5. Longest Palindromic Substring (经典题)_given a string, return the length of its longest p-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有效算法。通过两种方法——暴力枚举中心点并逐步向外扩展和使用马拉车算法（Manacher's Algorithm），文章详细解释了其背后的原理及实现过程。

5. Longest Palindromic Substring

Medium

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Given a string s, return the longest palindromic substring in s.

Example 1:

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

Example 2:

Input: s = "cbbd"
Output: "bb"

Example 3:

Input: s = "a"
Output: "a"

Example 4:

Input: s = "ac"
Output: "a"

Constraints:

1 <= s.length <= 1000
s consist of only digits and English letters (lower-case and/or upper-case)

class Solution:
    max_len = start = end = 0

    def longestPalindrome(self, s: str) -> str:
        """
        解题思路1：暴力枚举中心点，向两边展开
        时间复杂度：O(n^2),空间复杂度：O(1)
        """
        if s is None or len(s) <= 0:
            return ""

        def expand(cur_len: int, l: int, r: int):
            while l >= 0 and r < len(s) and s[l] == s[r]:
                cur_len += 2
                if cur_len > self.max_len:
                    self.max_len = cur_len
                    self.start = l
                    self.end = r
                l -= 1
                r += 1

        for i in range(len(s) - 1):
            if s[i] == s[i + 1]:
                # 中心对称
                expand(0, i, i + 1)
            expand(1, i - 1, i + 1)

        self.end += 1
        return s[self.start:self.end]

还有个解法：Manacher's Algorithm 马拉车算法：时间复杂度O(n)