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update 1143.最长公共子序列：优化文本错字，调整 rust 代码位置

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Yuhao Ju
2022-12-30 15:24:06 +08:00
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parent 2afe2709a7
commit 0fed2a54a3
1 changed files with 21 additions and 22 deletions
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43
problems/1143.最长公共子序列.md
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@ -49,7 +49,7 @@ dp[i][j]：长度为[0, i - 1]的字符串text1与长度为[0, j - 1]的字符
有同学会问：为什么要定义长度为[0, i - 1]的字符串text1，定义为长度为[0, i]的字符串text1不香么？
这样定义是为了后面代码实现方便，如果非要定义为为长度为[0, i]的字符串text1也可以，我在 [动态规划：718. 最长重复子数组](https://programmercarl.com/0718.最长重复子数组.html) 中的「拓展」里 详细讲解了区别所在，其实就是简化了dp数组第一行和第一列的初始化逻辑。
这样定义是为了后面代码实现方便，如果非要定义为长度为[0, i]的字符串text1也可以，我在 [动态规划：718. 最长重复子数组](https://programmercarl.com/0718.最长重复子数组.html) 中的「拓展」里 详细讲解了区别所在，其实就是简化了dp数组第一行和第一列的初始化逻辑。
2. 确定递推公式
@ -240,27 +240,6 @@ func max(a,b int)int {
```
Rust:
```rust
pub fn longest_common_subsequence(text1: String, text2: String) -> i32 {
let (n, m) = (text1.len(), text2.len());
let (s1, s2) = (text1.as_bytes(), text2.as_bytes());
let mut dp = vec![0; m + 1];
let mut last = vec![0; m + 1];
for i in 1..=n {
dp.swap_with_slice(&mut last);
for j in 1..=m {
dp[j] = if s1[i - 1] == s2[j - 1] {
last[j - 1] + 1
} else {
last[j].max(dp[j - 1])
};
}
}
dp[m]
}
```
Javascript：
```javascript
const longestCommonSubsequence = (text1, text2) => {
@ -304,6 +283,26 @@ function longestCommonSubsequence(text1: string, text2: string): number {
};
```
Rust:
```rust
pub fn longest_common_subsequence(text1: String, text2: String) -> i32 {
let (n, m) = (text1.len(), text2.len());
let (s1, s2) = (text1.as_bytes(), text2.as_bytes());
let mut dp = vec![0; m + 1];
let mut last = vec![0; m + 1];
for i in 1..=n {
dp.swap_with_slice(&mut last);
for j in 1..=m {
dp[j] = if s1[i - 1] == s2[j - 1] {
last[j - 1] + 1
} else {
last[j].max(dp[j - 1])
};
}
}
dp[m]
}
```