题目

给定一个字符串 s，找到 s 中最长的回文子串。你可以假设 s 的最大长度为 1000。

示例 1：
输入: “babad”
输出: “bab”
注意: “aba” 也是一个有效答案。

示例 2：
输入: “cbbd”
输出: “bb”

方案一（中心拓展）

class Solution:
    def expandAroundCenter(self, s, left, right):
        while left >= 0 and right < len(s) and s[left] == s[right]:
            left -= 1
            right += 1
        return left + 1, right - 1
    def longestPalindrome(self, s: str) -> str:
        start, end = 0, 0
        for i in range(len(s)):
            left1, right1 = self.expandAroundCenter(s, i, i)
            left2, right2 = self.expandAroundCenter(s, i, i + 1)
            if right1 - left1 > end - start:
                start, end = left1, right1
            if right2 - left2 > end - start:
                start, end = left2, right2
        return s[start: end + 1]

• leetcode答案

  方案二（动态规划）

  class Solution:
    def longestPalindrome(self, s: str) -> str:
        n = len(s)
        if n < 2:
            return s
  
        # 动态规划算法
        # dp[i][j] 表示 s[i:j+1] 是否是回文串
        dp = [[False] * n for _ in range(n)]
  
        for i in range(n):
            dp[i][i] = True
  
        start, end = 0, 0
        # 递推公式，dp[i][j] = dp[i-1][j-1] and (s[i] == s[j])
        # 注意填表顺序
        for j in range(1, n):
            for i in range(j):
                if s[i] == s[j]:
                    if j - i < 3:
                        dp[i][j] = True
                    else:
                        dp[i][j] = dp[i+1][j-1]
  
                if dp[i][j]:
                    if j - i > end - start:
                        start = i
                        end = j
  
        return s[start:end+1]
  

  原文

  https://leetcode-cn.com/explore/interview/card/2020-top-interview-questions/289/dynamic-programming/1293/
  https://leetcode-cn.com/problems/longest-palindromic-substring/solution/zhong-xin-kuo-san-dong-tai-gui-hua-by-liweiwei1419/
  https://leetcode-cn.com/problems/longest-palindromic-substring/solution/zui-chang-hui-wen-zi-chuan-by-leetcode-solution/