最直观的做法，时间，空间复杂度都是O(2^n)

class Solution:
    def longestPalindrome(self, s: str) -> str:
        dp = [[0 for i in range(len(s))] for j in range(len(s))]
        longestSubStr = "" # 存储最长回文子串
        longestLen = 0 # 最长回文子串的长度、
        for j in range(len(s)):
            for i in range(j+1):
                if j-i <= 1:
                    if s[i]==s[j]:
                        dp[i][j]=1
                        if longestLen<j-i+1:
                            longestLen = j-i+1
                            longestSubStr = s[i:j+1]
                elif j-i>1:
                    if s[i]==s[j] and dp[i+1][j-1] == 1:
                        dp[i][j] = 1
                        if longestLen < j-i+1:
                            longestLen = j-i+1
                            longestSubStr = s[i:j+1]
        return longestSubStr