3. Longest Substring Without Repeating Characters 3行 class Solution: def lengthOfLongestSubstring(self, s: str) -> int: b, m, d = 0, 0, {} for i, l in enumerate(s): b, m, d[l] = max(b, d.get(l, -1) + 1), max(m, i - b), i return max(m, len(s) - b) b...
🐍 Shortest-LeetCode-Python-Solutions Leet Code 刷题笔记 - - 不求最快最省,但求最短最优雅 🌿,Shorter is better here. 前言 代码精炼是 Python 的核心,同时能够反应对于语言的熟练程度,本项目目的在于汇总 leet code 最短最优雅的解法,拒绝长篇大论,缩短学习周期,掌握各种技巧,助您在面试中写出令人眼前...
You may assume that the maximum length of S is 1000, and there exists one unique longest palindromic substring. 所谓回文字符串,就是一个字符串,从左到右读和从右到左读是完全一样的。比如”a” , “aaabbaaa” 流川疯 2019/01/18 1.1K0 LeetCode 题目解答——Easy 部分 node.js编程算法网站 [...
* s[j] is unique but not in the "middle" in the string.*/publicString shortestPalindrome(String s) {intj = 0;for(inti = s.length() - 1; i >= 0; --i)//iterate i all the way to 0, such that we try to use s itself to match itself.if(s.charAt(i) == s.charAt(j))/...
You may assume that no string in words is a substring of another string in words. Example: Example 1: Input: words = ["alex","loves","leetcode"] Output: "alexlovesleetcode" Explanation: All permutations of "alex","loves","leetcode" would also be accepted. ...
3. Longest Substring Without Repeating Characters 3行 class Solution: def lengthOfLongestSubstring(self, s: str) -> int: i, r, d = 0, 0, {} for j, c in enumerate(s): i, r, d[c] = max(i, d.get(c, -1) + 1), max(r, j - i), j return max(r, len(s) - i) 双...