题解 | #最长回文子串#

判断回文，首先要理解什么是回文，会稳定定义是什么，详解里面说的很清楚。

# -*- coding:utf-8 -*-

class Solution:
    def getLongestPalindrome(self, A, n):
#         print([False] * n)
        # write code here
        if n < 2:return len(A)
        maxlen = 1
        dp = [[False] * n for _ in range(n)] //定义n*n矩阵，这里有大坑
        for i in range(n):
            dp[i][i] = True

        for right in range(n):
            for left in range(right):
                if A[left] != A[right]://如果两种字符不相等，肯定不是回文
                    continue

                if right == left://默认相等的情况
                    dp[left][right] = True
                elif right - left <= 2://类似aa这种也是回文子串
                    dp[left][right] = True
                else:
                    dp[left][right] = dp[left + 1][right - 1]//判断是否是回文子串

                if dp[left][right] and right - left + 1 > maxlen://计算最大回文的长度
                        maxlen = right - left + 1
        return maxlen