'''
对于一个字符串，请设计一个高效算法，计算其中最长回文子串的长度。
给定字符串A以及它的长度n，请返回最长回文子串
https://leetcode-cn.com/problems/longest-palindromic-substring/solution/zhong-xin-kuo-san-dong-tai-gui-hua-by-liweiwei1419/
'''

# 动态规划法
class Solution:
    def longestPalindrome(self, s: str) -> str:
        size = len(s)
        if size < 2:
            return s
        # dp[i][j] 表示以A[i]开头和A[j]结尾的字符串是否为回文串
        dp = [[False for _ in range(size)] for _ in range(size)]

        max_len = 1
        start = 0

        for i in range(size):
            dp[i][i] = True

        for i in range(size-2, -1, -1):
            for j in range(i+1, size):
                if s[i] == s[j]:
                    if j - i < 3:
                        dp[i][j] = True
                    else:
                        dp[i][j] = dp[i + 1][j - 1]
                else:
                    dp[i][j] = False

                if dp[i][j]:
                    cur_len = j - i + 1
                    if cur_len > max_len:
                        max_len = cur_len
                        start = i
        return s[start:start + max_len]

# 中心扩散
class Solution2:
    def longestPalindrome(self, s: str) -> str:
        size = len(s)
        if size < 2:
            return s

        # 至少是 1
        max_len = 1
        res = s[0]

        for i in range(size):
            palindrome_odd, odd_len = self.__center_spread(s, size, i, i)
            palindrome_even, even_len = self.__center_spread(s, size, i, i + 1)

            # 当前找到的最长回文子串
            cur_max_sub = palindrome_odd if odd_len >= even_len else palindrome_even
            if len(cur_max_sub) > max_len:
                max_len = len(cur_max_sub)
                res = cur_max_sub

        return res

    def __center_spread(self, s, size, left, right):
        """
        left = right 的时候，此时回文中心是一个字符，回文串的长度是奇数
        right = left + 1 的时候，此时回文中心是一个空隙，回文串的长度是偶数
        """
        i = left
        j = right

        while i >= 0 and j < size and s[i] == s[j]:
            i -= 1
            j += 1
        return s[i + 1:j], j - i - 1
