Longest Palindromic Substring 解答

Question

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.

Solution

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

class Solution(object):
    def spand(self, s, start, end):
        length = len(s)
        while start >= 0 and end < length:
            if s[start] == s[end]:
                start -= 1
                end += 1
            else:
                break
        return s[start + 1: end]
    
    def longestPalindrome(self, s):
        """
        :type s: str
        :rtype: str
        """
        length = len(s)
        result = s[0]
        for i in range(length - 1):
            # sub-length is odd 
            result1 = self.spand(s, i, i)
            if len(result) < len(result1):
                result = result1
            # sub-length is even
            if s[i] == s[i + 1]:
                result2 = self.spand(s, i, i + 1)
                if len(result) < len(result2):
                    result = result2
        return result