最长回文子串

思路：遍历所有元素，从当前元素向两边扩展，如果两边相同那么可以拓展，回文串长度加2，如此循环。

class Solution(object):
    def longestPalindrome(self, s):
        """
        :type s: str
        :rtype: str
        """
        max_len = 0
        max_list = ''
        for i in range(len(s)):
            odd_list, odd_len = self.palindrome(s, i, i)
            even_list, even_len = self.palindrome(s, i, i + 1)
            if odd_len > even_len:
                if odd_len > max_len:
                    max_len = odd_len
                    max_list = odd_list
            else:
                if even_len > max_len:
                    max_len = even_len
                    max_list = even_list
        return max_list

    def palindrome(self, s, left, right):
        while left >= 0 and right <= len(s) - 1 and s[left] == s[right]:
            left -= 1
            right += 1
        return s[left+1:right], right - left -1

思路2：dp[left][right] = (right-left<=2) or dp[left+1][right-1]

class Solution(object):
    def longestPalindrome(self, s):
        """
        :type s: str
        :rtype: str
        """
        n = len(s)
        max_len = 0
        max_list = ''
        dp = [[False]*n  for _ in range(n)]
        for r in range(n):
            for l in range(r+1):
                if s[l] == s[r] and (r-l <= 2 or dp[l+1][r-1]):
                    dp[l][r] = True
                    if r-l+1 >= max_len:
                        max_len = r-l+1
                        max_list = s[l:r+1]
        return max_list