From 0fed2a54a319160b17fe5cea69dd4fedd1e7e053 Mon Sep 17 00:00:00 2001
From: Yuhao Ju <juguaguaji@gmail.com>
Date: Fri, 30 Dec 2022 15:24:06 +0800
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---
 problems/1143.最长公共子序列.md | 43 +++++++++++++-------------
 1 file changed, 21 insertions(+), 22 deletions(-)

diff --git a/problems/1143.最长公共子序列.md b/problems/1143.最长公共子序列.md
index b655b5cd..f2e6e7e2 100644
--- a/problems/1143.最长公共子序列.md
+++ b/problems/1143.最长公共子序列.md
@@ -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]
+}
+```