class]{}-[func]{editDistanceDPComp}
/* Edit distance: Space-optimized dynamic programming */
function editDistanceDPComp(s: string, t: string): number {
const n = s.length,
m = t.length;
const dp = new Array(m + 1).fill(0);
// State transition: first row
for (let j = 1; j <= m; j++) {
dp[j] = j;
}
// State transition: rest of the rows
for (let i = 1; i <= n; i++) {
// State transition: first column
let leftup = dp[0]; // Temporarily store dp[i-1, j-1]
dp[0] = i;
// State transition: rest of the columns
for (let j = 1; j <= m; j++) {
const temp = dp[j];
if (s.charAt(i - 1) === t.charAt(j - 1)) {
// If two characters are equal, skip both characters
dp[j] = leftup;
} else {
// Minimum edit steps = minimum edit steps of insert, delete, replace + 1
dp[j] = Math.min(dp[j - 1], dp[j], leftup) + 1;
}
leftup = temp; // Update for next round's dp[i-1, j-1]
}
}
return dp[m];
}
```
=== "Dart"
```dart title="edit_distance.dart"
[class]{}-[func]{editDistanceDPComp}
/* Edit distance: Space-optimized dynamic programming */
int editDistanceDPComp(String s, String t) {
int n = s.length, m = t.length;
List<int> dp = List.filled(m + 1, 0);
// State transition: first row
for (int j = 1; j <= m; j++) {
dp[j] = j;
}
// State transition: rest of the rows
for (int i = 1; i <= n; i++) {
// State transition: first column
int leftup = dp[0]; // Temporarily store dp[i-1, j-1]
dp[0] = i;
// State transition: rest of the columns
for (int j = 1; j <= m; j++) {
int temp = dp[j];
if (s[i - 1] == t[j - 1]) {
// If two characters are equal, skip both characters
dp[j] = leftup;
} else {
// Minimum edit steps = minimum edit steps of insert, delete, replace + 1
dp[j] = min(min(dp[j - 1], dp[j]), leftup) + 1;
}
leftup = temp; // Update for next round's dp[i-1, j-1]
}
}
return dp[m];
}
```
=== "Rust"
```rust title="edit_distance.rs"
[class]{}-[func]{edit_distance_dp_comp}
/* Edit distance: Space-optimized dynamic programming */
fn edit_distance_dp_comp(s: &str, t: &str) -> i32 {
let (n, m) = (s.len(), t.len());
let mut dp = vec![0; m + 1];
// State transition: first row
for j in 1..m {
dp[j] = j as i32;
}
// State transition: rest of the rows
for i in 1..=n {
// State transition: first column
let mut leftup = dp[0]; // Temporarily store dp[i-1, j-1]
dp[0] = i as i32;
// State transition: rest of the columns
for j in 1..=m {
let temp = dp[j];
if s.chars().nth(i - 1) == t.chars().nth(j - 1) {
// If two characters are equal, skip both characters
dp[j] = leftup;
} else {
// Minimum edit steps = minimum edit steps of insert, delete, replace + 1
dp[j] = std::cmp::min(std::cmp::min(dp[j - 1], dp[j]), leftup) + 1;
}
leftup = temp; // Update for next round's dp[i-1, j-1]
}
}
dp[m]
}
```
=== "C"
```c title="edit_distance.c"
[class]{}-[func]{editDistanceDPComp}
/* Edit distance: Space-optimized dynamic programming */
int editDistanceDPComp(char *s, char *t, int n, int m) {
int *dp = calloc(m + 1, sizeof(int));
// State transition: first row
for (int j = 1; j <= m; j++) {
dp[j] = j;
}
// State transition: rest of the rows
for (int i = 1; i <= n; i++) {
// State transition: first column
int leftup = dp[0]; // Temporarily store dp[i-1, j-1]
dp[0] = i;
// State transition: rest of the columns
for (int j = 1; j <= m; j++) {
int temp = dp[j];
if (s[i - 1] == t[j - 1]) {
// If two characters are equal, skip both characters
dp[j] = leftup;
} else {
// Minimum edit steps = minimum edit steps of insert, delete, replace + 1
dp[j] = myMin(myMin(dp[j - 1], dp[j]), leftup) + 1;
}
leftup = temp; // Update for next round's dp[i-1, j-1]
}
}
int res = dp[m];
// Free memory
free(dp);
return res;
}
```
=== "Kotlin"
```kotlin title="edit_distance.kt"
[class]{}-[func]{editDistanceDPComp}
/* Edit distance: Space-optimized dynamic programming */
fun editDistanceDPComp(s: String, t: String): Int {
val n = s.length
val m = t.length
val dp = IntArray(m + 1)
// State transition: first row
for (j in 1..m) {
dp[j] = j
}
// State transition: rest of the rows
for (i in 1..n) {
// State transition: first column
var leftup = dp[0] // Temporarily store dp[i-1, j-1]
dp[0] = i
// State transition: rest of the columns
for (j in 1..m) {
val temp = dp[j]
if (s[i - 1] == t[j - 1]) {
// If two characters are equal, skip both characters
dp[j] = leftup
} else {
// Minimum edit steps = minimum edit steps of insert, delete, replace + 1
dp[j] = min(min(dp[j - 1], dp[j]), leftup) + 1
}
leftup = temp // Update for next round's dp[i-1, j-1]
}
}
return dp[m]
}
```
=== "Ruby"
```ruby title="edit_distance.rb"
[class]{}-[func]{edit_distance_dp_comp}
```
=== "Zig"
```zig title="edit_distance.zig"
[class]{}-[func]{editDistanceDPComp}
### Edit distance: space-optimized DP ###
def edit_distance_dp_comp(s, t)
n, m = s.length, t.length
dp = Array.new(m + 1, 0)
# State transition: first row
(1...(m + 1)).each { |j| dp[j] = j }
# State transition: rest of the rows
for i in 1...(n + 1)
# State transition: first column
leftup = dp.first # Temporarily store dp[i-1, j-1]
dp[0] += 1
# State transition: rest of the columns
for j in 1...(m + 1)
temp = dp[j]
if s[i - 1] == t[j - 1]
# If two characters are equal, skip both characters
dp[j] = leftup
else
# Minimum edit steps = minimum edit steps of insert, delete, replace + 1
dp[j] = [dp[j - 1], dp[j], leftup].min + 1
end
leftup = temp # Update for next round's dp[i-1, j-1]
end
end
dp[m]
end
```
en/docs/chapter_dynamic_programming/index.md
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icon: material/table-pivot
---
# Chapter 14. &nbsp; Dynamic programming
# Chapter 14. &nbsp; Dynamic Programming
![Dynamic programming](../assets/covers/chapter_dynamic_programming.jpg){ class="cover-image" }
!!! abstract
Streams merge into rivers, and rivers merge into the sea.
Dynamic programming weaves smaller problems’ solutions into larger ones, guiding us step by step toward the far shore—where the ultimate answer awaits.
Streams converge into rivers, rivers converge into the sea.
Dynamic programming gathers solutions to small problems into answers to large problems, step by step guiding us to the shore of problem-solving.
## Chapter contents
- [14.1 &nbsp; Introduction to dynamic programming](intro_to_dynamic_programming.md)
- [14.2 &nbsp; Characteristics of DP problems](dp_problem_features.md)
- [14.3 &nbsp; DP problem-solving approach](dp_solution_pipeline.md)
- [14.4 &nbsp; 0-1 Knapsack problem](knapsack_problem.md)
- [14.5 &nbsp; Unbounded knapsack problem](unbounded_knapsack_problem.md)
- [14.6 &nbsp; Edit distance problem](edit_distance_problem.md)
- [14.1 &nbsp; Introduction to Dynamic Programming](intro_to_dynamic_programming.md)
- [14.2 &nbsp; Characteristics of Dynamic Programming Problems](dp_problem_features.md)
- [14.3 &nbsp; Dynamic Programming Problem-Solving Approach](dp_solution_pipeline.md)
- [14.4 &nbsp; 0-1 Knapsack Problem](knapsack_problem.md)
- [14.5 &nbsp; Unbounded Knapsack Problem](unbounded_knapsack_problem.md)
- [14.6 &nbsp; Edit Distance Problem](edit_distance_problem.md)
- [14.7 &nbsp; Summary](summary.md)
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en/docs/chapter_dynamic_programming/knapsack_problem.md
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