classSolution{publicintlengthOfLongestSubstring(String s){intans=0; Map<Character, Integer> map =newHashMap<>();for(intl=0, r =0; r < s.length(); ++r) {charc=s.charAt(r);if(map.containsKey(c) && l <= map.get(c)) { l = map.get(c) +1; }else{ ans = Math.max(r - ...
1classSolution:2#@return an integer3deflengthOfLongestSubstring(self, s):4length=len(s)5if(length==0):6return07elif(length==1):8return19else:10#work with the data one by one11tmp=""12max_len=013str_len=""14foriinrange(length):15index=tmp.find(s[i])16if(index>-1):17if(len(...
题目地址:https://leetcode.com/problems/longest-substring-without-repeating-characters/description/ 题目描述 Given a string, find the length of thelongest substringwithout repeating characters. Example 1: Input: "abcabcbb" Output: 3 Explanation: The answer is "abc", which the length is 3. 1. ...
题目地址: https://leetcode.com/problems/longest-substring-with-at-least-k-repeating-characters/description/ 题目描述: Find the length of the longest substring T of a given string (consists of lowercase letters only) such that every character in T appears no less ...
代码(Python3) class Solution: def longestContinuousSubstring(self, s: str) -> int: # ans 维护最长的字母序连续字符串的长度 ans: int = 0 # cnt 表示当前字母序连续字符串的长度, # 初始为字母序连续字符串仅由第一个字母组成,长度为 1 cnt: int = 1 # 遍历 s 中的每个字母 for i in range...
This gives us anO(n2) Dynamic Programming solution which usesO(n2) space (could be improved to useO(n) space). Please read more about Longest Common Substringhere. 方法二:动态规划 首先从算法导论种找到相关概念 所谓动态规划,也就是将原问题的解拆分成子问题的解的组合,再利用自底向上的方式使每个...
算法很重要,但是每天也需要学学python,于是就想用python刷leetcode 的算法题,和我一起开始零基础python刷leetcode之旅吧。如有不对的地方,希望指正,万分感谢~~ 题目 3. Longest Substring Without Repeating Characters 最长的不重复子字符串的长度 题目解析 ...
代码实现(Python3) 方法一: classSolution:deflongestPalindrome(self,s):""" :type s: str :rtype: str """iflen(set(s))==1:returnselifs==s[::-1]:# s is already a palindromereturns Len=1start=0foriinrange(1,len(s)):# index values of stringp1,p2=i-Len,i+1ifp1>=1:temp=s[...
code: classSolution:deflengthOfLongestSubstring(self,s:str)->int:# Input: s = "", Output: 0, T=O(n)ifs=="":return0# store all possible characterstotal_str_list=[]# store each letter of substrtemp_str_list=[]temp_str_list.append(s[0])foriinrange(1,len(s)):ifs[i]intemp_str...
参考:LeetCode:Longest Palindromic Substring 最长回文子串 - tenos中的方法4 动态规划 AC代码: 代码语言:javascript 代码运行次数:0 classSolution{public:stringlongestPalindrome(string s){constint len=s.size();if(len<=1)returns;bool dp[len][len];//dp[i][j]表示s[i..j]是否是回文memset(dp,0,...