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en/docs/chapter_dynamic_programming/dp_problem_features.md
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@@ -44,15 +44,15 @@ According to the state transition equation, and the initial states $dp[1] = cost
```python title="min_cost_climbing_stairs_dp.py"
def min_cost_climbing_stairs_dp(cost: list[int]) -> int:
"""爬楼梯最小代价：动态规划"""
"""Climbing stairs with minimum cost: Dynamic programming"""
n = len(cost) - 1
if n == 1 or n == 2:
return cost[n]
# 初始化 dp 表，用于存储子问题的解
# Initialize dp table, used to store subproblem solutions
dp = [0] * (n + 1)
# 初始状态：预设最小子问题的解
# Initial state: preset the smallest subproblem solution
dp[1], dp[2] = cost[1], cost[2]
# 状态转移：从较小子问题逐步求解较大子问题
# State transition: gradually solve larger subproblems from smaller ones
for i in range(3, n + 1):
dp[i] = min(dp[i - 1], dp[i - 2]) + cost[i]
return dp[n]
@@ -61,38 +61,23 @@ According to the state transition equation, and the initial states $dp[1] = cost
=== "C++"
```cpp title="min_cost_climbing_stairs_dp.cpp"
/* 爬楼梯最小代价：动态规划 */
int minCostClimbingStairsDP(vector<int> &cost) {
int n = cost.size() - 1;
if (n == 1 || n == 2)
return cost[n];
// 初始化 dp 表，用于存储子问题的解
vector<int> dp(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 (int i = 3; i <= n; i++) {
dp[i] = min(dp[i - 1], dp[i - 2]) + cost[i];
}
return dp[n];
}
[class]{}-[func]{minCostClimbingStairsDP}
```
=== "Java"
```java title="min_cost_climbing_stairs_dp.java"
/* 爬楼梯最小代价：动态规划 */
/* Climbing stairs with minimum cost: Dynamic programming */
int minCostClimbingStairsDP(int[] cost) {
int n = cost.length - 1;
if (n == 1 || n == 2)
return cost[n];
// 初始化 dp 表，用于存储子问题的解
// Initialize dp table, used to store subproblem solutions
int[] dp = new int[n + 1];
// 初始状态：预设最小子问题的解
// Initial state: preset the smallest subproblem solution
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
// 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];
}
@@ -103,202 +88,55 @@ According to the state transition equation, and the initial states $dp[1] = cost
=== "C#"
```csharp title="min_cost_climbing_stairs_dp.cs"
/* 爬楼梯最小代价：动态规划 */
int MinCostClimbingStairsDP(int[] cost) {
int n = cost.Length - 1;
if (n == 1 || n == 2)
return cost[n];
// 初始化 dp 表，用于存储子问题的解
int[] dp = new int[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 (int i = 3; i <= n; i++) {
dp[i] = Math.Min(dp[i - 1], dp[i - 2]) + cost[i];
}
return dp[n];
}
[class]{min_cost_climbing_stairs_dp}-[func]{MinCostClimbingStairsDP}
```
=== "Go"
```go title="min_cost_climbing_stairs_dp.go"
/* 爬楼梯最小代价：动态规划 */
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
}
// 初始化 dp 表，用于存储子问题的解
dp := make([]int, 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 i := 3; i <= n; i++ {
dp[i] = min(dp[i-1], dp[i-2]) + cost[i]
}
return dp[n]
}
[class]{}-[func]{minCostClimbingStairsDP}
```
=== "Swift"
```swift title="min_cost_climbing_stairs_dp.swift"
/* 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] = 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 i in 3 ... n {
dp[i] = min(dp[i - 1], dp[i - 2]) + cost[i]
}
return dp[n]
}
[class]{}-[func]{minCostClimbingStairsDP}
```
=== "JS"
```javascript title="min_cost_climbing_stairs_dp.js"
/* 爬楼梯最小代价：动态规划 */
function minCostClimbingStairsDP(cost) {
const n = cost.length - 1;
if (n === 1 || n === 2) {
return cost[n];
}
// 初始化 dp 表，用于存储子问题的解
const dp = new Array(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 (let i = 3; i <= n; i++) {
dp[i] = Math.min(dp[i - 1], dp[i - 2]) + cost[i];
}
return dp[n];
}
[class]{}-[func]{minCostClimbingStairsDP}
```
=== "TS"
```typescript title="min_cost_climbing_stairs_dp.ts"
/* 爬楼梯最小代价：动态规划 */
function minCostClimbingStairsDP(cost: Array<number>): number {
const n = cost.length - 1;
if (n === 1 || n === 2) {
return cost[n];
}
// 初始化 dp 表，用于存储子问题的解
const dp = new Array(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 (let i = 3; i <= n; i++) {
dp[i] = Math.min(dp[i - 1], dp[i - 2]) + cost[i];
}
return dp[n];
}
[class]{}-[func]{minCostClimbingStairsDP}
```
=== "Dart"
```dart title="min_cost_climbing_stairs_dp.dart"
/* 爬楼梯最小代价：动态规划 */
int minCostClimbingStairsDP(List<int> cost) {
int n = cost.length - 1;
if (n == 1 || n == 2) return cost[n];
// 初始化 dp 表，用于存储子问题的解
List<int> dp = List.filled(n + 1, 0);
// 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 (int i = 3; i <= n; i++) {
dp[i] = min(dp[i - 1], dp[i - 2]) + cost[i];
}
return dp[n];
}
[class]{}-[func]{minCostClimbingStairsDP}
```
=== "Rust"
```rust title="min_cost_climbing_stairs_dp.rs"
/* 爬楼梯最小代价：动态规划 */
fn min_cost_climbing_stairs_dp(cost: &[i32]) -> i32 {
let n = cost.len() - 1;
if n == 1 || n == 2 {
return cost[n];
}
// 初始化 dp 表，用于存储子问题的解
let mut dp = vec![-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 i in 3..=n {
dp[i] = cmp::min(dp[i - 1], dp[i - 2]) + cost[i];
}
dp[n]
}
[class]{}-[func]{min_cost_climbing_stairs_dp}
```
=== "C"
```c title="min_cost_climbing_stairs_dp.c"
/* 爬楼梯最小代价：动态规划 */
int minCostClimbingStairsDP(int cost[], int costSize) {
int n = costSize - 1;
if (n == 1 || n == 2)
return cost[n];
// 初始化 dp 表，用于存储子问题的解
int *dp = calloc(n + 1, sizeof(int));
// 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 (int i = 3; i <= n; i++) {
dp[i] = myMin(dp[i - 1], dp[i - 2]) + cost[i];
}
int res = dp[n];
// 释放内存
free(dp);
return res;
}
[class]{}-[func]{minCostClimbingStairsDP}
```
=== "Kotlin"
```kotlin title="min_cost_climbing_stairs_dp.kt"
/* 爬楼梯最小代价：动态规划 */
fun minCostClimbingStairsDP(cost: IntArray): Int {
val n = cost.size - 1
if (n == 1 || n == 2) return cost[n]
// 初始化 dp 表，用于存储子问题的解
val dp = IntArray(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 (i in 3..n) {
dp[i] = min(dp[i - 1], dp[i - 2]) + cost[i]
}
return dp[n]
}
[class]{}-[func]{minCostClimbingStairsDP}
```
=== "Ruby"
@@ -310,30 +148,9 @@ According to the state transition equation, and the initial states $dp[1] = cost
=== "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];
}
[class]{}-[func]{minCostClimbingStairsDP}
```
??? pythontutor "Code Visualization"
<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>
<div style="margin-top: 5px;"><a href="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=800&codeDivWidth=600&cumulative=false&curInstr=4&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false" target="_blank" rel="noopener noreferrer">Full Screen ></a></div>
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" }
@@ -346,7 +163,7 @@ This problem can also be space-optimized, compressing one dimension to zero, red
```python title="min_cost_climbing_stairs_dp.py"
def min_cost_climbing_stairs_dp_comp(cost: list[int]) -> int:
"""爬楼梯最小代价：空间优化后的动态规划"""
"""Climbing stairs with minimum cost: Space-optimized dynamic programming"""
n = len(cost) - 1
if n == 1 or n == 2:
return cost[n]
@@ -359,25 +176,13 @@ This problem can also be space-optimized, compressing one dimension to zero, red
=== "C++"
```cpp title="min_cost_climbing_stairs_dp.cpp"
/* 爬楼梯最小代价：空间优化后的动态规划 */
int minCostClimbingStairsDPComp(vector<int> &cost) {
int n = cost.size() - 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;
}
[class]{}-[func]{minCostClimbingStairsDPComp}
```
=== "Java"
```java title="min_cost_climbing_stairs_dp.java"
/* 爬楼梯最小代价：空间优化后的动态规划 */
/* Climbing stairs with minimum cost: Space-optimized dynamic programming */
int minCostClimbingStairsDPComp(int[] cost) {
int n = cost.length - 1;
if (n == 1 || n == 2)
@@ -395,175 +200,55 @@ This problem can also be space-optimized, compressing one dimension to zero, red
=== "C#"
```csharp title="min_cost_climbing_stairs_dp.cs"
/* 爬楼梯最小代价：空间优化后的动态规划 */
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;
}
[class]{min_cost_climbing_stairs_dp}-[func]{MinCostClimbingStairsDPComp}
```
=== "Go"
```go title="min_cost_climbing_stairs_dp.go"
/* 爬楼梯最小代价：空间优化后的动态规划 */
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
}
// 初始状态：预设最小子问题的解
a, b := cost[1], cost[2]
// 状态转移：从较小子问题逐步求解较大子问题
for i := 3; i <= n; i++ {
tmp := b
b = min(a, tmp) + cost[i]
a = tmp
}
return b
}
[class]{}-[func]{minCostClimbingStairsDPComp}
```
=== "Swift"
```swift title="min_cost_climbing_stairs_dp.swift"
/* 爬楼梯最小代价：空间优化后的动态规划 */
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 3 ... n {
(a, b) = (b, min(a, b) + cost[i])
}
return b
}
[class]{}-[func]{minCostClimbingStairsDPComp}
```
=== "JS"
```javascript title="min_cost_climbing_stairs_dp.js"
/* 爬楼梯最小代价：状态压缩后的动态规划 */
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;
}
[class]{}-[func]{minCostClimbingStairsDPComp}
```
=== "TS"
```typescript title="min_cost_climbing_stairs_dp.ts"
/* 爬楼梯最小代价：状态压缩后的动态规划 */
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;
}
[class]{}-[func]{minCostClimbingStairsDPComp}
```
=== "Dart"
```dart title="min_cost_climbing_stairs_dp.dart"
/* 爬楼梯最小代价：空间优化后的动态规划 */
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;
}
[class]{}-[func]{minCostClimbingStairsDPComp}
```
=== "Rust"
```rust title="min_cost_climbing_stairs_dp.rs"
/* 爬楼梯最小代价：空间优化后的动态规划 */
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
}
[class]{}-[func]{min_cost_climbing_stairs_dp_comp}
```
=== "C"
```c title="min_cost_climbing_stairs_dp.c"
/* 爬楼梯最小代价：空间优化后的动态规划 */
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;
}
[class]{}-[func]{minCostClimbingStairsDPComp}
```
=== "Kotlin"
```kotlin title="min_cost_climbing_stairs_dp.kt"
/* 爬楼梯最小代价：空间优化后的动态规划 */
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
}
[class]{}-[func]{minCostClimbingStairsDPComp}
```
=== "Ruby"
@@ -575,29 +260,9 @@ This problem can also be space-optimized, compressing one dimension to zero, red
=== "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;
}
[class]{}-[func]{minCostClimbingStairsDPComp}
```
??? pythontutor "Code Visualization"
<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>
<div style="margin-top: 5px;"><a href="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=800&codeDivWidth=600&cumulative=false&curInstr=5&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false" target="_blank" rel="noopener noreferrer">Full Screen ></a></div>
## 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**.
@@ -644,15 +309,15 @@ In the end, returning $dp[n, 1] + dp[n, 2]$ will do, the sum of the two represen
```python title="climbing_stairs_constraint_dp.py"
def climbing_stairs_constraint_dp(n: int) -> int:
"""带约束爬楼梯：动态规划"""
"""Constrained climbing stairs: Dynamic programming"""
if n == 1 or n == 2:
return 1
# 初始化 dp 表，用于存储子问题的解
# Initialize dp table, used to store subproblem solutions
dp = [[0] * 3 for _ in range(n + 1)]
# 初始状态：预设最小子问题的解
# Initial state: preset the smallest subproblem solution
dp[1][1], dp[1][2] = 1, 0
dp[2][1], dp[2][2] = 0, 1
# 状态转移：从较小子问题逐步求解较大子问题
# State transition: gradually solve larger subproblems from smaller ones
for i in range(3, n + 1):
dp[i][1] = dp[i - 1][2]
dp[i][2] = dp[i - 2][1] + dp[i - 2][2]
@@ -662,43 +327,25 @@ 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"
/* 带约束爬楼梯：动态规划 */
int climbingStairsConstraintDP(int n) {
if (n == 1 || n == 2) {
return 1;
}
// 初始化 dp 表，用于存储子问题的解
vector<vector<int>> dp(n + 1, vector<int>(3, 0));
// 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 (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];
}
[class]{}-[func]{climbingStairsConstraintDP}
```
=== "Java"
```java title="climbing_stairs_constraint_dp.java"
/* 带约束爬楼梯：动态规划 */
/* Constrained climbing stairs: Dynamic programming */
int climbingStairsConstraintDP(int n) {
if (n == 1 || n == 2) {
return 1;
}
// 初始化 dp 表，用于存储子问题的解
// Initialize dp table, used to store subproblem solutions
int[][] dp = new int[n + 1][3];
// 初始状态：预设最小子问题的解
// Initial state: preset the smallest subproblem solution
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
// 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];
@@ -710,226 +357,55 @@ 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"
/* 带约束爬楼梯：动态规划 */
int ClimbingStairsConstraintDP(int n) {
if (n == 1 || n == 2) {
return 1;
}
// 初始化 dp 表，用于存储子问题的解
int[,] dp = new int[n + 1, 3];
// 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 (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];
}
[class]{climbing_stairs_constraint_dp}-[func]{ClimbingStairsConstraintDP}
```
=== "Go"
```go title="climbing_stairs_constraint_dp.go"
/* 带约束爬楼梯：动态规划 */
func climbingStairsConstraintDP(n int) int {
if n == 1 || n == 2 {
return 1
}
// 初始化 dp 表，用于存储子问题的解
dp := make([][3]int, 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 := 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]
}
[class]{}-[func]{climbingStairsConstraintDP}
```
=== "Swift"
```swift title="climbing_stairs_constraint_dp.swift"
/* 带约束爬楼梯：动态规划 */
func climbingStairsConstraintDP(n: Int) -> Int {
if n == 1 || n == 2 {
return 1
}
// 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 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]
}
[class]{}-[func]{climbingStairsConstraintDP}
```
=== "JS"
```javascript title="climbing_stairs_constraint_dp.js"
/* 带约束爬楼梯：动态规划 */
function climbingStairsConstraintDP(n) {
if (n === 1 || n === 2) {
return 1;
}
// 初始化 dp 表，用于存储子问题的解
const dp = Array.from(new Array(n + 1), () => new Array(3));
// 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 (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];
}
[class]{}-[func]{climbingStairsConstraintDP}
```
=== "TS"
```typescript title="climbing_stairs_constraint_dp.ts"
/* 带约束爬楼梯：动态规划 */
function climbingStairsConstraintDP(n: number): number {
if (n === 1 || n === 2) {
return 1;
}
// 初始化 dp 表，用于存储子问题的解
const dp = Array.from({ length: n + 1 }, () => new Array(3));
// 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 (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];
}
[class]{}-[func]{climbingStairsConstraintDP}
```
=== "Dart"
```dart title="climbing_stairs_constraint_dp.dart"
/* 带约束爬楼梯：动态规划 */
int climbingStairsConstraintDP(int n) {
if (n == 1 || n == 2) {
return 1;
}
// 初始化 dp 表，用于存储子问题的解
List<List<int>> dp = List.generate(n + 1, (index) => List.filled(3, 0));
// 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 (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];
}
[class]{}-[func]{climbingStairsConstraintDP}
```
=== "Rust"
```rust title="climbing_stairs_constraint_dp.rs"
/* 带约束爬楼梯：动态规划 */
fn climbing_stairs_constraint_dp(n: usize) -> i32 {
if n == 1 || n == 2 {
return 1;
};
// 初始化 dp 表，用于存储子问题的解
let mut dp = vec![vec![-1; 3]; 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 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]
}
[class]{}-[func]{climbing_stairs_constraint_dp}
```
=== "C"
```c title="climbing_stairs_constraint_dp.c"
/* 带约束爬楼梯：动态规划 */
int climbingStairsConstraintDP(int n) {
if (n == 1 || n == 2) {
return 1;
}
// 初始化 dp 表，用于存储子问题的解
int **dp = malloc((n + 1) * sizeof(int *));
for (int i = 0; i <= n; i++) {
dp[i] = calloc(3, sizeof(int));
}
// 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 (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];
// 释放内存
for (int i = 0; i <= n; i++) {
free(dp[i]);
}
free(dp);
return res;
}
[class]{}-[func]{climbingStairsConstraintDP}
```
=== "Kotlin"
```kotlin title="climbing_stairs_constraint_dp.kt"
/* 带约束爬楼梯：动态规划 */
fun climbingStairsConstraintDP(n: Int): Int {
if (n == 1 || n == 2) {
return 1
}
// 初始化 dp 表，用于存储子问题的解
val dp = Array(n + 1) { IntArray(3) }
// 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 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]
}
[class]{}-[func]{climbingStairsConstraintDP}
```
=== "Ruby"
@@ -941,32 +417,9 @@ In the end, returning $dp[n, 1] + dp[n, 2]$ will do, the sum of the two represen
=== "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];
}
[class]{}-[func]{climbingStairsConstraintDP}
```
??? pythontutor "Code Visualization"
<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>
<div style="margin-top: 5px;"><a href="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=800&codeDivWidth=600&cumulative=false&curInstr=4&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false" target="_blank" rel="noopener noreferrer">Full Screen ></a></div>
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".
!!! question "Stair climbing with obstacle generation"

1073
en/docs/chapter_dynamic_programming/dp_solution_pipeline.md
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700
en/docs/chapter_dynamic_programming/edit_distance_problem.md
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@@ -81,22 +81,22 @@ Observing the state transition equation, solving $dp[i, j]$ depends on the solut
```python title="edit_distance.py"
def edit_distance_dp(s: str, t: str) -> int:
"""编辑距离：动态规划"""
"""Edit distance: Dynamic programming"""
n, m = len(s), len(t)
dp = [[0] * (m + 1) for _ in range(n + 1)]
# 状态转移：首行首列
# State transition: first row and first column
for i in range(1, n + 1):
dp[i][0] = i
for j in range(1, m + 1):
dp[0][j] = j
# 状态转移：其余行和列
# State transition: the 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]:
# 若两字符相等，则直接跳过此两字符
# If the two characters are equal, skip these two characters
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
# The minimum number of edits = the minimum number of edits from three operations (insert, remove, replace) + 1
dp[i][j] = min(dp[i][j - 1], dp[i - 1][j], dp[i - 1][j - 1]) + 1
return dp[n][m]
```
@@ -104,55 +104,31 @@ Observing the state transition equation, solving $dp[i, j]$ depends on the solut
=== "C++"
```cpp title="edit_distance.cpp"
/* 编辑距离：动态规划 */
int editDistanceDP(string s, string t) {
int n = s.length(), m = t.length();
vector<vector<int>> dp(n + 1, vector<int>(m + 1, 0));
// 状态转移：首行首列
for (int i = 1; i <= n; i++) {
dp[i][0] = i;
}
for (int j = 1; j <= m; j++) {
dp[0][j] = j;
}
// 状态转移：其余行和列
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; 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];
}
[class]{}-[func]{editDistanceDP}
```
=== "Java"
```java title="edit_distance.java"
/* 编辑距离：动态规划 */
/* 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];
// 状态转移：首行首列
// 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;
}
// 状态转移：其余行和列
// State transition: the 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)) {
// 若两字符相等，则直接跳过此两字符
// If the two characters are equal, skip these two characters
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
// The minimum number of edits = the minimum number of edits from three operations (insert, remove, replace) + 1
dp[i][j] = Math.min(Math.min(dp[i][j - 1], dp[i - 1][j]), dp[i - 1][j - 1]) + 1;
}
}
@@ -164,291 +140,55 @@ Observing the state transition equation, solving $dp[i, j]$ depends on the solut
=== "C#"
```csharp title="edit_distance.cs"
/* 编辑距离：动态规划 */
int EditDistanceDP(string s, string t) {
int n = s.Length, m = t.Length;
int[,] dp = new int[n + 1, m + 1];
// 状态转移：首行首列
for (int i = 1; i <= n; i++) {
dp[i, 0] = i;
}
for (int j = 1; j <= m; j++) {
dp[0, j] = j;
}
// 状态转移：其余行和列
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; 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] = Math.Min(Math.Min(dp[i, j - 1], dp[i - 1, j]), dp[i - 1, j - 1]) + 1;
}
}
}
return dp[n, m];
}
[class]{edit_distance}-[func]{EditDistanceDP}
```
=== "Go"
```go title="edit_distance.go"
/* 编辑距离：动态规划 */
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)
}
// 状态转移：首行首列
for i := 1; i <= n; i++ {
dp[i][0] = i
}
for j := 1; j <= m; j++ {
dp[0][j] = j
}
// 状态转移：其余行和列
for i := 1; i <= n; i++ {
for j := 1; j <= m; 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] = MinInt(MinInt(dp[i][j-1], dp[i-1][j]), dp[i-1][j-1]) + 1
}
}
}
return dp[n][m]
}
[class]{}-[func]{editDistanceDP}
```
=== "Swift"
```swift title="edit_distance.swift"
/* 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 1 ... n {
dp[i][0] = i
}
for j in 1 ... m {
dp[0][j] = j
}
// 状态转移：其余行和列
for i in 1 ... n {
for j in 1 ... m {
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]
}
[class]{}-[func]{editDistanceDP}
```
=== "JS"
```javascript title="edit_distance.js"
/* 编辑距离：动态规划 */
function editDistanceDP(s, t) {
const n = s.length,
m = t.length;
const dp = Array.from({ length: n + 1 }, () => new Array(m + 1).fill(0));
// 状态转移：首行首列
for (let i = 1; i <= n; i++) {
dp[i][0] = i;
}
for (let j = 1; j <= m; j++) {
dp[0][j] = j;
}
// 状态转移：其余行和列
for (let i = 1; i <= n; i++) {
for (let j = 1; j <= m; j++) {
if (s.charAt(i - 1) === t.charAt(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] =
Math.min(dp[i][j - 1], dp[i - 1][j], dp[i - 1][j - 1]) + 1;
}
}
}
return dp[n][m];
}
[class]{}-[func]{editDistanceDP}
```
=== "TS"
```typescript title="edit_distance.ts"
/* 编辑距离：动态规划 */
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)
);
// 状态转移：首行首列
for (let i = 1; i <= n; i++) {
dp[i][0] = i;
}
for (let j = 1; j <= m; j++) {
dp[0][j] = j;
}
// 状态转移：其余行和列
for (let i = 1; i <= n; i++) {
for (let j = 1; j <= m; j++) {
if (s.charAt(i - 1) === t.charAt(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] =
Math.min(dp[i][j - 1], dp[i - 1][j], dp[i - 1][j - 1]) + 1;
}
}
}
return dp[n][m];
}
[class]{}-[func]{editDistanceDP}
```
=== "Dart"
```dart title="edit_distance.dart"
/* 编辑距离：动态规划 */
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));
// 状态转移：首行首列
for (int i = 1; i <= n; i++) {
dp[i][0] = i;
}
for (int j = 1; j <= m; j++) {
dp[0][j] = j;
}
// 状态转移：其余行和列
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; 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];
}
[class]{}-[func]{editDistanceDP}
```
=== "Rust"
```rust title="edit_distance.rs"
/* 编辑距离：动态规划 */
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];
// 状态转移：首行首列
for i in 1..=n {
dp[i][0] = i as i32;
}
for j in 1..m {
dp[0][j] = j as i32;
}
// 状态转移：其余行和列
for i in 1..=n {
for j in 1..=m {
if s.chars().nth(i - 1) == t.chars().nth(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] =
std::cmp::min(std::cmp::min(dp[i][j - 1], dp[i - 1][j]), dp[i - 1][j - 1]) + 1;
}
}
}
dp[n][m]
}
[class]{}-[func]{edit_distance_dp}
```
=== "C"
```c title="edit_distance.c"
/* 编辑距离：动态规划 */
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));
}
// 状态转移：首行首列
for (int i = 1; i <= n; i++) {
dp[i][0] = i;
}
for (int j = 1; j <= m; j++) {
dp[0][j] = j;
}
// 状态转移：其余行和列
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; 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] = myMin(myMin(dp[i][j - 1], dp[i - 1][j]), dp[i - 1][j - 1]) + 1;
}
}
}
int res = dp[n][m];
// 释放内存
for (int i = 0; i <= n; i++) {
free(dp[i]);
}
return res;
}
[class]{}-[func]{editDistanceDP}
```
=== "Kotlin"
```kotlin title="edit_distance.kt"
/* 编辑距离：动态规划 */
fun editDistanceDP(s: String, t: String): Int {
val n = s.length
val m = t.length
val dp = Array(n + 1) { IntArray(m + 1) }
// 状态转移：首行首列
for (i in 1..n) {
dp[i][0] = i
}
for (j in 1..m) {
dp[0][j] = j
}
// 状态转移：其余行和列
for (i in 1..n) {
for (j in 1..m) {
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]
}
[class]{}-[func]{editDistanceDP}
```
=== "Ruby"
@@ -460,39 +200,9 @@ Observing the state transition equation, solving $dp[i, j]$ depends on the solut
=== "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];
}
[class]{}-[func]{editDistanceDP}
```
??? pythontutor "Code Visualization"
<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>
<div style="margin-top: 5px;"><a href="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=800&codeDivWidth=600&cumulative=false&curInstr=6&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false" target="_blank" rel="noopener noreferrer">Full Screen ></a></div>
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.
=== "<1>"
@@ -552,90 +262,63 @@ For this reason, we can use a variable `leftup` to temporarily store the solutio
```python title="edit_distance.py"
def edit_distance_dp_comp(s: str, t: str) -> int:
"""编辑距离：空间优化后的动态规划"""
"""Edit distance: Space-optimized dynamic programming"""
n, m = len(s), len(t)
dp = [0] * (m + 1)
# 状态转移：首行
# State transition: first row
for j in range(1, m + 1):
dp[j] = j
# 状态转移：其余行
# State transition: the rest of the rows
for i in range(1, n + 1):
# 状态转移：首列
leftup = dp[0] # 暂存 dp[i-1, j-1]
# State transition: first column
leftup = dp[0] # Temporarily store dp[i-1, j-1]
dp[0] += 1
# 状态转移：其余列
# State transition: the rest of the columns
for j in range(1, m + 1):
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
# If the two characters are equal, skip these two characters
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
# The minimum number of edits = the minimum number of edits from three operations (insert, remove, replace) + 1
dp[j] = min(dp[j - 1], dp[j], leftup) + 1
leftup = temp # 更新为下一轮的 dp[i-1, j-1]
leftup = temp # Update for the next round of dp[i-1, j-1]
return dp[m]
```
=== "C++"
```cpp title="edit_distance.cpp"
/* 编辑距离：空间优化后的动态规划 */
int editDistanceDPComp(string s, string t) {
int n = s.length(), m = t.length();
vector<int> dp(m + 1, 0);
// 状态转移：首行
for (int j = 1; j <= m; j++) {
dp[j] = j;
}
// 状态转移：其余行
for (int i = 1; i <= n; i++) {
// 状态转移：首列
int leftup = dp[0]; // 暂存 dp[i-1, j-1]
dp[0] = i;
// 状态转移：其余列
for (int j = 1; j <= m; j++) {
int 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];
}
[class]{}-[func]{editDistanceDPComp}
```
=== "Java"
```java title="edit_distance.java"
/* 编辑距离：空间优化后的动态规划 */
/* 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];
// 状态转移：首行
// State transition: first row
for (int j = 1; j <= m; j++) {
dp[j] = j;
}
// 状态转移：其余行
// State transition: the rest of the rows
for (int i = 1; i <= n; i++) {
// 状态转移：首列
int leftup = dp[0]; // 暂存 dp[i-1, j-1]
// State transition: first column
int leftup = dp[0]; // Temporarily store dp[i-1, j-1]
dp[0] = i;
// 状态转移：其余列
// State transition: the rest of the columns
for (int j = 1; j <= m; j++) {
int temp = dp[j];
if (s.charAt(i - 1) == t.charAt(j - 1)) {
// 若两字符相等，则直接跳过此两字符
// If the two characters are equal, skip these two characters
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
// The minimum number of edits = the minimum number of edits from three operations (insert, remove, replace) + 1
dp[j] = Math.min(Math.min(dp[j - 1], dp[j]), leftup) + 1;
}
leftup = temp; // 更新为下一轮的 dp[i-1, j-1]
leftup = temp; // Update for the next round of dp[i-1, j-1]
}
}
return dp[m];
@@ -645,305 +328,55 @@ For this reason, we can use a variable `leftup` to temporarily store the solutio
=== "C#"
```csharp title="edit_distance.cs"
/* 编辑距离：空间优化后的动态规划 */
int EditDistanceDPComp(string s, string t) {
int n = s.Length, m = t.Length;
int[] dp = new int[m + 1];
// 状态转移：首行
for (int j = 1; j <= m; j++) {
dp[j] = j;
}
// 状态转移：其余行
for (int i = 1; i <= n; i++) {
// 状态转移：首列
int leftup = dp[0]; // 暂存 dp[i-1, j-1]
dp[0] = i;
// 状态转移：其余列
for (int j = 1; j <= m; j++) {
int 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] = Math.Min(Math.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];
}
[class]{edit_distance}-[func]{EditDistanceDPComp}
```
=== "Go"
```go title="edit_distance.go"
/* 编辑距离：空间优化后的动态规划 */
func editDistanceDPComp(s string, t string) int {
n := len(s)
m := len(t)
dp := make([]int, m+1)
// 状态转移：首行
for j := 1; j <= m; j++ {
dp[j] = j
}
// 状态转移：其余行
for i := 1; i <= n; i++ {
// 状态转移：首列
leftUp := dp[0] // 暂存 dp[i-1, j-1]
dp[0] = i
// 状态转移：其余列
for j := 1; j <= m; j++ {
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] = MinInt(MinInt(dp[j-1], dp[j]), leftUp) + 1
}
leftUp = temp // 更新为下一轮的 dp[i-1, j-1]
}
}
return dp[m]
}
[class]{}-[func]{editDistanceDPComp}
```
=== "Swift"
```swift title="edit_distance.swift"
/* 编辑距离：空间优化后的动态规划 */
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 1 ... m {
dp[j] = j
}
// 状态转移：其余行
for i in 1 ... n {
// 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 1 ... m {
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]
}
[class]{}-[func]{editDistanceDPComp}
```
=== "JS"
```javascript title="edit_distance.js"
/* 编辑距离：状态压缩后的动态规划 */
function editDistanceDPComp(s, t) {
const n = s.length,
m = t.length;
const dp = new Array(m + 1).fill(0);
// 状态转移：首行
for (let j = 1; j <= m; j++) {
dp[j] = j;
}
// 状态转移：其余行
for (let i = 1; i <= n; i++) {
// 状态转移：首列
let leftup = dp[0]; // 暂存 dp[i-1, j-1]
dp[0] = i;
// 状态转移：其余列
for (let j = 1; j <= m; j++) {
const temp = dp[j];
if (s.charAt(i - 1) === t.charAt(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] = Math.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];
}
[class]{}-[func]{editDistanceDPComp}
```
=== "TS"
```typescript title="edit_distance.ts"
/* 编辑距离：状态压缩后的动态规划 */
function editDistanceDPComp(s: string, t: string): number {
const n = s.length,
m = t.length;
const dp = new Array(m + 1).fill(0);
// 状态转移：首行
for (let j = 1; j <= m; j++) {
dp[j] = j;
}
// 状态转移：其余行
for (let i = 1; i <= n; i++) {
// 状态转移：首列
let leftup = dp[0]; // 暂存 dp[i-1, j-1]
dp[0] = i;
// 状态转移：其余列
for (let j = 1; j <= m; j++) {
const temp = dp[j];
if (s.charAt(i - 1) === t.charAt(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] = Math.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];
}
[class]{}-[func]{editDistanceDPComp}
```
=== "Dart"
```dart title="edit_distance.dart"
/* 编辑距离：空间优化后的动态规划 */
int editDistanceDPComp(String s, String t) {
int n = s.length, m = t.length;
List<int> dp = List.filled(m + 1, 0);
// 状态转移：首行
for (int j = 1; j <= m; j++) {
dp[j] = j;
}
// 状态转移：其余行
for (int i = 1; i <= n; i++) {
// 状态转移：首列
int leftup = dp[0]; // 暂存 dp[i-1, j-1]
dp[0] = i;
// 状态转移：其余列
for (int j = 1; j <= m; j++) {
int 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];
}
[class]{}-[func]{editDistanceDPComp}
```
=== "Rust"
```rust title="edit_distance.rs"
/* 编辑距离：空间优化后的动态规划 */
fn edit_distance_dp_comp(s: &str, t: &str) -> i32 {
let (n, m) = (s.len(), t.len());
let mut dp = vec![0; m + 1];
// 状态转移：首行
for j in 1..m {
dp[j] = j as i32;
}
// 状态转移：其余行
for i in 1..=n {
// 状态转移：首列
let mut leftup = dp[0]; // 暂存 dp[i-1, j-1]
dp[0] = i as i32;
// 状态转移：其余列
for j in 1..=m {
let temp = dp[j];
if s.chars().nth(i - 1) == t.chars().nth(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] = std::cmp::min(std::cmp::min(dp[j - 1], dp[j]), leftup) + 1;
}
leftup = temp; // 更新为下一轮的 dp[i-1, j-1]
}
}
dp[m]
}
[class]{}-[func]{edit_distance_dp_comp}
```
=== "C"
```c title="edit_distance.c"
/* 编辑距离：空间优化后的动态规划 */
int editDistanceDPComp(char *s, char *t, int n, int m) {
int *dp = calloc(m + 1, sizeof(int));
// 状态转移：首行
for (int j = 1; j <= m; j++) {
dp[j] = j;
}
// 状态转移：其余行
for (int i = 1; i <= n; i++) {
// 状态转移：首列
int leftup = dp[0]; // 暂存 dp[i-1, j-1]
dp[0] = i;
// 状态转移：其余列
for (int j = 1; j <= m; j++) {
int 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] = myMin(myMin(dp[j - 1], dp[j]), leftup) + 1;
}
leftup = temp; // 更新为下一轮的 dp[i-1, j-1]
}
}
int res = dp[m];
// 释放内存
free(dp);
return res;
}
[class]{}-[func]{editDistanceDPComp}
```
=== "Kotlin"
```kotlin title="edit_distance.kt"
/* 编辑距离：空间优化后的动态规划 */
fun editDistanceDPComp(s: String, t: String): Int {
val n = s.length
val m = t.length
val dp = IntArray(m + 1)
// 状态转移：首行
for (j in 1..m) {
dp[j] = j
}
// 状态转移：其余行
for (i in 1..n) {
// 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 1..m) {
val 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]
}
[class]{}-[func]{editDistanceDPComp}
```
=== "Ruby"
@@ -955,38 +388,5 @@ For this reason, we can use a variable `leftup` to temporarily store the solutio
=== "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];
}
[class]{}-[func]{editDistanceDPComp}
```
??? pythontutor "Code Visualization"
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