题解 | #最长回文子串#动态规划

#
# 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)
        ans = 1
        dp = [[0 for _ in range(n)] for _ in range(n)]
        for i in range(n):
            dp[i][i] = 1
        for L in range(2, n + 1):  # 遍历长度 L. !!! 一定要注意先遍历短的字符串
            for l in range(n):  # 遍历左边的索引 l
                r = l + L - 1  # 右边的索引 r
                if r > n - 1:
                    break
                if A[l] != A[r]:
                    dp[l][r] = 0
                elif r - l + 1 <= 2:
                    dp[l][r] = 1
                else:
                    dp[l][r] = dp[l + 1][r - 1]
                
                if dp[l][r] == 1:
                    ans = max(ans, r - l + 1)
        return ans