#  状态转移方程 P(i,j)=P(i+1,j−1)∧(Si==Sj)class Solution:def longestPalindrome(self, s: str) -> str:n = len(s)if n < 2:return smax_len = 1begin = 0# dp[i][j] 表示s[i..j]是否是回文串dp = [[False] * n for _ in range(n)]for i in range(n):dp[i][i] = True# 开始递推# 先枚举子串长度for L in range(2, n+1):# 枚举左边界for i in range(n):# 由L和i可以确定右边界，即j-i+1 = Lj = L + i - 1# 如果右边界越界，退出当前循环if j >= n:break  # 跳出当前循环if s[i] != s[j]:dp[i][j] = Falseelse:if j-i<3:dp[i][j] = Trueelse:dp[i][j] = dp[i+1][j-1]# 只要 dp[i][j]==True 成立，就表示子串s[i][j]是回文，此时记录回文长度和起始位置if dp[i][j] and j-i+1>max_len:max_len = j-i+1begin = ireturn s[begin:begin+max_len]