DAY26：leetcode #5 Longest Palindromic Substring

Given a string s, find the longest palindromic substring in s. You may assume that the maximum length of s is 1000.

Example:

Input: "babad"Output: "bab"Note: "aba" is also a valid answer.

Example:

Input: "cbbd"Output: "bb"

Subscribe to see which companies asked this question

class Solution(object):    def findPalindrome(self, s, index):        step = 0        i = index - 1        j = index + 1        while i >= 0 and j< len(s):            if s[i] == s[j]:                i -= 1                j += 1                step += 1            else:                break        return step    def longestPalindrome(self, s):        """        :type s: str        :rtype: str        """        s_temp = ['#']        for s_t in s:            s_temp.append(s_t)            s_temp.append('#')        steps = [0]*len(s_temp)        C = L = R = 0         for i in range(len(s_temp)):            i_l = 2*C - i             if i_l < 0 or steps[i_l] >= R - i:                steps[i] = self.findPalindrome(s_temp, i)                C = i                L = i - steps[i]                R = i + steps[i]            else:                steps[i] = steps[i_l]        m_index = steps.index(max(steps))        result = []        L = R = m_index        i = 0        while i <= steps[m_index]:            if L == R:                if s_temp[L] != '#':                    result.append(s_temp[L])            else:                if s_temp[L] != '#':                    result.insert(0,s_temp[L])                if s_temp[R] != '#':                    result.append(s_temp[R])            i += 1            L -= 1            R += 1        return ''.join(result)

译文在此：[译+改]最长回文子串(Longest Palindromic Substring) Part II

这道题我采用的思路是Manacher算法，其主要思想先将字符与字符之间插入#号，这样从中间往两边找的时候，就不用奇偶序列分别考虑了。

然后在从前到后遍历的过程中，利用前面的中心轴对称性质，可以简化掉一部分计算。详见链接文章。