【重点】【DP】5.最长回文子串|516.最长回文子序列

Allenlzcoder

已于 2025-06-11 01:18:42 修改

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分类专栏：力扣Top100 labuladong的算法小抄笔记文章标签： dp 回文串中心扩展法

于 2023-12-21 00:17:07 首次发布

本文链接：https://blog.youkuaiyun.com/Allenlzcoder/article/details/135120491

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文章对比了两种解决最长回文子串问题的方法：二维动态规划和中心扩展法，同时提及了求解最长回文子序列的动态规划解题思路。

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两个求解目标类似的题目，对比记忆！

5.最长回文子串

题目

Python

class Solution:
    def longestPalindrome(self, s: str) -> str:
        n = len(s)
        s_array = list(s)
        start, end = 0, 0
        dp = [[False]*n for _ in range(n)] # dp[i][j]表示s[i,j]是否是回文子串
        for i in range(n):
            dp[i][i] = True
        
        for i in range(n-1, -1, -1):
            for j in range(i+1, n):
                if s_array[i] == s_array[j] and (j-i == 1 or dp[i+1][j-1]):
                    dp[i][j] = True
                    if j - i > end - start:
                        start, end = i, j
                
        return s[start: end+1]

Java

法1：二维DP

最基础方法！必须掌握！
O(N^2) + O(N^2)

class Solution {
    public String longestPalindrome(String s) {
        int n = s.length();
        if (n == 1) {
            return s;
        }
        String res = s.substring(0, 1);
        boolean[][] dp = new boolean[n][n];
        for (int i = 0; i < n; ++i) {
            dp[i][i] = true;
        }
        for (int i = n - 1; i >= 0; --i) {
            for (int j = i + 1; j < n; ++j) {
                dp[i][j] = (j - i == 1 || dp[i + 1][j - 1]) && (s.charAt(i) == s.charAt(j));
                if (dp[i][j] && (j - i + 1 > res.length())) {
                    res = s.substring(i, j + 1);
                } 
            }
        }

        return res;
    }
}

法2：中心扩展法

O(N^2) + O(1)

class Solution {
    public String longestPalindrome(String s) {
        if (s.length() < 2) {
            return s;
        }
        String res = "";
        for (int i = 0; i < s.length(); ++i) {
            String res1 = palindrome(s, i, i);
            String res2 = palindrome(s, i, i + 1);
            res = res1.length() > res.length() ? res1 : res;
            res = res2.length() > res.length() ? res2 : res;
        }

        return res;
    }

    public String palindrome(String s, int i, int j) {
        while (i >= 0 && j < s.length() && (s.charAt(i) == s.charAt(j))) {
            --i;
            ++j;
        }

        return s.substring(i + 1, j);
    }
}

516.最长回文子序列

题目

Python

class Solution:
    def longestPalindromeSubseq(self, s: str) -> int:
        n = len(s)
        dp = [[0] * n for _ in range(n)]
        for i in range(n):
            dp[i][i] = 1
        for i in range(n-1, -1, -1):
            for j in range(i+1, n):
                if s[i] == s[j]:
                    dp[i][j] = (0 if j - i == 0 else dp[i+1][j-1]) + 2
                else:
                    dp[i][j] = max(dp[i+1][j], dp[i][j-1])

        return dp[0][n-1]

Java

class Solution {
    public int longestPalindromeSubseq(String s) {
        int n = s.length();
        int[][] dp = new int[n][n];
        for (int i = 0; i < n; ++i) {
            dp[i][i] = 1;
        }
        for (int i = n - 1; i >= 0; --i) {
            for (int j = i + 1; j < n; ++j) {
                if (s.charAt(i) == s.charAt(j)) {
                    dp[i][j] = (j - i == 1 ? 0 : dp[i + 1][j - 1]) + 2;
                } else {
                    dp[i][j] = Math.max(dp[i + 1][j], dp[i][j - 1]);
                }
            }
        }

        return dp[0][n - 1];
    }
}