题解 | 最长回文子串

#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
# 
# @param A string字符串 
# @return int整型
#
class Solution:
    def getLongestPalindrome(self , A: str) -> int:
        # write code here
        n = len(A)
        if n == 0:
            return 0 
        dp = [[False]*n for _ in range(n)]
        max_length = 1

        for i in range(n):
            dp[i][i] = True
        
        for i in range(n-1):
            if A[i] == A[i+1]:
                dp[i][i+1] = True
                max_length = 2
        # 计算回文大于等于3的回文长度
        for length in range(3,n+1):
            for i in range(n-length+1):
                j = i+length-1
                if A[i]==A[j] and dp[i+1][j-1]:
                    dp[i][j] = True
                    max_length = length

        return max_length