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feat: add solutions to lc problem: No.1682
No.1682.Longest Palindromic Subsequence II
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‎solution/1600-1699/1682.Longest Palindromic Subsequence II/README.md

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<!-- 这里可写通用的实现逻辑 -->
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**方法一：记忆化搜索**
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我们设计一个函数 $dfs(i, j, x)$ 表示字符串 $s$ 中下标范围 $[i, j]$ 内，且以字符 $x$ 结尾的最长“好的回文子序列”的长度。答案为 $dfs(0, n - 1, 26)$。
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函数 $dfs(i, j, x)$ 的计算过程如下：
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- 如果 $i >= j$，则 $dfs(i, j, x) = 0$；
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- 如果 $s[i] = s[j]$，且 $s[i] \neq x$，那么 $dfs(i, j, x) = dfs(i + 1, j - 1, s[i]) + 2$；
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- 如果 $s[i] \neq s[j]$，那么 $dfs(i, j, x) = max(dfs(i + 1, j, x), dfs(i, j - 1, x))$。
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过程中，我们可以使用记忆化搜索的方式，避免重复计算。
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时间复杂度 $O(n^2 \times C)$。其中 $n$ 为字符串 $s$ 的长度，而 $C$ 为字符集大小。本题中 $C = 26$。
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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 longestPalindromeSubseq(self, s: str) -> int:
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@cache
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def dfs(i, j, x):
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if i >= j:
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return 0
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if s[i] == s[j] and s[i] != x:
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return dfs(i + 1, j - 1, s[i]) + 2
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return max(dfs(i + 1, j, x), dfs(i, j - 1, x))
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ans = dfs(0, len(s) - 1, '')
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dfs.cache_clear()
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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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private int[][][] f;
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private String s;
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public int longestPalindromeSubseq(String s) {
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int n = s.length();
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this.s = s;
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f = new int[n][n][27];
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for (var a : f) {
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for (var b : a) {
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Arrays.fill(b, -1);
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}
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}
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return dfs(0, n - 1, 26);
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}
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private int dfs(int i, int j, int x) {
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if (i >= j) {
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return 0;
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}
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if (f[i][j][x] != -1) {
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return f[i][j][x];
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}
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int ans = 0;
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if (s.charAt(i) == s.charAt(j) && s.charAt(i) - 'a' != x) {
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ans = dfs(i + 1, j - 1, s.charAt(i) - 'a') + 2;
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} else {
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ans = Math.max(dfs(i + 1, j, x), dfs(i, j - 1, x));
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}
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f[i][j][x] = ans;
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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 f[251][251][27];
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int longestPalindromeSubseq(string s) {
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int n = s.size();
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memset(f, -1, sizeof f);
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function<int(int, int, int)> dfs = [&](int i, int j, int x) -> int {
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if (i >= j) return 0;
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if (f[i][j][x] != -1) return f[i][j][x];
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int ans = 0;
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if (s[i] == s[j] && s[i] - 'a' != x) ans = dfs(i + 1, j - 1, s[i] - 'a') + 2;
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else ans = max(dfs(i + 1, j, x), dfs(i, j - 1, x));
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f[i][j][x] = ans;
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return ans;
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};
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return dfs(0, n - 1, 26);
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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 longestPalindromeSubseq(s string) int {
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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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for j := range f[i] {
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f[i][j] = make([]int, 27)
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for k := range f[i][j] {
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f[i][j][k] = -1
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}
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}
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}
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var dfs func(i, j, x int) int
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dfs = func(i, j, x int) int {
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if i >= j {
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return 0
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}
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if f[i][j][x] != -1 {
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return f[i][j][x]
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}
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ans := 0
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if s[i] == s[j] && int(s[i]-'a') != x {
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ans = dfs(i+1, j-1, int(s[i]-'a')) + 2
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} else {
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ans = max(dfs(i+1, j, x), dfs(i, j-1, x))
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}
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f[i][j][x] = ans
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return ans
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}
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return dfs(0, n-1, 26)
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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/1600-1699/1682.Longest Palindromic Subsequence II/README_EN.md

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### **Python3**
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```python
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class Solution:
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def longestPalindromeSubseq(self, s: str) -> int:
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@cache
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def dfs(i, j, x):
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if i >= j:
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return 0
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if s[i] == s[j] and s[i] != x:
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return dfs(i + 1, j - 1, s[i]) + 2
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return max(dfs(i + 1, j, x), dfs(i, j - 1, x))
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ans = dfs(0, len(s) - 1, '')
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dfs.cache_clear()
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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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private int[][][] f;
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private String s;
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public int longestPalindromeSubseq(String s) {
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int n = s.length();
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this.s = s;
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f = new int[n][n][27];
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for (var a : f) {
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for (var b : a) {
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Arrays.fill(b, -1);
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}
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}
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return dfs(0, n - 1, 26);
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}
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private int dfs(int i, int j, int x) {
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if (i >= j) {
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return 0;
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}
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if (f[i][j][x] != -1) {
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return f[i][j][x];
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}
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int ans = 0;
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if (s.charAt(i) == s.charAt(j) && s.charAt(i) - 'a' != x) {
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ans = dfs(i + 1, j - 1, s.charAt(i) - 'a') + 2;
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} else {
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ans = Math.max(dfs(i + 1, j, x), dfs(i, j - 1, x));
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}
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f[i][j][x] = ans;
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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 f[251][251][27];
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int longestPalindromeSubseq(string s) {
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int n = s.size();
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memset(f, -1, sizeof f);
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function<int(int, int, int)> dfs = [&](int i, int j, int x) -> int {
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if (i >= j) return 0;
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if (f[i][j][x] != -1) return f[i][j][x];
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int ans = 0;
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if (s[i] == s[j] && s[i] - 'a' != x) ans = dfs(i + 1, j - 1, s[i] - 'a') + 2;
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else ans = max(dfs(i + 1, j, x), dfs(i, j - 1, x));
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f[i][j][x] = ans;
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return ans;
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};
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return dfs(0, n - 1, 26);
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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 longestPalindromeSubseq(s string) int {
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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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for j := range f[i] {
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f[i][j] = make([]int, 27)
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for k := range f[i][j] {
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f[i][j][k] = -1
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}
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}
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}
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var dfs func(i, j, x int) int
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dfs = func(i, j, x int) int {
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if i >= j {
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return 0
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}
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if f[i][j][x] != -1 {
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return f[i][j][x]
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}
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ans := 0
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if s[i] == s[j] && int(s[i]-'a') != x {
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ans = dfs(i+1, j-1, int(s[i]-'a')) + 2
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} else {
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ans = max(dfs(i+1, j, x), dfs(i, j-1, x))
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}
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f[i][j][x] = ans
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return ans
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}
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return dfs(0, n-1, 26)
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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/1600-1699/1682.Longest Palindromic Subsequence II/Solution.cpp

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class Solution {
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public:
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int f[251][251][27];
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int longestPalindromeSubseq(string s) {
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int n = s.size();
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memset(f, -1, sizeof f);
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function<int(int, int, int)> dfs = [&](int i, int j, int x) -> int {
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if (i >= j) return 0;
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if (f[i][j][x] != -1) return f[i][j][x];
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int ans = 0;
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if (s[i] == s[j] && s[i] - 'a' != x) ans = dfs(i + 1, j - 1, s[i] - 'a') + 2;
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else ans = max(dfs(i + 1, j, x), dfs(i, j - 1, x));
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f[i][j][x] = ans;
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return ans;
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};
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return dfs(0, n - 1, 26);
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}
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};

‎solution/1600-1699/1682.Longest Palindromic Subsequence II/Solution.go

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func longestPalindromeSubseq(s string) int {
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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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for j := range f[i] {
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f[i][j] = make([]int, 27)
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for k := range f[i][j] {
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f[i][j][k] = -1
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}
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}
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}
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var dfs func(i, j, x int) int
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dfs = func(i, j, x int) int {
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if i >= j {
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return 0
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}
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if f[i][j][x] != -1 {
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return f[i][j][x]
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}
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ans := 0
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if s[i] == s[j] && int(s[i]-'a') != x {
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ans = dfs(i+1, j-1, int(s[i]-'a')) + 2
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} else {
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ans = max(dfs(i+1, j, x), dfs(i, j-1, x))
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}
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f[i][j][x] = ans
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return ans
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}
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return dfs(0, n-1, 26)
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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/1600-1699/1682.Longest Palindromic Subsequence II/Solution.java

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class Solution {
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private int[][][] f;
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private String s;
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public int longestPalindromeSubseq(String s) {
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int n = s.length();
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this.s = s;
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f = new int[n][n][27];
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for (var a : f) {
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for (var b : a) {
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Arrays.fill(b, -1);
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}
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}
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return dfs(0, n - 1, 26);
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}
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private int dfs(int i, int j, int x) {
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if (i >= j) {
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return 0;
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}
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if (f[i][j][x] != -1) {
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return f[i][j][x];
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}
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int ans = 0;
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if (s.charAt(i) == s.charAt(j) && s.charAt(i) - 'a' != x) {
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ans = dfs(i + 1, j - 1, s.charAt(i) - 'a') + 2;
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} else {
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ans = Math.max(dfs(i + 1, j, x), dfs(i, j - 1, x));
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}
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f[i][j][x] = ans;
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return ans;
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}
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}

‎solution/1600-1699/1682.Longest Palindromic Subsequence II/Solution.py

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class Solution:
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def longestPalindromeSubseq(self, s: str) -> int:
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@cache
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def dfs(i, j, x):
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if i >= j:
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return 0
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if s[i] == s[j] and s[i] != x:
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return dfs(i + 1, j - 1, s[i]) + 2
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return max(dfs(i + 1, j, x), dfs(i, j - 1, x))
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ans = dfs(0, len(s) - 1, '')
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dfs.cache_clear()
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return ans

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