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feat: add solutions to lcof2 problem: No.095
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‎lcof2/剑指 Offer II 095. 最长公共子序列/README.md

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<!-- 这里可写通用的实现逻辑 -->
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动态规划法。
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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）。
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递推公式如下：
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![](https://cdn.jsdelivr.net/gh/doocs/leetcode@main/lcof2/剑指%20Offer%20II%20095.%20最长公共子序列/images/gif.gif)
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def longestCommonSubsequence(self, text1: str, text2: str) -> int:
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m, n = len(text1), len(text2)
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dp = [[0] * (n + 1) for _ in range(m + 1)]
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for i in range(1, m + 1):
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for j in range(1, n + 1):
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if text1[i - 1] == text2[j - 1]:
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dp[i][j] = dp[i - 1][j - 1] + 1
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else:
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dp[i][j] = max(dp[i - 1][j], dp[i][j - 1])
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return dp[-1][-1]
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public int longestCommonSubsequence(String text1, String text2) {
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int m = text1.length(), n = text2.length();
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int[][] dp = new int[m + 1][n + 1];
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for (int i = 1; i <= m; ++i) {
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for (int j = 1; j <= n; ++j) {
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if (text1.charAt(i - 1) == text2.charAt(j - 1)) {
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dp[i][j] = dp[i - 1][j - 1] + 1;
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} else {
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dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
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}
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}
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}
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return dp[m][n];
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int longestCommonSubsequence(string text1, string text2) {
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int m = text1.size(), n = text2.size();
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vector<vector<int>> dp(m + 1, vector<int>(n + 1));
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for (int i = 1; i <= m; ++i)
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{
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for (int j = 1; j <= n; ++j)
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{
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if (text1[i - 1] == text2[j - 1])
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{
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dp[i][j] = dp[i - 1][j - 1] + 1;
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}
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else
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{
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dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);
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}
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}
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}
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return dp[m][n];
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}
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};
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```
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### **Go**
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```go
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func longestCommonSubsequence(text1 string, text2 string) int {
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m, n := len(text1), len(text2)
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dp := make([][]int, m+1)
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for i := 0; i <= m; i++ {
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dp[i] = make([]int, n+1)
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}
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for i := 1; i <= m; i++ {
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for j := 1; j <= n; j++ {
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if text1[i-1] == text2[j-1] {
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dp[i][j] = dp[i-1][j-1] + 1
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} else {
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dp[i][j] = max(dp[i-1][j], dp[i][j-1])
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}
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}
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}
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return dp[m][n]
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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```
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### **...**

‎lcof2/剑指 Offer II 095. 最长公共子序列/Solution.cpp

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class Solution {
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public:
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int longestCommonSubsequence(string text1, string text2) {
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int m = text1.size(), n = text2.size();
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vector<vector<int>> dp(m + 1, vector<int>(n + 1));
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for (int i = 1; i <= m; ++i)
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{
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for (int j = 1; j <= n; ++j)
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{
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if (text1[i - 1] == text2[j - 1])
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{
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dp[i][j] = dp[i - 1][j - 1] + 1;
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}
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else
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{
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dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);
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}
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}
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}
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return dp[m][n];
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}
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};

‎lcof2/剑指 Offer II 095. 最长公共子序列/Solution.go

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func longestCommonSubsequence(text1 string, text2 string) int {
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m, n := len(text1), len(text2)
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dp := make([][]int, m+1)
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for i := 0; i <= m; i++ {
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dp[i] = make([]int, n+1)
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}
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for i := 1; i <= m; i++ {
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for j := 1; j <= n; j++ {
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if text1[i-1] == text2[j-1] {
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dp[i][j] = dp[i-1][j-1] + 1
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} else {
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dp[i][j] = max(dp[i-1][j], dp[i][j-1])
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}
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}
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}
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return dp[m][n]
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}

‎lcof2/剑指 Offer II 095. 最长公共子序列/Solution.java

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class Solution {
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public int longestCommonSubsequence(String text1, String text2) {
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int m = text1.length(), n = text2.length();
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int[][] dp = new int[m + 1][n + 1];
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for (int i = 1; i <= m; ++i) {
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for (int j = 1; j <= n; ++j) {
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if (text1.charAt(i - 1) == text2.charAt(j - 1)) {
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dp[i][j] = dp[i - 1][j - 1] + 1;
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} else {
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dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
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}
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}
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}
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return dp[m][n];
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}
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}

‎lcof2/剑指 Offer II 095. 最长公共子序列/Solution.py

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

‎lcof2/剑指 Offer II 095. 最长公共子序列/images/gif.gif

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

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int longestCommonSubsequence(string text1, string text2) {
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int m = text1.size(), n = text2.size();
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vector<vector<int>> dp(m + 1, vector<int>(n + 1));
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for (int i = 1; i <= m; ++i) {
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for (int j = 1; j <= n; ++j) {
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if (text1[i - 1] == text2[j - 1]) {
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for (int i = 1; i <= m; ++i)
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{
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for (int j = 1; j <= n; ++j)
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{
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if (text1[i - 1] == text2[j - 1])
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{
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dp[i][j] = dp[i - 1][j - 1] + 1;
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} else {
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}
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else
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{
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dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);
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}
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}

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

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int longestCommonSubsequence(string text1, string text2) {
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int m = text1.size(), n = text2.size();
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vector<vector<int>> dp(m + 1, vector<int>(n + 1));
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for (int i = 1; i <= m; ++i) {
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for (int j = 1; j <= n; ++j) {
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if (text1[i - 1] == text2[j - 1]) {
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for (int i = 1; i <= m; ++i)
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{
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for (int j = 1; j <= n; ++j)
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{
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if (text1[i - 1] == text2[j - 1])
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{
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dp[i][j] = dp[i - 1][j - 1] + 1;
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} else {
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}
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else
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{
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dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);
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}
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}

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

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int longestCommonSubsequence(string text1, string text2) {
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int m = text1.size(), n = text2.size();
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vector<vector<int>> dp(m + 1, vector<int>(n + 1));
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for (int i = 1; i <= m; ++i) {
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for (int j = 1; j <= n; ++j) {
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if (text1[i - 1] == text2[j - 1]) {
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for (int i = 1; i <= m; ++i)
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{
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for (int j = 1; j <= n; ++j)
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{
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if (text1[i - 1] == text2[j - 1])
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{
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dp[i][j] = dp[i - 1][j - 1] + 1;
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} else {
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}
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else
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{
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dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);
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}
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}

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