[TOC]

P005 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, and there exists one unique longest palindromic substring.

思路分析

最长回文最简单最直观的思路就是：

• 从某个中点(middle)向两边扫描之道不是回文
  • 中点紧邻的两个点记为left和right
  • left 和 right有可能相等(字符串长度为偶数时)
  • 直到不是回文的时候将该轮循环的子串和上轮做比较取较长者
• 遍历所有的可能性---O(n^2)

代码

java

public class Solution005 {
    private String longestPalindrome(String s, int left, int right) {
        while (left >= 0 && right < s.length() && s.charAt(left) == s.charAt(right)) {
            left--;
            right++;
        }
        return s.substring(left + 1, right);
    }

    public String longestPalindrome(String s) {
        if (s == null || s.length() == 0)
            return "";
        String ret = "";
        for (int i = 0; i < s.length() * 2 - 1; i++) {
            int left = i / 2;
            int right = i / 2;
            if ((i & 1) == 1)// 奇数
                right++;
            String tmp = this.longestPalindrome(s, left, right);
            if (ret.length() < tmp.length())
                ret = tmp;
        }
        return ret;
    }

}

python

class Solution005(object):
    
    def longestStr(self, s, left, right):
        l = len(s)
        while left >= 0 and right < l and s[left] == s[right]:
            left -= 1
            right += 1
        return s[left + 1:right]
            
        
    
    def longestPalindrome(self, s):
        """
        :type s: str
        :rtype: str
        """
        if not s : return ""
        
        ret = "";i = 0
        for i in range(len(s) * 2 - 1):
            left = i / 2
            right = i / 2
            if (i & 1) == 1:right += 1
            
            tmp = self.longestStr(s, left, right)
            
            if len(tmp) > len(ret):ret = tmp
        
        return ret