Longest Palindromic Substring 最长回文子串 学习笔记 1. Brute method 第一种方法:直接循环求解, o(n2) o(n^2) 代码语言:javascript 代码运行次数:0 运行 AI代码解释 class Solution: def longestPalindrome(self, s): """ :type s: str :rtype: str "
class Solution: def longestPalindrome(self, s: str) -> str: #def longestPalindrome(s: str) -> str: n = len(s) if n < 2: return s # 如果字符串长度小于2,它本身就是最长的回文子串 # dp[i][j]表示s[i:j+1]是否是回文串 dp = [[False] * n for _ in range(n)] start, max_...
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常用的方法有中心扩展法与Manacher算法,其中Manacher算法时间复杂度可以达到O(N)O(N), 空间复杂度O(N)O(N),需要重点掌握。 下面是Python的coding代码: classSolution:deflongestPalindrome(self, s):ifnots:return''t ='#'+'#'.join(list(s))+'#'p = [0foriinrange(len(t))] mx =id= resLen = ...
classSolution(object):deflongestPalindrome(self, s):""":type s: str :rtype: str"""size=len(s)ifsize == 1:returnsifsize == 2:ifs[0] == s[1]:returnsreturns[0] maxp= 1ans=s[0] i=0whilei <size: j= i + 1whilej <size:ifs[i] ==s[j]: ...
Longest Palindromic Substring 最长回文子串 标签: Leetcode 题目地址:https://leetcode-cn.com/problems/longest-palindromic-substring/ 题目描述 给定一个字符串 s,找到 s 中最长的回文子串。你可以假设 s 的最大长度为 1000。 示例 1: 输入: “ba......
class Solution { public: string longestPalindrome(string s) { string manaStr = "$#"; for (int i=0;i<s.size();i++) //首先构造出新的字符串 { manaStr += s[i]; manaStr += '#'; } vector<int> rd(manaStr.size(), 0);//用一个辅助数组来记录最大的回文串长度,注意这里记录的是...
class Solution: def longestPalindrome(self, s: str) -> str: l, ms = len(s), s[0] if s else '' # 内循环少一次 if l < 2: return s for i in range(l): for j in range(i + len(ms), l):# 关键点一: + len(ms)
class Solution(object): def longestPalindrome(self, s): ans = '' max_len = 0 n = len(s) DP = [[False] * n for _ in range(n)] # 字符串长度为1 for i in range(n): DP[i][i] = True max_len = 1 ans = s[i]
class Solution { public: string preProcess(string s) { int n = s.length(); if (n == 0) return "^$"; string ret = "^"; for (int i = 0; i < n; i++) ret += "#" + s.substr(i, 1); ret += "#$"; return ret; } string longestPalindrome(string s) { string T = ...