leetcode-005 Longest Palindromic Substring-优快云博客

本文介绍了一种寻找字符串中最长回文子串的方法。通过从每个可能的中点出发，向两边扩展来检查是否形成回文，最终找到最长的回文子串。这种方法的时间复杂度为O(n^2)。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

P005 Longest Palindromic Substring
- 思路分析
- 代码
  - java
  - python

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