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krahets
2023-07-19 01:44:39 +08:00
parent ec6f3fd337
commit 1184c791c5
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chapter_dynamic_programming/dp_problem_features.md
+48 -3
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@@ -143,7 +143,23 @@ $$
=== "Swift"
```swift title="min_cost_climbing_stairs_dp.swift"
[class]{}-[func]{minCostClimbingStairsDP}
/* 爬楼梯最小代价：动态规划 */
func minCostClimbingStairsDP(cost: [Int]) -> Int {
let n = cost.count - 1
if n == 1 || n == 2 {
return cost[n]
}
// 初始化 dp 表，用于存储子问题的解
var dp = Array(repeating: 0, count: n + 1)
// 初始状态：预设最小子问题的解
dp[1] = 1
dp[2] = 2
// 状态转移：从较小子问题逐步求解较大子问题
for i in stride(from: 3, through: n, by: 1) {
dp[i] = min(dp[i - 1], dp[i - 2]) + cost[i]
}
return dp[n]
}
```
=== "Zig"
@@ -275,7 +291,18 @@ $$
=== "Swift"
```swift title="min_cost_climbing_stairs_dp.swift"
[class]{}-[func]{minCostClimbingStairsDPComp}
/* 爬楼梯最小代价：状态压缩后的动态规划 */
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 stride(from: 3, through: n, by: 1) {
(a, b) = (b, min(a, b) + cost[i])
}
return b
}
```
=== "Zig"
@@ -465,7 +492,25 @@ $$
=== "Swift"
```swift title="climbing_stairs_constraint_dp.swift"
[class]{}-[func]{climbingStairsConstraintDP}
/* 带约束爬楼梯：动态规划 */
func climbingStairsConstraintDP(n: Int) -> Int {
if n == 1 || n == 2 {
return n
}
// 初始化 dp 表，用于存储子问题的解
var dp = Array(repeating: Array(repeating: 0, count: 3), count: n + 1)
// 初始状态：预设最小子问题的解
dp[1][1] = 1
dp[1][2] = 0
dp[2][1] = 0
dp[2][2] = 1
// 状态转移：从较小子问题逐步求解较大子问题
for i in stride(from: 3, through: n, by: 1) {
dp[i][1] = dp[i - 1][2]
dp[i][2] = dp[i - 2][1] + dp[i - 2][2]
}
return dp[n][1] + dp[n][2]
}
```
=== "Zig"
chapter_dynamic_programming/dp_solution_pipeline.md
+82 -4
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@@ -216,7 +216,22 @@ $$
=== "Swift"
```swift title="min_path_sum.swift"
[class]{}-[func]{minPathSumDFS}
/* 最小路径和：暴力搜索 */
func minPathSumDFS(grid: [[Int]], i: Int, j: Int) -> Int {
// 若为左上角单元格，则终止搜索
if i == 0, j == 0 {
return grid[0][0]
}
// 若行列索引越界，则返回 +∞ 代价
if i < 0 || j < 0 {
return .max
}
// 计算从左上角到 (i-1, j) 和 (i, j-1) 的最小路径代价
let left = minPathSumDFS(grid: grid, i: i - 1, j: j)
let up = minPathSumDFS(grid: grid, i: i, j: j - 1)
// 返回从左上角到 (i, j) 的最小路径代价
return min(left, up) + grid[i][j]
}
```
=== "Zig"
@@ -389,7 +404,27 @@ $$
=== "Swift"
```swift title="min_path_sum.swift"
[class]{}-[func]{minPathSumDFSMem}
/* 最小路径和：记忆化搜索 */
func minPathSumDFSMem(grid: [[Int]], mem: inout [[Int]], i: Int, j: Int) -> Int {
// 若为左上角单元格，则终止搜索
if i == 0, j == 0 {
return grid[0][0]
}
// 若行列索引越界，则返回 +∞ 代价
if i < 0 || j < 0 {
return .max
}
// 若已有记录，则直接返回
if mem[i][j] != -1 {
return mem[i][j]
}
// 左边和上边单元格的最小路径代价
let left = minPathSumDFSMem(grid: grid, mem: &mem, i: i - 1, j: j)
let up = minPathSumDFSMem(grid: grid, mem: &mem, i: i, j: j - 1)
// 记录并返回左上角到 (i, j) 的最小路径代价
mem[i][j] = min(left, up) + grid[i][j]
return mem[i][j]
}
```
=== "Zig"
@@ -565,7 +600,29 @@ $$
=== "Swift"
```swift title="min_path_sum.swift"
[class]{}-[func]{minPathSumDP}
/* 最小路径和：动态规划 */
func minPathSumDP(grid: [[Int]]) -> Int {
let n = grid.count
let m = grid[0].count
// 初始化 dp 表
var dp = Array(repeating: Array(repeating: 0, count: m), count: n)
dp[0][0] = grid[0][0]
// 状态转移：首行
for j in stride(from: 1, to: m, by: 1) {
dp[0][j] = dp[0][j - 1] + grid[0][j]
}
// 状态转移：首列
for i in stride(from: 1, to: n, by: 1) {
dp[i][0] = dp[i - 1][0] + grid[i][0]
}
// 状态转移：其余行列
for i in stride(from: 1, to: n, by: 1) {
for j in stride(from: 1, to: m, by: 1) {
dp[i][j] = min(dp[i][j - 1], dp[i - 1][j]) + grid[i][j]
}
}
return dp[n - 1][m - 1]
}
```
=== "Zig"
@@ -771,7 +828,28 @@ $$
=== "Swift"
```swift title="min_path_sum.swift"
[class]{}-[func]{minPathSumDPComp}
/* 最小路径和：状态压缩后的动态规划 */
func minPathSumDPComp(grid: [[Int]]) -> Int {
let n = grid.count
let m = grid[0].count
// 初始化 dp 表
var dp = Array(repeating: 0, count: m)
// 状态转移：首行
dp[0] = grid[0][0]
for j in stride(from: 1, to: m, by: 1) {
dp[j] = dp[j - 1] + grid[0][j]
}
// 状态转移：其余行
for i in stride(from: 1, to: n, by: 1) {
// 状态转移：首列
dp[0] = dp[0] + grid[i][0]
// 状态转移：其余列
for j in stride(from: 1, to: m, by: 1) {
dp[j] = min(dp[j - 1], dp[j]) + grid[i][j]
}
}
return dp[m - 1]
}
```
=== "Zig"
chapter_dynamic_programming/edit_distance_problem.md
+55 -2
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@@ -213,7 +213,32 @@ $$
=== "Swift"
```swift title="edit_distance.swift"
[class]{}-[func]{editDistanceDP}
/* 编辑距离：动态规划 */
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)
// 状态转移：首行首列
for i in stride(from: 1, through: n, by: 1) {
dp[i][0] = i
}
for j in stride(from: 1, through: m, by: 1) {
dp[0][j] = j
}
// 状态转移：其余行列
for i in stride(from: 1, through: n, by: 1) {
for j in stride(from: 1, through: m, by: 1) {
if s.utf8CString[i - 1] == t.utf8CString[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]
}
```
=== "Zig"
@@ -458,7 +483,35 @@ $$
=== "Swift"
```swift title="edit_distance.swift"
[class]{}-[func]{editDistanceDPComp}
/* 编辑距离：状态压缩后的动态规划 */
func editDistanceDPComp(s: String, t: String) -> Int {
let n = s.utf8CString.count
let m = t.utf8CString.count
var dp = Array(repeating: 0, count: m + 1)
// 状态转移：首行
for j in stride(from: 1, through: m, by: 1) {
dp[j] = j
}
// 状态转移：其余行
for i in stride(from: 1, through: n, by: 1) {
// 状态转移：首列
var leftup = dp[0] // 暂存 dp[i-1, j-1]
dp[0] = i
// 状态转移：其余列
for j in stride(from: 1, through: m, by: 1) {
let temp = dp[j]
if s.utf8CString[i - 1] == t.utf8CString[j - 1] {
// 若两字符相等，则直接跳过此两字符
dp[j] = leftup
} else {
// 最少编辑步数 = 插入、删除、替换这三种操作的最少编辑步数 + 1
dp[j] = min(min(dp[j - 1], dp[j]), leftup) + 1
}
leftup = temp // 更新为下一轮的 dp[i-1, j-1]
}
}
return dp[m]
}
```
=== "Zig"
chapter_dynamic_programming/intro_to_dynamic_programming.md
+88 -8
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@@ -170,9 +170,31 @@ status: new
=== "Swift"
```swift title="climbing_stairs_backtrack.swift"
[class]{}-[func]{backtrack}
/* 回溯 */
func backtrack(choices: [Int], state: Int, n: Int, res: inout [Int]) {
// 当爬到第 n 阶时，方案数量加 1
if state == n {
res[0] += 1
}
// 遍历所有选择
for choice in choices {
// 剪枝：不允许越过第 n 阶
if state + choice > n {
break
}
backtrack(choices: choices, state: state + choice, n: n, res: &res)
}
}
[class]{}-[func]{climbingStairsBacktrack}
/* 爬楼梯：回溯 */
func climbingStairsBacktrack(n: Int) -> Int {
let choices = [1, 2] // 可选择向上爬 1 或 2 阶
let state = 0 // 从第 0 阶开始爬
var res: [Int] = []
res.append(0) // 使用 res[0] 记录方案数量
backtrack(choices: choices, state: state, n: n, res: &res)
return res[0]
}
```
=== "Zig"
@@ -353,9 +375,21 @@ $$
=== "Swift"
```swift title="climbing_stairs_dfs.swift"
[class]{}-[func]{dfs}
/* 搜索 */
func dfs(i: Int) -> Int {
// 已知 dp[1] 和 dp[2] ，返回之
if i == 1 || i == 2 {
return i
}
// dp[i] = dp[i-1] + dp[i-2]
let count = dfs(i: i - 1) + dfs(i: i - 2)
return count
}
[class]{}-[func]{climbingStairsDFS}
/* 爬楼梯：搜索 */
func climbingStairsDFS(n: Int) -> Int {
dfs(i: n)
}
```
=== "Zig"
@@ -540,9 +574,29 @@ $$
=== "Swift"
```swift title="climbing_stairs_dfs_mem.swift"
[class]{}-[func]{dfs}
/* 记忆化搜索 */
func dfs(i: Int, mem: inout [Int]) -> Int {
// 已知 dp[1] 和 dp[2] ，返回之
if i == 1 || i == 2 {
return i
}
// 若存在记录 dp[i] ，则直接返回之
if mem[i] != -1 {
return mem[i]
}
// dp[i] = dp[i-1] + dp[i-2]
let count = dfs(i: i - 1, mem: &mem) + dfs(i: i - 2, mem: &mem)
// 记录 dp[i]
mem[i] = count
return count
}
[class]{}-[func]{climbingStairsDFSMem}
/* 爬楼梯：记忆化搜索 */
func climbingStairsDFSMem(n: Int) -> Int {
// mem[i] 记录爬到第 i 阶的方案总数，-1 代表无记录
var mem = Array(repeating: -1, count: n + 1)
return dfs(i: n, mem: &mem)
}
```
=== "Zig"
@@ -699,7 +753,22 @@ $$
=== "Swift"
```swift title="climbing_stairs_dp.swift"
[class]{}-[func]{climbingStairsDP}
/* 爬楼梯：动态规划 */
func climbingStairsDP(n: Int) -> Int {
if n == 1 || n == 2 {
return n
}
// 初始化 dp 表，用于存储子问题的解
var dp = Array(repeating: 0, count: 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 i in stride(from: 3, through: n, by: 1) {
dp[i] = dp[i - 1] + dp[i - 2]
}
return dp[n]
}
```
=== "Zig"
@@ -833,7 +902,18 @@ $$
=== "Swift"
```swift title="climbing_stairs_dp.swift"
[class]{}-[func]{climbingStairsDPComp}
/* 爬楼梯：状态压缩后的动态规划 */
func climbingStairsDPComp(n: Int) -> Int {
if n == 1 || n == 2 {
return n
}
var a = 1
var b = 2
for _ in stride(from: 3, through: n, by: 1) {
(a, b) = (b, a + b)
}
return b
}
```
=== "Zig"
chapter_dynamic_programming/knapsack_problem.md
+73 -4
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@@ -172,7 +172,22 @@ $$
=== "Swift"
```swift title="knapsack.swift"
[class]{}-[func]{knapsackDFS}
/* 0-1 背包：暴力搜索 */
func knapsackDFS(wgt: [Int], val: [Int], i: Int, c: Int) -> Int {
// 若已选完所有物品或背包无容量，则返回价值 0
if i == 0 || c == 0 {
return 0
}
// 若超过背包容量，则只能不放入背包
if wgt[i - 1] > c {
return knapsackDFS(wgt: wgt, val: val, i: i - 1, c: c)
}
// 计算不放入和放入物品 i 的最大价值
let no = knapsackDFS(wgt: wgt, val: val, i: i - 1, c: c)
let yes = knapsackDFS(wgt: wgt, val: val, i: i - 1, c: c - 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)
}
```
=== "Zig"
@@ -343,7 +358,27 @@ $$
=== "Swift"
```swift title="knapsack.swift"
[class]{}-[func]{knapsackDFSMem}
/* 0-1 背包：记忆化搜索 */
func knapsackDFSMem(wgt: [Int], val: [Int], mem: inout [[Int]], i: Int, c: Int) -> Int {
// 若已选完所有物品或背包无容量，则返回价值 0
if i == 0 || c == 0 {
return 0
}
// 若已有记录，则直接返回
if mem[i][c] != -1 {
return mem[i][c]
}
// 若超过背包容量，则只能不放入背包
if wgt[i - 1] > c {
return knapsackDFSMem(wgt: wgt, val: val, mem: &mem, i: i - 1, c: c)
}
// 计算不放入和放入物品 i 的最大价值
let no = knapsackDFSMem(wgt: wgt, val: val, mem: &mem, i: i - 1, c: c)
let yes = knapsackDFSMem(wgt: wgt, val: val, mem: &mem, i: i - 1, c: c - wgt[i - 1]) + val[i - 1]
// 记录并返回两种方案中价值更大的那一个
mem[i][c] = max(no, yes)
return mem[i][c]
}
```
=== "Zig"
@@ -507,7 +542,25 @@ $$
=== "Swift"
```swift title="knapsack.swift"
[class]{}-[func]{knapsackDP}
/* 0-1 背包：动态规划 */
func knapsackDP(wgt: [Int], val: [Int], cap: Int) -> Int {
let n = wgt.count
// 初始化 dp 表
var dp = Array(repeating: Array(repeating: 0, count: cap + 1), count: n + 1)
// 状态转移
for i in stride(from: 1, through: n, by: 1) {
for c in stride(from: 1, through: cap, by: 1) {
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 - wgt[i - 1]] + val[i - 1])
}
}
}
return dp[n][cap]
}
```
=== "Zig"
@@ -727,7 +780,23 @@ $$
=== "Swift"
```swift title="knapsack.swift"
[class]{}-[func]{knapsackDPComp}
/* 0-1 背包：状态压缩后的动态规划 */
func knapsackDPComp(wgt: [Int], val: [Int], cap: Int) -> Int {
let n = wgt.count
// 初始化 dp 表
var dp = Array(repeating: 0, count: cap + 1)
// 状态转移
for i in stride(from: 1, through: n, by: 1) {
// 倒序遍历
for c in stride(from: cap, through: 1, by: -1) {
if wgt[i - 1] <= c {
// 不选和选物品 i 这两种方案的较大值
dp[c] = max(dp[c], dp[c - wgt[i - 1]] + val[i - 1])
}
}
}
return dp[cap]
}
```
=== "Zig"
chapter_dynamic_programming/unbounded_knapsack_problem.md
+127 -7
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@@ -154,7 +154,25 @@ $$
=== "Swift"
```swift title="unbounded_knapsack.swift"
[class]{}-[func]{unboundedKnapsackDP}
/* 完全背包：动态规划 */
func unboundedKnapsackDP(wgt: [Int], val: [Int], cap: Int) -> Int {
let n = wgt.count
// 初始化 dp 表
var dp = Array(repeating: Array(repeating: 0, count: cap + 1), count: n + 1)
// 状态转移
for i in stride(from: 1, through: n, by: 1) {
for c in stride(from: 1, through: cap, by: 1) {
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 - wgt[i - 1]] + val[i - 1])
}
}
}
return dp[n][cap]
}
```
=== "Zig"
@@ -329,7 +347,25 @@ $$
=== "Swift"
```swift title="unbounded_knapsack.swift"
[class]{}-[func]{unboundedKnapsackDPComp}
/* 完全背包：状态压缩后的动态规划 */
func unboundedKnapsackDPComp(wgt: [Int], val: [Int], cap: Int) -> Int {
let n = wgt.count
// 初始化 dp 表
var dp = Array(repeating: 0, count: cap + 1)
// 状态转移
for i in stride(from: 1, through: n, by: 1) {
for c in stride(from: 1, through: cap, by: 1) {
if wgt[i - 1] > c {
// 若超过背包容量，则不选物品 i
dp[c] = dp[c]
} else {
// 不选和选物品 i 这两种方案的较大值
dp[c] = max(dp[c], dp[c - wgt[i - 1]] + val[i - 1])
}
}
}
return dp[cap]
}
```
=== "Zig"
@@ -368,7 +404,7 @@ $$
!!! question
给定 $n$ 种硬币，第 $i$ 个硬币的面值为 $coins[i - 1]$ ，为目标金额 $amt$ ，**每种硬币可以重复选取**，问能够凑出目标金额的最少硬币个数。如果无法凑出目标金额则返回 $-1$ 。
给定 $n$ 种硬币，第 $i$ 个硬币的面值为 $coins[i - 1]$ ，目标金额为 $amt$ ，**每种硬币可以重复选取**，问能够凑出目标金额的最少硬币个数。如果无法凑出目标金额则返回 $-1$ 。
如下图所示，凑出 $11$ 元最少需要 $3$ 枚硬币，方案为 $1 + 2 + 5 = 11$ 。
@@ -547,7 +583,30 @@ $$
=== "Swift"
```swift title="coin_change.swift"
[class]{}-[func]{coinChangeDP}
/* 零钱兑换：动态规划 */
func coinChangeDP(coins: [Int], amt: Int) -> Int {
let n = coins.count
let MAX = amt + 1
// 初始化 dp 表
var dp = Array(repeating: Array(repeating: 0, count: amt + 1), count: n + 1)
// 状态转移：首行首列
for a in stride(from: 1, through: amt, by: 1) {
dp[0][a] = MAX
}
// 状态转移：其余行列
for i in stride(from: 1, through: n, by: 1) {
for a in stride(from: 1, through: amt, by: 1) {
if coins[i - 1] > a {
// 若超过背包容量，则不选硬币 i
dp[i][a] = dp[i - 1][a]
} else {
// 不选和选硬币 i 这两种方案的较小值
dp[i][a] = min(dp[i - 1][a], dp[i][a - coins[i - 1]] + 1)
}
}
}
return dp[n][amt] != MAX ? dp[n][amt] : -1
}
```
=== "Zig"
@@ -768,7 +827,27 @@ $$
=== "Swift"
```swift title="coin_change.swift"
[class]{}-[func]{coinChangeDPComp}
/* 零钱兑换：状态压缩后的动态规划 */
func coinChangeDPComp(coins: [Int], amt: Int) -> Int {
let n = coins.count
let MAX = amt + 1
// 初始化 dp 表
var dp = Array(repeating: MAX, count: amt + 1)
dp[0] = 0
// 状态转移
for i in stride(from: 1, through: n, by: 1) {
for a in stride(from: 1, through: amt, by: 1) {
if coins[i - 1] > a {
// 若超过背包容量，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 - coins[i - 1]] + 1)
}
}
}
return dp[amt] != MAX ? dp[amt] : -1
}
```
=== "Zig"
@@ -962,7 +1041,29 @@ $$
=== "Swift"
```swift title="coin_change_ii.swift"
[class]{}-[func]{coinChangeIIDP}
/* 零钱兑换 II：动态规划 */
func coinChangeIIDP(coins: [Int], amt: Int) -> Int {
let n = coins.count
// 初始化 dp 表
var dp = Array(repeating: Array(repeating: 0, count: amt + 1), count: n + 1)
// 初始化首列
for i in stride(from: 0, through: n, by: 1) {
dp[i][0] = 1
}
// 状态转移
for i in stride(from: 1, through: n, by: 1) {
for a in stride(from: 1, through: amt, by: 1) {
if coins[i - 1] > a {
// 若超过背包容量，则不选硬币 i
dp[i][a] = dp[i - 1][a]
} else {
// 不选和选硬币 i 这两种方案之和
dp[i][a] = dp[i - 1][a] + dp[i][a - coins[i - 1]]
}
}
}
return dp[n][amt]
}
```
=== "Zig"
@@ -1125,7 +1226,26 @@ $$
=== "Swift"
```swift title="coin_change_ii.swift"
[class]{}-[func]{coinChangeIIDPComp}
/* 零钱兑换 II：状态压缩后的动态规划 */
func coinChangeIIDPComp(coins: [Int], amt: Int) -> Int {
let n = coins.count
// 初始化 dp 表
var dp = Array(repeating: 0, count: amt + 1)
dp[0] = 1
// 状态转移
for i in stride(from: 1, through: n, by: 1) {
for a in stride(from: 1, through: amt, by: 1) {
if coins[i - 1] > a {
// 若超过背包容量，则不选硬币 i
dp[a] = dp[a]
} else {
// 不选和选硬币 i 这两种方案之和
dp[a] = dp[a] + dp[a - coins[i - 1]]
}
}
}
return dp[amt]
}
```
=== "Zig"