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feat: add solutions to lc/lcof2 problems
* lc No.0712.Minimum ASCII Delete Sum for Two Strings * lcof2 No.095.Longest Common Subsequence
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‎README.md

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@@ -74,6 +74,8 @@
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- [俄罗斯套娃信封问题](/solution/0300-0399/0354.Russian%20Doll%20Envelopes/README.md) - 线性 DP、最长上升子序列模型、贪心优化
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- [堆叠长方体的最大高度](/solution/1600-1699/1691.Maximum%20Height%20by%20Stacking%20Cuboids/README.md) - 排序、线性 DP、最长上升子序列模型
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- [无矛盾的最佳球队](/solution/1600-1699/1626.Best%20Team%20With%20No%20Conflicts/README.md) - 排序、线性 DP、最长上升子序列模型
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- [最长公共子序列](/solution/1100-1199/1143.Longest%20Common%20Subsequence/README.md) - 线性 DP、最长公共子序列模型
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- [两个字符串的最小 ASCII 删除和](/solution/0700-0799/0712.Minimum%20ASCII%20Delete%20Sum%20for%20Two%20Strings/README.md) - 线性 DP、最长公共子序列模型
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<!-- 背包问题、状态机模型、状压DP、区间DP、树形DP、数位DP 待补充 -->
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### 4. 高级数据结构

‎README_EN.md

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@@ -71,6 +71,8 @@ Complete solutions to [LeetCode](https://leetcode.com/problemset/all/), [LCOF](h
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- [Russian Doll Envelopes](/solution/0300-0399/0354.Russian%20Doll%20Envelopes/README_EN.md) - Linear problem, LIS
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- [Maximum Height by Stacking Cuboids](/solution/1600-1699/1691.Maximum%20Height%20by%20Stacking%20Cuboids/README_EN.md) - Sort, Linear problem, LIS
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- [Best Team With No Conflicts](/solution/1600-1699/1626.Best%20Team%20With%20No%20Conflicts/README_EN.md) - Sort, Linear problem, LIS
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- [Longest Common Subsequence](/solution/1100-1199/1143.Longest%20Common%20Subsequence/README_EN.md) - Linear problem, LCS
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- [Minimum ASCII Delete Sum for Two Strings](/solution/0700-0799/0712.Minimum%20ASCII%20Delete%20Sum%20for%20Two%20Strings/README_EN.md) - Linear problem, LCS
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### 4. Advanced Data Structures
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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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**方法一：动态规划**
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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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时间复杂度 O(mn)。
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<!-- tabs:start -->
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### **Python3**
@@ -164,6 +166,31 @@ func max(a, b int) int {
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}
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```
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### **JavaScript**
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```js
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/**
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* @param {string} text1
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* @param {string} text2
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* @return {number}
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*/
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var longestCommonSubsequence = function (text1, text2) {
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const m = text1.length;
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const n = text2.length;
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const dp = new Array(m + 1).fill(0).map(() => new Array(n + 1).fill(0));
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for (let i = 1; i <= m; ++i) {
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for (let 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] = 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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```

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

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/**
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* @param {string} text1
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* @param {string} text2
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* @return {number}
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*/
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var longestCommonSubsequence = function (text1, text2) {
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const m = text1.length;
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const n = text2.length;
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const dp = new Array(m + 1).fill(0).map(() => new Array(n + 1).fill(0));
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for (let i = 1; i <= m; ++i) {
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for (let 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] = 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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};

‎solution/0700-0799/0712.Minimum ASCII Delete Sum for Two Strings/README.md

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<!-- 这里可写通用的实现逻辑 -->
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**方法一：动态规划**
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类似[1143. 最长公共子序列](/solution/1100-1199/1143.Longest%20Common%20Subsequence/README.md)。
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定义 `dp[i][j]` 表示使得 `s1[0:i-1]` 和 `s2[0:j-1]` 两个字符串相等所需删除的字符的 ASCII 值的最小值。
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时间复杂度 O(mn)。
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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 minimumDeleteSum(self, s1: str, s2: str) -> int:
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m, n = len(s1), len(s2)
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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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dp[i][0] = dp[i - 1][0] + ord(s1[i - 1])
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for j in range(1, n + 1):
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dp[0][j] = dp[0][j - 1] + ord(s2[j - 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 s1[i - 1] == s2[j - 1]:
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dp[i][j] = dp[i - 1][j - 1]
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else:
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dp[i][j] = min(dp[i - 1][j] + ord(s1[i - 1]),
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dp[i][j - 1] + ord(s2[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 minimumDeleteSum(String s1, String s2) {
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int m = s1.length(), n = s2.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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dp[i][0] = dp[i - 1][0] + s1.codePointAt(i - 1);
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}
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for (int j = 1; j <= n; ++j) {
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dp[0][j] = dp[0][j - 1] + s2.codePointAt(j - 1);
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}
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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 (s1.charAt(i - 1) == s2.charAt(j - 1)) {
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dp[i][j] = dp[i - 1][j - 1];
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} else {
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dp[i][j] = Math.min(dp[i - 1][j] + s1.codePointAt(i - 1), dp[i][j - 1] + s2.codePointAt(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 minimumDeleteSum(string s1, string s2) {
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int m = s1.size(), n = s2.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) dp[i][0] = dp[i - 1][0] + s1[i - 1];
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for (int j = 1; j <= n; ++j) dp[0][j] = dp[0][j - 1] + s2[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 (s1[i - 1] == s2[j - 1]) dp[i][j] = dp[i - 1][j - 1];
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else dp[i][j] = min(dp[i - 1][j] + s1[i - 1], dp[i][j - 1] + s2[j - 1]);
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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 minimumDeleteSum(s1 string, s2 string) int {
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m, n := len(s1), len(s2)
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dp := make([][]int, m+1)
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for i := range dp {
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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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dp[i][0] = dp[i-1][0] + int(s1[i-1])
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}
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for j := 1; j <= n; j++ {
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dp[0][j] = dp[0][j-1] + int(s2[j-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 s1[i-1] == s2[j-1] {
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dp[i][j] = dp[i-1][j-1]
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} else {
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dp[i][j] = min(dp[i-1][j]+int(s1[i-1]), dp[i][j-1]+int(s2[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 min(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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### **...**

‎solution/0700-0799/0712.Minimum ASCII Delete Sum for Two Strings/README_EN.md

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## Solutions
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Dynamic programming.
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<!-- tabs:start -->
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### **Python3**
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```python
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class Solution:
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def minimumDeleteSum(self, s1: str, s2: str) -> int:
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m, n = len(s1), len(s2)
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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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dp[i][0] = dp[i - 1][0] + ord(s1[i - 1])
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for j in range(1, n + 1):
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dp[0][j] = dp[0][j - 1] + ord(s2[j - 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 s1[i - 1] == s2[j - 1]:
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dp[i][j] = dp[i - 1][j - 1]
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else:
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dp[i][j] = min(dp[i - 1][j] + ord(s1[i - 1]),
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dp[i][j - 1] + ord(s2[j - 1]))
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return dp[-1][-1]
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```
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### **Java**
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```java
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class Solution {
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public int minimumDeleteSum(String s1, String s2) {
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int m = s1.length(), n = s2.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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dp[i][0] = dp[i - 1][0] + s1.codePointAt(i - 1);
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}
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for (int j = 1; j <= n; ++j) {
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dp[0][j] = dp[0][j - 1] + s2.codePointAt(j - 1);
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}
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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 (s1.charAt(i - 1) == s2.charAt(j - 1)) {
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dp[i][j] = dp[i - 1][j - 1];
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} else {
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dp[i][j] = Math.min(dp[i - 1][j] + s1.codePointAt(i - 1), dp[i][j - 1] + s2.codePointAt(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 minimumDeleteSum(string s1, string s2) {
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int m = s1.size(), n = s2.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) dp[i][0] = dp[i - 1][0] + s1[i - 1];
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for (int j = 1; j <= n; ++j) dp[0][j] = dp[0][j - 1] + s2[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 (s1[i - 1] == s2[j - 1]) dp[i][j] = dp[i - 1][j - 1];
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else dp[i][j] = min(dp[i - 1][j] + s1[i - 1], dp[i][j - 1] + s2[j - 1]);
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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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### **...**

‎solution/0700-0799/0712.Minimum ASCII Delete Sum for Two Strings/Solution.cpp

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class Solution {
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public:
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int minimumDeleteSum(string s1, string s2) {
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int m = s1.size(), n = s2.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) dp[i][0] = dp[i - 1][0] + s1[i - 1];
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for (int j = 1; j <= n; ++j) dp[0][j] = dp[0][j - 1] + s2[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 (s1[i - 1] == s2[j - 1]) dp[i][j] = dp[i - 1][j - 1];
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else dp[i][j] = min(dp[i - 1][j] + s1[i - 1], dp[i][j - 1] + s2[j - 1]);
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}
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}
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return dp[m][n];
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}
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};

‎solution/0700-0799/0712.Minimum ASCII Delete Sum for Two Strings/Solution.go

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func minimumDeleteSum(s1 string, s2 string) int {
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m, n := len(s1), len(s2)
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dp := make([][]int, m+1)
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for i := range dp {
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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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dp[i][0] = dp[i-1][0] + int(s1[i-1])
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}
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for j := 1; j <= n; j++ {
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dp[0][j] = dp[0][j-1] + int(s2[j-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 s1[i-1] == s2[j-1] {
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dp[i][j] = dp[i-1][j-1]
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} else {
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dp[i][j] = min(dp[i-1][j]+int(s1[i-1]), dp[i][j-1]+int(s2[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 min(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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}

‎solution/0700-0799/0712.Minimum ASCII Delete Sum for Two Strings/Solution.java

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class Solution {
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public int minimumDeleteSum(String s1, String s2) {
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int m = s1.length(), n = s2.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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dp[i][0] = dp[i - 1][0] + s1.codePointAt(i - 1);
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}
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for (int j = 1; j <= n; ++j) {
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dp[0][j] = dp[0][j - 1] + s2.codePointAt(j - 1);
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}
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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 (s1.charAt(i - 1) == s2.charAt(j - 1)) {
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dp[i][j] = dp[i - 1][j - 1];
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} else {
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dp[i][j] = Math.min(dp[i - 1][j] + s1.codePointAt(i - 1), dp[i][j - 1] + s2.codePointAt(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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}

‎solution/0700-0799/0712.Minimum ASCII Delete Sum for Two Strings/Solution.py

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class Solution:
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def minimumDeleteSum(self, s1: str, s2: str) -> int:
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m, n = len(s1), len(s2)
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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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dp[i][0] = dp[i - 1][0] + ord(s1[i - 1])
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for j in range(1, n + 1):
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dp[0][j] = dp[0][j - 1] + ord(s2[j - 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 s1[i - 1] == s2[j - 1]:
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dp[i][j] = dp[i - 1][j - 1]
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else:
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dp[i][j] = min(dp[i - 1][j] + ord(s1[i - 1]),
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dp[i][j - 1] + ord(s2[j - 1]))
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return dp[-1][-1]

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

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![](https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/1100-1199/1143.Longest%20Common%20Subsequence/images/CodeCogsEqn.gif)
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时间复杂度 O(mn)。
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<!-- tabs:start -->
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### **Python3**

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