class Solution:
    def longestPalindrome(self, s: str) -> str:
        n = len(s)
        if n < 2 or s == s[::-1]: #如果数组长度小于2或者正反相等，那么return
            return s
        max_len = 1
        start = 0
        for i in range(1,n): # i是左指针 从1开始遍历
            even = s[i-max_len:i+1]     #获取偶数切片
            odd = s[i-max_len-1:i+1]    #获取奇数切片 
            if i-max_len-1>=0 and odd == odd[::-1]: #确保起始下标为正，且切片是回文
                start = i-max_len-1     # 更新起始下标
                max_len += 2            # 判断奇数回文序列时 i-max_len-1 多减了一个1所以要加回来
            if i-max_len>=0 and even == even[::-1]: #确保起始下标为正，且切片是回文
                start = i-max_len       # 更新起始下标
                max_len += 1            # 更新max_len
        return s[start:start+max_len]