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feat: add solutions to lc problem: No.1771
No.1771.Maximize Palindrome Length From Subsequences
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‎solution/1700-1799/1771.Maximize Palindrome Length From Subsequences/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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我们首先将字符串 `word1` 和 `word2` 连接起来，得到字符串 $s$，然后我们可以将问题转化为求字符串 $s$ 的最长回文子序列的长度。只不过这里在算最后的答案时，需要保证回文字符串中，至少有一个字符来自 `word1`，另一个字符来自 `word2`。
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我们定义 $f[i][j]$ 表示字符串 $s$ 中下标范围在 $[i, j]$ 内的子串的最长回文子序列的长度。
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如果 $s[i] = s[j]$，那么 $s[i]$ 和 $s[j]$ 一定在最长回文子序列中，此时 $f[i][j] = f[i + 1][j - 1] + 2$，这时候我们还需要判断 $s[i]$ 和 $s[j]$ 是否来自 `word1` 和 `word2`，如果是，我们将答案的最大值更新为 $ans=\max(ans, f[i][j])$。
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如果 $s[i] \neq s[j]$，那么 $s[i]$ 和 $s[j]$ 一定不会同时出现在最长回文子序列中，此时 $f[i][j] = max(f[i + 1][j], f[i][j - 1])$。
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最后我们返回答案即可。
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时间复杂度为 $O(n^2)$，其中 $n$ 是字符串 $s$ 的长度。
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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 longestPalindrome(self, word1: str, word2: str) -> int:
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s = word1 + word2
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n = len(s)
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f = [[0] * n for _ in range(n)]
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for i in range(n):
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f[i][i] = 1
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ans = 0
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for i in range(n - 1, -1, -1):
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for j in range(i + 1, n):
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if s[i] == s[j]:
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f[i][j] = f[i + 1][j - 1] + 2
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if i < len(word1) and j >= len(word1):
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ans = max(ans, f[i][j])
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else:
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f[i][j] = max(f[i + 1][j], f[i][j - 1])
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return ans
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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 longestPalindrome(String word1, String word2) {
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String s = word1 + word2;
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int n = s.length();
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int[][] f = new int[n][n];
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for (int i = 0; i < n; ++i) {
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f[i][i] = 1;
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}
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int ans = 0;
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for (int i = n - 2; i >= 0; --i) {
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for (int j = i + 1; j < n; ++j) {
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if (s.charAt(i) == s.charAt(j)) {
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f[i][j] = f[i + 1][j - 1] + 2;
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if (i < word1.length() && j >= word1.length()) {
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ans = Math.max(ans, f[i][j]);
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}
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} else {
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f[i][j] = Math.max(f[i + 1][j], f[i][j - 1]);
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}
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}
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}
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return ans;
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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 longestPalindrome(string word1, string word2) {
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string s = word1 + word2;
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int n = s.size();
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int f[n][n];
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memset(f, 0, sizeof f);
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for (int i = 0; i < n; ++i) f[i][i] = 1;
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int ans = 0;
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for (int i = n - 2; ~i; --i) {
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for (int j = i + 1; j < n; ++j) {
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if (s[i] == s[j]) {
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f[i][j] = f[i + 1][j - 1] + 2;
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if (i < word1.size() && j >= word1.size()) {
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ans = max(ans, f[i][j]);
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}
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} else {
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f[i][j] = max(f[i + 1][j], f[i][j - 1]);
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}
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}
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}
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return ans;
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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 longestPalindrome(word1 string, word2 string) (ans int) {
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s := word1 + word2
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n := len(s)
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f := make([][]int, n)
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for i := range f {
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f[i] = make([]int, n)
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f[i][i] = 1
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}
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for i := n - 2; i >= 0; i-- {
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for j := i + 1; j < n; j++ {
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if s[i] == s[j] {
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f[i][j] = f[i+1][j-1] + 2
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if i < len(word1) && j >= len(word1) && ans < f[i][j] {
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ans = f[i][j]
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}
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} else {
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f[i][j] = max(f[i+1][j], f[i][j-1])
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}
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}
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}
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return ans
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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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### **...**

‎solution/1700-1799/1771.Maximize Palindrome Length From Subsequences/README_EN.md

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### **Python3**
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```python
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class Solution:
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def longestPalindrome(self, word1: str, word2: str) -> int:
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s = word1 + word2
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n = len(s)
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f = [[0] * n for _ in range(n)]
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for i in range(n):
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f[i][i] = 1
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ans = 0
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for i in range(n - 1, -1, -1):
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for j in range(i + 1, n):
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if s[i] == s[j]:
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f[i][j] = f[i + 1][j - 1] + 2
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if i < len(word1) and j >= len(word1):
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ans = max(ans, f[i][j])
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else:
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f[i][j] = max(f[i + 1][j], f[i][j - 1])
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return ans
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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 longestPalindrome(String word1, String word2) {
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String s = word1 + word2;
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int n = s.length();
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int[][] f = new int[n][n];
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for (int i = 0; i < n; ++i) {
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f[i][i] = 1;
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}
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int ans = 0;
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for (int i = n - 2; i >= 0; --i) {
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for (int j = i + 1; j < n; ++j) {
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if (s.charAt(i) == s.charAt(j)) {
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f[i][j] = f[i + 1][j - 1] + 2;
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if (i < word1.length() && j >= word1.length()) {
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ans = Math.max(ans, f[i][j]);
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}
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} else {
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f[i][j] = Math.max(f[i + 1][j], f[i][j - 1]);
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}
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}
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}
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return ans;
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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 longestPalindrome(string word1, string word2) {
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string s = word1 + word2;
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int n = s.size();
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int f[n][n];
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memset(f, 0, sizeof f);
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for (int i = 0; i < n; ++i) f[i][i] = 1;
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int ans = 0;
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for (int i = n - 2; ~i; --i) {
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for (int j = i + 1; j < n; ++j) {
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if (s[i] == s[j]) {
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f[i][j] = f[i + 1][j - 1] + 2;
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if (i < word1.size() && j >= word1.size()) {
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ans = max(ans, f[i][j]);
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}
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} else {
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f[i][j] = max(f[i + 1][j], f[i][j - 1]);
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}
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}
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}
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return ans;
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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 longestPalindrome(word1 string, word2 string) (ans int) {
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s := word1 + word2
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n := len(s)
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f := make([][]int, n)
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for i := range f {
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f[i] = make([]int, n)
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f[i][i] = 1
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}
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for i := n - 2; i >= 0; i-- {
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for j := i + 1; j < n; j++ {
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if s[i] == s[j] {
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f[i][j] = f[i+1][j-1] + 2
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if i < len(word1) && j >= len(word1) && ans < f[i][j] {
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ans = f[i][j]
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}
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} else {
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f[i][j] = max(f[i+1][j], f[i][j-1])
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}
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}
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}
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return ans
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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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### **...**

‎solution/1700-1799/1771.Maximize Palindrome Length From Subsequences/Solution.cpp

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class Solution {
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public:
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int longestPalindrome(string word1, string word2) {
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string s = word1 + word2;
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int n = s.size();
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int f[n][n];
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memset(f, 0, sizeof f);
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for (int i = 0; i < n; ++i) f[i][i] = 1;
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int ans = 0;
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for (int i = n - 2; ~i; --i) {
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for (int j = i + 1; j < n; ++j) {
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if (s[i] == s[j]) {
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f[i][j] = f[i + 1][j - 1] + 2;
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if (i < word1.size() && j >= word1.size()) {
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ans = max(ans, f[i][j]);
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}
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} else {
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f[i][j] = max(f[i + 1][j], f[i][j - 1]);
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}
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}
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}
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return ans;
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}
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};

‎solution/1700-1799/1771.Maximize Palindrome Length From Subsequences/Solution.go

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func longestPalindrome(word1 string, word2 string) (ans int) {
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s := word1 + word2
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n := len(s)
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f := make([][]int, n)
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for i := range f {
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f[i] = make([]int, n)
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f[i][i] = 1
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}
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for i := n - 2; i >= 0; i-- {
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for j := i + 1; j < n; j++ {
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if s[i] == s[j] {
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f[i][j] = f[i+1][j-1] + 2
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if i < len(word1) && j >= len(word1) && ans < f[i][j] {
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ans = f[i][j]
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}
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} else {
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f[i][j] = max(f[i+1][j], f[i][j-1])
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}
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}
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}
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return ans
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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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}

‎solution/1700-1799/1771.Maximize Palindrome Length From Subsequences/Solution.java

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class Solution {
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public int longestPalindrome(String word1, String word2) {
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String s = word1 + word2;
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int n = s.length();
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int[][] f = new int[n][n];
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for (int i = 0; i < n; ++i) {
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f[i][i] = 1;
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}
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int ans = 0;
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for (int i = n - 2; i >= 0; --i) {
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for (int j = i + 1; j < n; ++j) {
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if (s.charAt(i) == s.charAt(j)) {
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f[i][j] = f[i + 1][j - 1] + 2;
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if (i < word1.length() && j >= word1.length()) {
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ans = Math.max(ans, f[i][j]);
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}
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} else {
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f[i][j] = Math.max(f[i + 1][j], f[i][j - 1]);
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}
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}
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}
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return ans;
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}
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}

‎solution/1700-1799/1771.Maximize Palindrome Length From Subsequences/Solution.py

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class Solution:
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def longestPalindrome(self, word1: str, word2: str) -> int:
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s = word1 + word2
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n = len(s)
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f = [[0] * n for _ in range(n)]
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for i in range(n):
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f[i][i] = 1
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ans = 0
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for i in range(n - 1, -1, -1):
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for j in range(i + 1, n):
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if s[i] == s[j]:
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f[i][j] = f[i + 1][j - 1] + 2
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if i < len(word1) and j >= len(word1):
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ans = max(ans, f[i][j])
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else:
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f[i][j] = max(f[i + 1][j], f[i][j - 1])
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return ans

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