问longestSubstring python解决方案对于s.split(c)中的t -->意味着什么EN我一直在研究leetCode，我遇到...
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 """ l = len(s) max_length = 0 palindromic ...
Given a string, find the length of the longest substring without repeating characters. For example, the longest substring without repeating letters for “abcabcbb” is “abc”, which the length is 3. For “bbbbb” the longest substring is “b”, with the length of 1. 思路 最长非重复字符...
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 - ...
return max(self.longestSubstring(t, k) for t in s.split(c)) return len(s) 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 参考资料: https://leetcode.com/problems/longest-substring-with-at-least-k-repeating-characters/discuss/87768/4-lines-Python ...
[Leetcode][python]Longest Substring Without Repeating Characters/无重复字符的最长子串,题目大意给定一个字符串,从中找出不含重复字符的最长子串的长度。例如,”abcabcbb”的不含重复字母的最长子串为”abc”,其长度是3。”bbbbb”的最长子串是”b”,长度为1。解题
classSolution(object):deflengthOfLongestSubstring(self, s):""":type s: str :rtype: int"""#当s长度为0或1时iflen(s) ==0:return0eliflen(s) == 1:return1#设置两个指针 一个从0开始指向子字符串的起始位置 另一个从1开始(s[0,1]就是第一个子字符串)向后遍历beginPointer =0 ...
我的时间复杂度比较高,新建立许多变量,导致空间占用较多。Leetcode上另一个python解法是在原无重复字符串上操作,比较简洁: 1classSolution:2deflengthOfLongestSubstring(self, s: str) ->int:3cur, longest =[], 04foriins: # 串中每个元素5ifiincur: # 如果出现在表里了则只取其之后的所有字符,这点是最...
下面的python代码是根据right指向的字符是否出现在set中而反复的进行循环,代码如下: class Solution(object): def lengthOfLongestSubstring(self, s): """ :type s: str :rtype: int """ left, right = 0, 0 chars = set() res = 0 while left < len(s) and right < len(s): ...
public int lengthOfLongestSubstring(String s) { int longest = 0; for(int i=0; i<s.length()-1; i++){ for(int j=i+1;j<=s.length();j++){ if(isn_repeated(s,i,j)) longest = Math.max(longest,j-i); else break; }