leetcode5 Longest Palindromic Substring

leetcode第5道题，

题目如下：

Given a string S, find the longest palindromic substring in S. You may assume that the maximum length of S is 1000, and there exists one unique longest palindromic substring.

题目解释：给定一个字符串S，找出最长的回文子串

这里提供两种解法：

1.最原始的做法，一个一个挑出子字符串判定是否为回文字符串（能实现，但超时了）

class Solution(object):    #思路：逐一比较子串，找出最长回文子串        #判断是否是回文字符    def isPalind(self,ss,start,end):        while start < end:            if ss[start] != ss[end]:                return False            start += 1            end -= 1        return True            def longestPalindrome(self, s):     #这个不就是先固定i,在依次移动右边位置，依次比较的嘛        len_s = len(s)        max_s,left,right = 0, 0, 0        for i in range(len_s):            j = i + 1            while j < len_s:                if self.isPalind(s, i, j):                    if (j -i + 1) > max_s:                        left,right = i, j                        max_s = j - i + 1                j += 1                return s[left : right + 1]

2.双指针两边扩展法

class Solution(object):         def longestPalindrome(self, s):          #思路：双指针两边扩展，一往左，一往右，找相等        # maxr,maxl,maxs = 0,0,0        m1,m2,maxs = 0,0,0        n = len(s)                for i in range(n):            #当回文字符串长度为偶数时，中间的相邻等，然后分别往左右扩展            maxl = i            maxr = i + 1            while maxl >= 0 and maxr < n:                if s[maxl] == s[maxr]:                    if (maxr - maxl + 1) > maxs:                        m1 ,m2 = maxl, maxr                        maxs = maxr - maxl + 1                    maxl -= 1                    maxr += 1                else:                    break                                #当回文字符串长度为奇数时，中间三数间隔为1等，再分别左右扩展            maxl = i - 1            maxr = i + 1            while maxl >=0 and maxr < n:                if s[maxl] == s[maxr]:                    if (maxr - maxl + 1) > maxs:                        m1 ,m2 = maxl, maxr                        maxs = maxr - maxl + 1                    maxl -= 1                    maxr += 1                else:                    break                            return s[m1 : m2 + 1]