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Sort the coding languages by applications. (#721)

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Yudong Jin
2023-09-04 03:19:08 +08:00
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docs/chapter_backtracking/backtracking_algorithm.md
+296 -296
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@@ -10,10 +10,10 @@
对于此题，我们前序遍历这颗树，并判断当前节点的值是否为 $7$ ，若是则将该节点的值加入到结果列表 `res` 之中。相关过程实现如下图和以下代码所示。
=== "Java"
=== "Python"
```java title="preorder_traversal_i_compact.java"
[class]{preorder_traversal_i_compact}-[func]{preOrder}
```python title="preorder_traversal_i_compact.py"
[class]{}-[func]{pre_order}
```
=== "C++"
@@ -22,10 +22,16 @@
[class]{}-[func]{preOrder}
```
=== "Python"
=== "Java"
```python title="preorder_traversal_i_compact.py"
[class]{}-[func]{pre_order}
```java title="preorder_traversal_i_compact.java"
[class]{preorder_traversal_i_compact}-[func]{preOrder}
```
=== "C#"
```csharp title="preorder_traversal_i_compact.cs"
[class]{preorder_traversal_i_compact}-[func]{preOrder}
```
=== "Go"
@@ -34,6 +40,12 @@
[class]{}-[func]{preOrderI}
```
=== "Swift"
```swift title="preorder_traversal_i_compact.swift"
[class]{}-[func]{preOrder}
```
=== "JS"
```javascript title="preorder_traversal_i_compact.js"
@@ -46,30 +58,6 @@
[class]{}-[func]{preOrder}
```
=== "C"
```c title="preorder_traversal_i_compact.c"
[class]{}-[func]{preOrder}
```
=== "C#"
```csharp title="preorder_traversal_i_compact.cs"
[class]{preorder_traversal_i_compact}-[func]{preOrder}
```
=== "Swift"
```swift title="preorder_traversal_i_compact.swift"
[class]{}-[func]{preOrder}
```
=== "Zig"
```zig title="preorder_traversal_i_compact.zig"
[class]{}-[func]{preOrder}
```
=== "Dart"
```dart title="preorder_traversal_i_compact.dart"
@@ -82,6 +70,18 @@
[class]{}-[func]{pre_order}
```
=== "C"
```c title="preorder_traversal_i_compact.c"
[class]{}-[func]{preOrder}
```
=== "Zig"
```zig title="preorder_traversal_i_compact.zig"
[class]{}-[func]{preOrder}
```
![在前序遍历中搜索节点](backtracking_algorithm.assets/preorder_find_nodes.png)
## 尝试与回退
@@ -98,10 +98,10 @@
在例题一代码的基础上，我们需要借助一个列表 `path` 记录访问过的节点路径。当访问到值为 $7$ 的节点时，则复制 `path` 并添加进结果列表 `res` 。遍历完成后，`res` 中保存的就是所有的解。
=== "Java"
=== "Python"
```java title="preorder_traversal_ii_compact.java"
[class]{preorder_traversal_ii_compact}-[func]{preOrder}
```python title="preorder_traversal_ii_compact.py"
[class]{}-[func]{pre_order}
```
=== "C++"
@@ -110,10 +110,16 @@
[class]{}-[func]{preOrder}
```
=== "Python"
=== "Java"
```python title="preorder_traversal_ii_compact.py"
[class]{}-[func]{pre_order}
```java title="preorder_traversal_ii_compact.java"
[class]{preorder_traversal_ii_compact}-[func]{preOrder}
```
=== "C#"
```csharp title="preorder_traversal_ii_compact.cs"
[class]{preorder_traversal_ii_compact}-[func]{preOrder}
```
=== "Go"
@@ -122,6 +128,12 @@
[class]{}-[func]{preOrderII}
```
=== "Swift"
```swift title="preorder_traversal_ii_compact.swift"
[class]{}-[func]{preOrder}
```
=== "JS"
```javascript title="preorder_traversal_ii_compact.js"
@@ -134,30 +146,6 @@
[class]{}-[func]{preOrder}
```
=== "C"
```c title="preorder_traversal_ii_compact.c"
[class]{}-[func]{preOrder}
```
=== "C#"
```csharp title="preorder_traversal_ii_compact.cs"
[class]{preorder_traversal_ii_compact}-[func]{preOrder}
```
=== "Swift"
```swift title="preorder_traversal_ii_compact.swift"
[class]{}-[func]{preOrder}
```
=== "Zig"
```zig title="preorder_traversal_ii_compact.zig"
[class]{}-[func]{preOrder}
```
=== "Dart"
```dart title="preorder_traversal_ii_compact.dart"
@@ -170,6 +158,18 @@
[class]{}-[func]{pre_order}
```
=== "C"
```c title="preorder_traversal_ii_compact.c"
[class]{}-[func]{preOrder}
```
=== "Zig"
```zig title="preorder_traversal_ii_compact.zig"
[class]{}-[func]{preOrder}
```
在每次“尝试”中，我们通过将当前节点添加进 `path` 来记录路径；而在“回退”前，我们需要将该节点从 `path` 中弹出，**以恢复本次尝试之前的状态**。
观察下图所示的过程，**我们可以将尝试和回退理解为“前进”与“撤销”**，两个操作是互为逆向的。
@@ -217,10 +217,10 @@
为了满足以上约束条件，**我们需要添加剪枝操作**：在搜索过程中，若遇到值为 $3$ 的节点，则提前返回，停止继续搜索。
=== "Java"
=== "Python"
```java title="preorder_traversal_iii_compact.java"
[class]{preorder_traversal_iii_compact}-[func]{preOrder}
```python title="preorder_traversal_iii_compact.py"
[class]{}-[func]{pre_order}
```
=== "C++"
@@ -229,10 +229,16 @@
[class]{}-[func]{preOrder}
```
=== "Python"
=== "Java"
```python title="preorder_traversal_iii_compact.py"
[class]{}-[func]{pre_order}
```java title="preorder_traversal_iii_compact.java"
[class]{preorder_traversal_iii_compact}-[func]{preOrder}
```
=== "C#"
```csharp title="preorder_traversal_iii_compact.cs"
[class]{preorder_traversal_iii_compact}-[func]{preOrder}
```
=== "Go"
@@ -241,6 +247,12 @@
[class]{}-[func]{preOrderIII}
```
=== "Swift"
```swift title="preorder_traversal_iii_compact.swift"
[class]{}-[func]{preOrder}
```
=== "JS"
```javascript title="preorder_traversal_iii_compact.js"
@@ -253,30 +265,6 @@
[class]{}-[func]{preOrder}
```
=== "C"
```c title="preorder_traversal_iii_compact.c"
[class]{}-[func]{preOrder}
```
=== "C#"
```csharp title="preorder_traversal_iii_compact.cs"
[class]{preorder_traversal_iii_compact}-[func]{preOrder}
```
=== "Swift"
```swift title="preorder_traversal_iii_compact.swift"
[class]{}-[func]{preOrder}
```
=== "Zig"
```zig title="preorder_traversal_iii_compact.zig"
[class]{}-[func]{preOrder}
```
=== "Dart"
```dart title="preorder_traversal_iii_compact.dart"
@@ -289,6 +277,18 @@
[class]{}-[func]{pre_order}
```
=== "C"
```c title="preorder_traversal_iii_compact.c"
[class]{}-[func]{preOrder}
```
=== "Zig"
```zig title="preorder_traversal_iii_compact.zig"
[class]{}-[func]{preOrder}
```
剪枝是一个非常形象的名词。如下图所示，在搜索过程中，**我们“剪掉”了不满足约束条件的搜索分支**，避免许多无意义的尝试，从而提高了搜索效率。
![根据约束条件剪枝](backtracking_algorithm.assets/preorder_find_constrained_paths.png)
@@ -299,30 +299,26 @@
在以下框架代码中，`state` 表示问题的当前状态，`choices` 表示当前状态下可以做出的选择。
=== "Java"
=== "Python"
```java title=""
/* 回溯算法框架 */
void backtrack(State state, List<Choice> choices, List<State> res) {
// 判断是否为解
if (isSolution(state)) {
// 记录解
recordSolution(state, res);
// 停止继续搜索
return;
}
// 遍历所有选择
for (Choice choice : choices) {
// 剪枝：判断选择是否合法
if (isValid(state, choice)) {
// 尝试：做出选择，更新状态
makeChoice(state, choice);
backtrack(state, choices, res);
// 回退：撤销选择，恢复到之前的状态
undoChoice(state, choice);
}
}
}
```python title=""
def backtrack(state: State, choices: list[choice], res: list[state]):
"""回溯算法框架"""
# 判断是否为解
if is_solution(state):
# 记录解
record_solution(state, res)
# 停止继续搜索
return
# 遍历所有选择
for choice in choices:
# 剪枝：判断选择是否合法
if is_valid(state, choice):
# 尝试：做出选择，更新状态
make_choice(state, choice)
backtrack(state, choices, res)
# 回退：撤销选择，恢复到之前的状态
undo_choice(state, choice)
```
=== "C++"
@@ -351,26 +347,56 @@
}
```
=== "Python"
=== "Java"
```python title=""
def backtrack(state: State, choices: list[choice], res: list[state]):
"""回溯算法框架"""
# 判断是否为解
if is_solution(state):
# 记录解
record_solution(state, res)
# 停止继续搜索
return
# 遍历所有选择
for choice in choices:
# 剪枝：判断选择是否合法
if is_valid(state, choice):
# 尝试：做出选择，更新状态
make_choice(state, choice)
backtrack(state, choices, res)
# 回退：撤销选择，恢复到之前的状态
undo_choice(state, choice)
```java title=""
/* 回溯算法框架 */
void backtrack(State state, List<Choice> choices, List<State> res) {
// 判断是否为解
if (isSolution(state)) {
// 记录解
recordSolution(state, res);
// 停止继续搜索
return;
}
// 遍历所有选择
for (Choice choice : choices) {
// 剪枝：判断选择是否合法
if (isValid(state, choice)) {
// 尝试：做出选择，更新状态
makeChoice(state, choice);
backtrack(state, choices, res);
// 回退：撤销选择，恢复到之前的状态
undoChoice(state, choice);
}
}
}
```
=== "C#"
```csharp title=""
/* 回溯算法框架 */
void backtrack(State state, List<Choice> choices, List<State> res) {
// 判断是否为解
if (isSolution(state)) {
// 记录解
recordSolution(state, res);
// 停止继续搜索
return;
}
// 遍历所有选择
foreach (Choice choice in choices) {
// 剪枝：判断选择是否合法
if (isValid(state, choice)) {
// 尝试：做出选择，更新状态
makeChoice(state, choice);
backtrack(state, choices, res);
// 回退：撤销选择，恢复到之前的状态
undoChoice(state, choice);
}
}
}
```
=== "Go"
@@ -399,6 +425,32 @@
}
```
=== "Swift"
```swift title=""
/* 回溯算法框架 */
func backtrack(state: inout State, choices: [Choice], res: inout [State]) {
// 判断是否为解
if isSolution(state: state) {
// 记录解
recordSolution(state: state, res: &res)
// 停止继续搜索
return
}
// 遍历所有选择
for choice in choices {
// 剪枝：判断选择是否合法
if isValid(state: state, choice: choice) {
// 尝试：做出选择，更新状态
makeChoice(state: &state, choice: choice)
backtrack(state: &state, choices: choices, res: &res)
// 回退：撤销选择，恢复到之前的状态
undoChoice(state: &state, choice: choice)
}
}
}
```
=== "JS"
```javascript title=""
@@ -451,90 +503,6 @@
}
```
=== "C"
```c title=""
/* 回溯算法框架 */
void backtrack(State *state, Choice *choices, int numChoices, State *res, int numRes) {
// 判断是否为解
if (isSolution(state)) {
// 记录解
recordSolution(state, res, numRes);
// 停止继续搜索
return;
}
// 遍历所有选择
for (int i = 0; i < numChoices; i++) {
// 剪枝：判断选择是否合法
if (isValid(state, &choices[i])) {
// 尝试：做出选择，更新状态
makeChoice(state, &choices[i]);
backtrack(state, choices, numChoices, res, numRes);
// 回退：撤销选择，恢复到之前的状态
undoChoice(state, &choices[i]);
}
}
}
```
=== "C#"
```csharp title=""
/* 回溯算法框架 */
void backtrack(State state, List<Choice> choices, List<State> res) {
// 判断是否为解
if (isSolution(state)) {
// 记录解
recordSolution(state, res);
// 停止继续搜索
return;
}
// 遍历所有选择
foreach (Choice choice in choices) {
// 剪枝：判断选择是否合法
if (isValid(state, choice)) {
// 尝试：做出选择，更新状态
makeChoice(state, choice);
backtrack(state, choices, res);
// 回退：撤销选择，恢复到之前的状态
undoChoice(state, choice);
}
}
}
```
=== "Swift"
```swift title=""
/* 回溯算法框架 */
func backtrack(state: inout State, choices: [Choice], res: inout [State]) {
// 判断是否为解
if isSolution(state: state) {
// 记录解
recordSolution(state: state, res: &res)
// 停止继续搜索
return
}
// 遍历所有选择
for choice in choices {
// 剪枝：判断选择是否合法
if isValid(state: state, choice: choice) {
// 尝试：做出选择，更新状态
makeChoice(state: &state, choice: choice)
backtrack(state: &state, choices: choices, res: &res)
// 回退：撤销选择，恢复到之前的状态
undoChoice(state: &state, choice: choice)
}
}
}
```
=== "Zig"
```zig title=""
```
=== "Dart"
```dart title=""
@@ -567,22 +535,54 @@
```
=== "C"
```c title=""
/* 回溯算法框架 */
void backtrack(State *state, Choice *choices, int numChoices, State *res, int numRes) {
// 判断是否为解
if (isSolution(state)) {
// 记录解
recordSolution(state, res, numRes);
// 停止继续搜索
return;
}
// 遍历所有选择
for (int i = 0; i < numChoices; i++) {
// 剪枝：判断选择是否合法
if (isValid(state, &choices[i])) {
// 尝试：做出选择，更新状态
makeChoice(state, &choices[i]);
backtrack(state, choices, numChoices, res, numRes);
// 回退：撤销选择，恢复到之前的状态
undoChoice(state, &choices[i]);
}
}
}
```
=== "Zig"
```zig title=""
```
接下来，我们基于框架代码来解决例题三。状态 `state` 为节点遍历路径，选择 `choices` 为当前节点的左子节点和右子节点，结果 `res` 是路径列表。
=== "Java"
=== "Python"
```java title="preorder_traversal_iii_template.java"
[class]{preorder_traversal_iii_template}-[func]{isSolution}
```python title="preorder_traversal_iii_template.py"
[class]{}-[func]{is_solution}
[class]{preorder_traversal_iii_template}-[func]{recordSolution}
[class]{}-[func]{record_solution}
[class]{preorder_traversal_iii_template}-[func]{isValid}
[class]{}-[func]{is_valid}
[class]{preorder_traversal_iii_template}-[func]{makeChoice}
[class]{}-[func]{make_choice}
[class]{preorder_traversal_iii_template}-[func]{undoChoice}
[class]{}-[func]{undo_choice}
[class]{preorder_traversal_iii_template}-[func]{backtrack}
[class]{}-[func]{backtrack}
```
=== "C++"
@@ -601,20 +601,36 @@
[class]{}-[func]{backtrack}
```
=== "Python"
=== "Java"
```python title="preorder_traversal_iii_template.py"
[class]{}-[func]{is_solution}
```java title="preorder_traversal_iii_template.java"
[class]{preorder_traversal_iii_template}-[func]{isSolution}
[class]{}-[func]{record_solution}
[class]{preorder_traversal_iii_template}-[func]{recordSolution}
[class]{}-[func]{is_valid}
[class]{preorder_traversal_iii_template}-[func]{isValid}
[class]{}-[func]{make_choice}
[class]{preorder_traversal_iii_template}-[func]{makeChoice}
[class]{}-[func]{undo_choice}
[class]{preorder_traversal_iii_template}-[func]{undoChoice}
[class]{}-[func]{backtrack}
[class]{preorder_traversal_iii_template}-[func]{backtrack}
```
=== "C#"
```csharp title="preorder_traversal_iii_template.cs"
[class]{preorder_traversal_iii_template}-[func]{isSolution}
[class]{preorder_traversal_iii_template}-[func]{recordSolution}
[class]{preorder_traversal_iii_template}-[func]{isValid}
[class]{preorder_traversal_iii_template}-[func]{makeChoice}
[class]{preorder_traversal_iii_template}-[func]{undoChoice}
[class]{preorder_traversal_iii_template}-[func]{backtrack}
```
=== "Go"
@@ -633,6 +649,22 @@
[class]{}-[func]{backtrackIII}
```
=== "Swift"
```swift title="preorder_traversal_iii_template.swift"
[class]{}-[func]{isSolution}
[class]{}-[func]{recordSolution}
[class]{}-[func]{isValid}
[class]{}-[func]{makeChoice}
[class]{}-[func]{undoChoice}
[class]{}-[func]{backtrack}
```
=== "JS"
```javascript title="preorder_traversal_iii_template.js"
@@ -665,70 +697,6 @@
[class]{}-[func]{backtrack}
```
=== "C"
```c title="preorder_traversal_iii_template.c"
[class]{}-[func]{isSolution}
[class]{}-[func]{recordSolution}
[class]{}-[func]{isValid}
[class]{}-[func]{makeChoice}
[class]{}-[func]{undoChoice}
[class]{}-[func]{backtrack}
```
=== "C#"
```csharp title="preorder_traversal_iii_template.cs"
[class]{preorder_traversal_iii_template}-[func]{isSolution}
[class]{preorder_traversal_iii_template}-[func]{recordSolution}
[class]{preorder_traversal_iii_template}-[func]{isValid}
[class]{preorder_traversal_iii_template}-[func]{makeChoice}
[class]{preorder_traversal_iii_template}-[func]{undoChoice}
[class]{preorder_traversal_iii_template}-[func]{backtrack}
```
=== "Swift"
```swift title="preorder_traversal_iii_template.swift"
[class]{}-[func]{isSolution}
[class]{}-[func]{recordSolution}
[class]{}-[func]{isValid}
[class]{}-[func]{makeChoice}
[class]{}-[func]{undoChoice}
[class]{}-[func]{backtrack}
```
=== "Zig"
```zig title="preorder_traversal_iii_template.zig"
[class]{}-[func]{isSolution}
[class]{}-[func]{recordSolution}
[class]{}-[func]{isValid}
[class]{}-[func]{makeChoice}
[class]{}-[func]{undoChoice}
[class]{}-[func]{backtrack}
```
=== "Dart"
```dart title="preorder_traversal_iii_template.dart"
@@ -761,6 +729,38 @@
[class]{}-[func]{backtrack}
```
=== "C"
```c title="preorder_traversal_iii_template.c"
[class]{}-[func]{isSolution}
[class]{}-[func]{recordSolution}
[class]{}-[func]{isValid}
[class]{}-[func]{makeChoice}
[class]{}-[func]{undoChoice}
[class]{}-[func]{backtrack}
```
=== "Zig"
```zig title="preorder_traversal_iii_template.zig"
[class]{}-[func]{isSolution}
[class]{}-[func]{recordSolution}
[class]{}-[func]{isValid}
[class]{}-[func]{makeChoice}
[class]{}-[func]{undoChoice}
[class]{}-[func]{backtrack}
```
根据题意，我们在找到值为 7 的节点后应该继续搜索，**因此需要将记录解之后的 `return` 语句删除**。下图对比了保留或删除 `return` 语句的搜索过程。
![保留与删除 return 的搜索过程对比](backtracking_algorithm.assets/backtrack_remove_return_or_not.png)
docs/chapter_backtracking/n_queens_problem.md
+40 -40
View File
@@ -40,12 +40,12 @@
请注意，$n$ 维方阵中 $row - col$ 的范围是 $[-n + 1, n - 1]$ ，$row + col$ 的范围是 $[0, 2n - 2]$ ，所以主对角线和次对角线的数量都为 $2n - 1$ ，即数组 `diag1` 和 `diag2` 的长度都为 $2n - 1$ 。
=== "Java"
=== "Python"
```java title="n_queens.java"
[class]{n_queens}-[func]{backtrack}
```python title="n_queens.py"
[class]{}-[func]{backtrack}
[class]{n_queens}-[func]{nQueens}
[class]{}-[func]{n_queens}
```
=== "C++"
@@ -56,12 +56,20 @@
[class]{}-[func]{nQueens}
```
=== "Python"
=== "Java"
```python title="n_queens.py"
[class]{}-[func]{backtrack}
```java title="n_queens.java"
[class]{n_queens}-[func]{backtrack}
[class]{}-[func]{n_queens}
[class]{n_queens}-[func]{nQueens}
```
=== "C#"
```csharp title="n_queens.cs"
[class]{n_queens}-[func]{backtrack}
[class]{n_queens}-[func]{nQueens}
```
=== "Go"
@@ -72,6 +80,14 @@
[class]{}-[func]{nQueens}
```
=== "Swift"
```swift title="n_queens.swift"
[class]{}-[func]{backtrack}
[class]{}-[func]{nQueens}
```
=== "JS"
```javascript title="n_queens.js"
@@ -88,38 +104,6 @@
[class]{}-[func]{nQueens}
```
=== "C"
```c title="n_queens.c"
[class]{}-[func]{backtrack}
[class]{}-[func]{nQueens}
```
=== "C#"
```csharp title="n_queens.cs"
[class]{n_queens}-[func]{backtrack}
[class]{n_queens}-[func]{nQueens}
```
=== "Swift"
```swift title="n_queens.swift"
[class]{}-[func]{backtrack}
[class]{}-[func]{nQueens}
```
=== "Zig"
```zig title="n_queens.zig"
[class]{}-[func]{backtrack}
[class]{}-[func]{nQueens}
```
=== "Dart"
```dart title="n_queens.dart"
@@ -136,6 +120,22 @@
[class]{}-[func]{n_queens}
```
=== "C"
```c title="n_queens.c"
[class]{}-[func]{backtrack}
[class]{}-[func]{nQueens}
```
=== "Zig"
```zig title="n_queens.zig"
[class]{}-[func]{backtrack}
[class]{}-[func]{nQueens}
```
逐行放置 $n$ 次，考虑列约束，则从第一行到最后一行分别有 $n$、$n-1$、$\dots$、$2$、$1$ 个选择，**因此时间复杂度为 $O(n!)$** 。实际上，根据对角线约束的剪枝也能够大幅地缩小搜索空间，因而搜索效率往往优于以上时间复杂度。
数组 `state` 使用 $O(n^2)$ 空间，数组 `cols`、`diags1` 和 `diags2` 皆使用 $O(n)$ 空间。最大递归深度为 $n$ ，使用 $O(n)$ 栈帧空间。因此，**空间复杂度为 $O(n^2)$** 。
docs/chapter_backtracking/permutations_problem.md
+80 -80
View File
@@ -43,12 +43,12 @@
想清楚以上信息之后，我们就可以在框架代码中做“完形填空”了。为了缩短代码行数，我们不单独实现框架代码中的各个函数，而是将他们展开在 `backtrack()` 函数中。
=== "Java"
=== "Python"
```java title="permutations_i.java"
[class]{permutations_i}-[func]{backtrack}
```python title="permutations_i.py"
[class]{}-[func]{backtrack}
[class]{permutations_i}-[func]{permutationsI}
[class]{}-[func]{permutations_i}
```
=== "C++"
@@ -59,12 +59,20 @@
[class]{}-[func]{permutationsI}
```
=== "Python"
=== "Java"
```python title="permutations_i.py"
[class]{}-[func]{backtrack}
```java title="permutations_i.java"
[class]{permutations_i}-[func]{backtrack}
[class]{}-[func]{permutations_i}
[class]{permutations_i}-[func]{permutationsI}
```
=== "C#"
```csharp title="permutations_i.cs"
[class]{permutations_i}-[func]{backtrack}
[class]{permutations_i}-[func]{permutationsI}
```
=== "Go"
@@ -75,6 +83,14 @@
[class]{}-[func]{permutationsI}
```
=== "Swift"
```swift title="permutations_i.swift"
[class]{}-[func]{backtrack}
[class]{}-[func]{permutationsI}
```
=== "JS"
```javascript title="permutations_i.js"
@@ -91,38 +107,6 @@
[class]{}-[func]{permutationsI}
```
=== "C"
```c title="permutations_i.c"
[class]{}-[func]{backtrack}
[class]{}-[func]{permutationsI}
```
=== "C#"
```csharp title="permutations_i.cs"
[class]{permutations_i}-[func]{backtrack}
[class]{permutations_i}-[func]{permutationsI}
```
=== "Swift"
```swift title="permutations_i.swift"
[class]{}-[func]{backtrack}
[class]{}-[func]{permutationsI}
```
=== "Zig"
```zig title="permutations_i.zig"
[class]{}-[func]{backtrack}
[class]{}-[func]{permutationsI}
```
=== "Dart"
```dart title="permutations_i.dart"
@@ -139,6 +123,22 @@
[class]{}-[func]{permutations_i}
```
=== "C"
```c title="permutations_i.c"
[class]{}-[func]{backtrack}
[class]{}-[func]{permutationsI}
```
=== "Zig"
```zig title="permutations_i.zig"
[class]{}-[func]{backtrack}
[class]{}-[func]{permutationsI}
```
## 考虑相等元素的情况
!!! question
@@ -167,12 +167,12 @@
在上一题的代码的基础上，我们考虑在每一轮选择中开启一个哈希表 `duplicated` ，用于记录该轮中已经尝试过的元素，并将重复元素剪枝。
=== "Java"
=== "Python"
```java title="permutations_ii.java"
[class]{permutations_ii}-[func]{backtrack}
```python title="permutations_ii.py"
[class]{}-[func]{backtrack}
[class]{permutations_ii}-[func]{permutationsII}
[class]{}-[func]{permutations_ii}
```
=== "C++"
@@ -183,12 +183,20 @@
[class]{}-[func]{permutationsII}
```
=== "Python"
=== "Java"
```python title="permutations_ii.py"
[class]{}-[func]{backtrack}
```java title="permutations_ii.java"
[class]{permutations_ii}-[func]{backtrack}
[class]{}-[func]{permutations_ii}
[class]{permutations_ii}-[func]{permutationsII}
```
=== "C#"
```csharp title="permutations_ii.cs"
[class]{permutations_ii}-[func]{backtrack}
[class]{permutations_ii}-[func]{permutationsII}
```
=== "Go"
@@ -199,6 +207,14 @@
[class]{}-[func]{permutationsII}
```
=== "Swift"
```swift title="permutations_ii.swift"
[class]{}-[func]{backtrack}
[class]{}-[func]{permutationsII}
```
=== "JS"
```javascript title="permutations_ii.js"
@@ -215,38 +231,6 @@
[class]{}-[func]{permutationsII}
```
=== "C"
```c title="permutations_ii.c"
[class]{}-[func]{backtrack}
[class]{}-[func]{permutationsII}
```
=== "C#"
```csharp title="permutations_ii.cs"
[class]{permutations_ii}-[func]{backtrack}
[class]{permutations_ii}-[func]{permutationsII}
```
=== "Swift"
```swift title="permutations_ii.swift"
[class]{}-[func]{backtrack}
[class]{}-[func]{permutationsII}
```
=== "Zig"
```zig title="permutations_ii.zig"
[class]{}-[func]{backtrack}
[class]{}-[func]{permutationsII}
```
=== "Dart"
```dart title="permutations_ii.dart"
@@ -263,6 +247,22 @@
[class]{}-[func]{permutations_ii}
```
=== "C"
```c title="permutations_ii.c"
[class]{}-[func]{backtrack}
[class]{}-[func]{permutationsII}
```
=== "Zig"
```zig title="permutations_ii.zig"
[class]{}-[func]{backtrack}
[class]{}-[func]{permutationsII}
```
假设元素两两之间互不相同，则 $n$ 个元素共有 $n!$ 种排列（阶乘）；在记录结果时，需要复制长度为 $n$ 的列表，使用 $O(n)$ 时间。**因此时间复杂度为 $O(n!n)$** 。
最大递归深度为 $n$ ，使用 $O(n)$ 栈帧空间。`selected` 使用 $O(n)$ 空间。同一时刻最多共有 $n$ 个 `duplicated` ，使用 $O(n^2)$ 空间。**因此空间复杂度为 $O(n^2)$** 。
docs/chapter_backtracking/subset_sum_problem.md
+120 -120
View File
@@ -17,12 +17,12 @@
而与全排列问题不同的是，**本题集合中的元素可以被无限次选取**，因此无须借助 `selected` 布尔列表来记录元素是否已被选择。我们可以对全排列代码进行小幅修改，初步得到解题代码。
=== "Java"
=== "Python"
```java title="subset_sum_i_naive.java"
[class]{subset_sum_i_naive}-[func]{backtrack}
```python title="subset_sum_i_naive.py"
[class]{}-[func]{backtrack}
[class]{subset_sum_i_naive}-[func]{subsetSumINaive}
[class]{}-[func]{subset_sum_i_naive}
```
=== "C++"
@@ -33,12 +33,20 @@
[class]{}-[func]{subsetSumINaive}
```
=== "Python"
=== "Java"
```python title="subset_sum_i_naive.py"
[class]{}-[func]{backtrack}
```java title="subset_sum_i_naive.java"
[class]{subset_sum_i_naive}-[func]{backtrack}
[class]{}-[func]{subset_sum_i_naive}
[class]{subset_sum_i_naive}-[func]{subsetSumINaive}
```
=== "C#"
```csharp title="subset_sum_i_naive.cs"
[class]{subset_sum_i_naive}-[func]{backtrack}
[class]{subset_sum_i_naive}-[func]{subsetSumINaive}
```
=== "Go"
@@ -49,6 +57,14 @@
[class]{}-[func]{subsetSumINaive}
```
=== "Swift"
```swift title="subset_sum_i_naive.swift"
[class]{}-[func]{backtrack}
[class]{}-[func]{subsetSumINaive}
```
=== "JS"
```javascript title="subset_sum_i_naive.js"
@@ -65,38 +81,6 @@
[class]{}-[func]{subsetSumINaive}
```
=== "C"
```c title="subset_sum_i_naive.c"
[class]{}-[func]{backtrack}
[class]{}-[func]{subsetSumINaive}
```
=== "C#"
```csharp title="subset_sum_i_naive.cs"
[class]{subset_sum_i_naive}-[func]{backtrack}
[class]{subset_sum_i_naive}-[func]{subsetSumINaive}
```
=== "Swift"
```swift title="subset_sum_i_naive.swift"
[class]{}-[func]{backtrack}
[class]{}-[func]{subsetSumINaive}
```
=== "Zig"
```zig title="subset_sum_i_naive.zig"
[class]{}-[func]{backtrack}
[class]{}-[func]{subsetSumINaive}
```
=== "Dart"
```dart title="subset_sum_i_naive.dart"
@@ -113,6 +97,22 @@
[class]{}-[func]{subset_sum_i_naive}
```
=== "C"
```c title="subset_sum_i_naive.c"
[class]{}-[func]{backtrack}
[class]{}-[func]{subsetSumINaive}
```
=== "Zig"
```zig title="subset_sum_i_naive.zig"
[class]{}-[func]{backtrack}
[class]{}-[func]{subsetSumINaive}
```
向以上代码输入数组 $[3, 4, 5]$ 和目标元素 $9$ ，输出结果为 $[3, 3, 3], [4, 5], [5, 4]$ 。**虽然成功找出了所有和为 $9$ 的子集，但其中存在重复的子集 $[4, 5]$ 和 $[5, 4]$** 。
这是因为搜索过程是区分选择顺序的，然而子集不区分选择顺序。如下图所示，先选 $4$ 后选 $5$ 与先选 $5$ 后选 $4$ 是两个不同的分支，但两者对应同一个子集。
@@ -150,12 +150,12 @@
- 在开启搜索前，先将数组 `nums` 排序。在遍历所有选择时，**当子集和超过 `target` 时直接结束循环**，因为后边的元素更大，其子集和都一定会超过 `target` 。
- 省去元素和变量 `total`，**通过在 `target` 上执行减法来统计元素和**，当 `target` 等于 $0$ 时记录解。
=== "Java"
=== "Python"
```java title="subset_sum_i.java"
[class]{subset_sum_i}-[func]{backtrack}
```python title="subset_sum_i.py"
[class]{}-[func]{backtrack}
[class]{subset_sum_i}-[func]{subsetSumI}
[class]{}-[func]{subset_sum_i}
```
=== "C++"
@@ -166,12 +166,20 @@
[class]{}-[func]{subsetSumI}
```
=== "Python"
=== "Java"
```python title="subset_sum_i.py"
[class]{}-[func]{backtrack}
```java title="subset_sum_i.java"
[class]{subset_sum_i}-[func]{backtrack}
[class]{}-[func]{subset_sum_i}
[class]{subset_sum_i}-[func]{subsetSumI}
```
=== "C#"
```csharp title="subset_sum_i.cs"
[class]{subset_sum_i}-[func]{backtrack}
[class]{subset_sum_i}-[func]{subsetSumI}
```
=== "Go"
@@ -182,6 +190,14 @@
[class]{}-[func]{subsetSumI}
```
=== "Swift"
```swift title="subset_sum_i.swift"
[class]{}-[func]{backtrack}
[class]{}-[func]{subsetSumI}
```
=== "JS"
```javascript title="subset_sum_i.js"
@@ -198,38 +214,6 @@
[class]{}-[func]{subsetSumI}
```
=== "C"
```c title="subset_sum_i.c"
[class]{}-[func]{backtrack}
[class]{}-[func]{subsetSumI}
```
=== "C#"
```csharp title="subset_sum_i.cs"
[class]{subset_sum_i}-[func]{backtrack}
[class]{subset_sum_i}-[func]{subsetSumI}
```
=== "Swift"
```swift title="subset_sum_i.swift"
[class]{}-[func]{backtrack}
[class]{}-[func]{subsetSumI}
```
=== "Zig"
```zig title="subset_sum_i.zig"
[class]{}-[func]{backtrack}
[class]{}-[func]{subsetSumI}
```
=== "Dart"
```dart title="subset_sum_i.dart"
@@ -246,6 +230,22 @@
[class]{}-[func]{subset_sum_i}
```
=== "C"
```c title="subset_sum_i.c"
[class]{}-[func]{backtrack}
[class]{}-[func]{subsetSumI}
```
=== "Zig"
```zig title="subset_sum_i.zig"
[class]{}-[func]{backtrack}
[class]{}-[func]{subsetSumI}
```
如下图所示，为将数组 $[3, 4, 5]$ 和目标元素 $9$ 输入到以上代码后的整体回溯过程。
![子集和 I 回溯过程](subset_sum_problem.assets/subset_sum_i.png)
@@ -270,12 +270,12 @@
### 代码实现
=== "Java"
=== "Python"
```java title="subset_sum_ii.java"
[class]{subset_sum_ii}-[func]{backtrack}
```python title="subset_sum_ii.py"
[class]{}-[func]{backtrack}
[class]{subset_sum_ii}-[func]{subsetSumII}
[class]{}-[func]{subset_sum_ii}
```
=== "C++"
@@ -286,12 +286,20 @@
[class]{}-[func]{subsetSumII}
```
=== "Python"
=== "Java"
```python title="subset_sum_ii.py"
[class]{}-[func]{backtrack}
```java title="subset_sum_ii.java"
[class]{subset_sum_ii}-[func]{backtrack}
[class]{}-[func]{subset_sum_ii}
[class]{subset_sum_ii}-[func]{subsetSumII}
```
=== "C#"
```csharp title="subset_sum_ii.cs"
[class]{subset_sum_ii}-[func]{backtrack}
[class]{subset_sum_ii}-[func]{subsetSumII}
```
=== "Go"
@@ -302,6 +310,14 @@
[class]{}-[func]{subsetSumII}
```
=== "Swift"
```swift title="subset_sum_ii.swift"
[class]{}-[func]{backtrack}
[class]{}-[func]{subsetSumII}
```
=== "JS"
```javascript title="subset_sum_ii.js"
@@ -318,38 +334,6 @@
[class]{}-[func]{subsetSumII}
```
=== "C"
```c title="subset_sum_ii.c"
[class]{}-[func]{backtrack}
[class]{}-[func]{subsetSumII}
```
=== "C#"
```csharp title="subset_sum_ii.cs"
[class]{subset_sum_ii}-[func]{backtrack}
[class]{subset_sum_ii}-[func]{subsetSumII}
```
=== "Swift"
```swift title="subset_sum_ii.swift"
[class]{}-[func]{backtrack}
[class]{}-[func]{subsetSumII}
```
=== "Zig"
```zig title="subset_sum_ii.zig"
[class]{}-[func]{backtrack}
[class]{}-[func]{subsetSumII}
```
=== "Dart"
```dart title="subset_sum_ii.dart"
@@ -366,6 +350,22 @@
[class]{}-[func]{subset_sum_ii}
```
=== "C"
```c title="subset_sum_ii.c"
[class]{}-[func]{backtrack}
[class]{}-[func]{subsetSumII}
```
=== "Zig"
```zig title="subset_sum_ii.zig"
[class]{}-[func]{backtrack}
[class]{}-[func]{subsetSumII}
```
下图展示了数组 $[4, 4, 5]$ 和目标元素 $9$ 的回溯过程，共包含四种剪枝操作。请你将图示与代码注释相结合，理解整个搜索过程，以及每种剪枝操作是如何工作的。
![子集和 II 回溯过程](subset_sum_problem.assets/subset_sum_ii.png)