解法1：暴力列举所有子数，再求回文数，时间复杂度O(n^3)
解法2：遍历所有字符，查找所有基于此字符的回文数，时间复杂度O(n^2)
解法3：manacher算法，时间复杂度O(n)。参考

class Solution(object):
    def longestPalindrome(self, s):
        """
        :type s: str
        :rtype: str
        manacher算法，先使用特殊符号间隔开所有字符,所得到的回文数就肯定是基数了
        """
        l = len(s)
        if l < 2: return s

        newS = ""
        for i in s:
            newS = newS + "#" + i
        newS = "$" + newS + "#"

        newSl = len(newS)
        p = []
        p.append(1)
        mx = 0
        id = 0
        max = 0
        maxi = 0

        # 以下计算出数组对应的i位置最长回文长度P数组
        for i in range(1, newSl):
            if mx > i :
                p.append(min(p[2*id-i], mx-i))
            else:
                p.append(1)

            while i + p[i] < newSl and newS[i + p[i]] == newS[i - p[i]]:
                p[i] = p[i] + 1

            if i+p[i] > mx:
                mx = i + p[i]
                id = i

                if max < p[i]:
                    max = p[i]
                    maxi = i

        #求出最大长度的回文数

        if max < 2:
            return s[2]
        else:
            left = maxi-(max-1)
            right = maxi+max-1
            r = newS[left:right]
            r = r.replace('#', '')
            return r