feat: add solutions to leetcode problem: No.1143. Longest Common Subsequence

Author: yanglbme

README.md

@@ -158,6 +158,7 @@
 - [打家劫舍 II](/solution/0200-0299/0213.House%20Robber%20II/README.md)
 - [最长上升子序列](/solution/0300-0399/0300.Longest%20Increasing%20Subsequence/README.md)
 - [俄罗斯套娃信封问题](/solution/0300-0399/0354.Russian%20Doll%20Envelopes/README.md)
+- [最长公共子序列](/solution/1100-1199/1143.Longest%20Common%20Subsequence/README.md)
 
 ### 回溯算法
 

README_EN.md

@@ -152,6 +152,7 @@ Complete solutions to [LeetCode](https://leetcode-cn.com/problemset/all/), [LCOF
 - [House Robber II](/solution/0200-0299/0213.House%20Robber%20II/README_EN.md)
 - [Longest Increasing Subsequence](/solution/0300-0399/0300.Longest%20Increasing%20Subsequence/README_EN.md)
 - [Russian Doll Envelopes](/solution/0300-0399/0354.Russian%20Doll%20Envelopes/README_EN.md)
+- [Longest Common Subsequence](/solution/1100-1199/1143.Longest%20Common%20Subsequence/README_EN.md)
 
 ### Backtracking
 

solution/1100-1199/1143.Longest Common Subsequence/README.md

@@ -49,22 +49,49 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
+动态规划法。
+
+定义 `dp[i][j]` 表示 `text1[0:i-1]` 和 `text2[0:j-1]` 的最长公共子序列（闭区间）。
+
+递推公式如下：
+
+![](./images/gif.gif)
+
 <!-- tabs:start -->
 
 ### **Python3**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+class Solution:
+    def longestCommonSubsequence(self, text1: str, text2: str) -> int:
+        m, n = len(text1), len(text2)
+        dp = [[0] * (n + 1) for _ in range(m + 1)]
+        for i in range(1, m + 1):
+            for j in range(1, n + 1):
+                dp[i][j] = dp[i - 1][j - 1] + 1 if text1[i - 1] == text2[j - 1] else max(dp[i - 1][j], dp[i][j - 1])
+        return dp[m][n]
 ```
 
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
-
+class Solution {
+    public int longestCommonSubsequence(String text1, String text2) {
+        int m = text1.length(), n = text2.length();
+        int[][] dp = new int[m + 1][n + 1];
+        for (int i = 1; i <= m; ++i) {
+            for (int j = 1; j <= n; ++j) {
+                char c1 = text1.charAt(i - 1), c2 = text2.charAt(j - 1);
+                dp[i][j] = c1 == c2 ? dp[i - 1][j - 1] + 1 : Math.max(dp[i - 1][j], dp[i][j - 1]);
+            }
+        }
+        return dp[m][n];
+    }
+}
 ```
 
 ### **...**

solution/1100-1199/1143.Longest Common Subsequence/README_EN.md

@@ -53,13 +53,32 @@
 ### **Python3**
 
 ```python
-
+class Solution:
+    def longestCommonSubsequence(self, text1: str, text2: str) -> int:
+        m, n = len(text1), len(text2)
+        dp = [[0] * (n + 1) for _ in range(m + 1)]
+        for i in range(1, m + 1):
+            for j in range(1, n + 1):
+                dp[i][j] = dp[i - 1][j - 1] + 1 if text1[i - 1] == text2[j - 1] else max(dp[i - 1][j], dp[i][j - 1])
+        return dp[m][n]
 ```
 
 ### **Java**
 
 ```java
-
+class Solution {
+    public int longestCommonSubsequence(String text1, String text2) {
+        int m = text1.length(), n = text2.length();
+        int[][] dp = new int[m + 1][n + 1];
+        for (int i = 1; i <= m; ++i) {
+            for (int j = 1; j <= n; ++j) {
+                char c1 = text1.charAt(i - 1), c2 = text2.charAt(j - 1);
+                dp[i][j] = c1 == c2 ? dp[i - 1][j - 1] + 1 : Math.max(dp[i - 1][j], dp[i][j - 1]);
+            }
+        }
+        return dp[m][n];
+    }
+}
 ```
 
 ### **...**

solution/1100-1199/1143.Longest Common Subsequence/Solution.java

@@ -0,0 +1,13 @@
+class Solution {
+    public int longestCommonSubsequence(String text1, String text2) {
+        int m = text1.length(), n = text2.length();
+        int[][] dp = new int[m + 1][n + 1];
+        for (int i = 1; i <= m; ++i) {
+            for (int j = 1; j <= n; ++j) {
+                char c1 = text1.charAt(i - 1), c2 = text2.charAt(j - 1);
+                dp[i][j] = c1 == c2 ? dp[i - 1][j - 1] + 1 : Math.max(dp[i - 1][j], dp[i][j - 1]);
+            }
+        }
+        return dp[m][n];
+    }
+}

solution/1100-1199/1143.Longest Common Subsequence/Solution.py

@@ -0,0 +1,8 @@
+class Solution:
+    def longestCommonSubsequence(self, text1: str, text2: str) -> int:
+        m, n = len(text1), len(text2)
+        dp = [[0] * (n + 1) for _ in range(m + 1)]
+        for i in range(1, m + 1):
+            for j in range(1, n + 1):
+                dp[i][j] = dp[i - 1][j - 1] + 1 if text1[i - 1] == text2[j - 1] else max(dp[i - 1][j], dp[i][j - 1])
+        return dp[m][n]