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.
Palindrome类型的题目的关键都是这个递推表达式：
dp[i][j] = (s[i] == s[j]) && dp[i + 1][j - 1]
逆向思维，于是我们想到由 dp[i][j] 可以推出以下：
dp[i - 1][j + 1], dp[i - 2][j + 2], dp[i - 3][j + 3]...
这题的一个思路是用DP构造2D Array，参见 Palindrome Subarray
另一个思路也是借助了DP的思想，时间复杂度仍是O(n2)，但是空间复杂度是O(1)
我们对于每个起点遍历，找以 1. 它为中心的最长对称子序列 2. （如果它和它的邻居相等）它和它的邻居为中心的最长对称子序列
代码如下
1 class Solution(object): 2 def spand(self, s, start, end): 3 length = len(s) 4 while start >= 0 and end < length: 5 if s[start] == s[end]: 6 start -= 1 7 end += 1 8 else: 9 break 10 return s[start + 1: end] 11 12 def longestPalindrome(self, s): 13 """ 14 :type s: str 15 :rtype: str 16 """ 17 length = len(s) 18 result = s[0] 19 for i in range(length - 1): 20 # sub-length is odd 21 result1 = self.spand(s, i, i) 22 if len(result) < len(result1): 23 result = result1 24 # sub-length is even 25 if s[i] == s[i + 1]: 26 result2 = self.spand(s, i, i + 1) 27 if len(result) < len(result2): 28 result = result2 29 return result
Longest Palindromic Substring 解答
原文：http://www.cnblogs.com/ireneyanglan/p/4934877.html