Given a string s and a non-empty string p, find all the start indices of p's anagrams in s. Strings consists of lowercase English letters only and the length of both strings s and p will not be larger than 20,100. The order of output does not matter.…

Input: s: "abab" p: "ab" Output: [0, 1, 2] Explanation: The substring with start index = 0 is "ab", which is an anagram of "ab". The substring with start index = 1 is "ba", which is an anagram of "ab&…

详见:https://leetcode.com/problems/find-all-anagrams-in-a-string/description/ C++: class Solution { public: vector<int> findAnagrams(string s, string p) { if(s.empty()) { return {}; } int ss=s.size(),ps=p.size(),i=0; vector<int> res,cnt(128,0);…

Given a string s and a non-empty string p, find all the start indices of p's anagrams in s. Strings consists of lowercase English letters only and the length of both strings s and p will not be larger than 20,100. The order of output does not matter.…

Given a string s and a non-empty string p, find all the start indices of p's anagrams in s. Strings consists of lowercase English letters only and the length of both strings s and p will not be larger than 20,100. The order of output does not matter.…

Given a string s and a non-empty string p, find all the start indices of p's anagrams in s. Strings consists of lowercase English letters only and the length of both strings s and p will not be larger than 20,100. The order of output does not matter.…

题目描述 找出字符串中第一个只出现一次的字符 如果无此字符 请输出'.' 输入描述: 输入一串字符,由小写字母组成 输出描述: 输出一个字符 输入例子: asdfasdfo 输出例子: o 思路:数组s记录出现的字母顺序.time数组记录出现的次数,每个char对应一个int型,,,所以,字母a出现的次数可以直接用time['a']表示. AC代码: #include "iostream" #include "string.h" #define MAX 201 us…

如题~ 此算法仅供参考,小菜基本不懂高深的算法,只能用最朴实的思想去表达. //找出字符串中第一个不重复的字符 // firstUniqueChar("vdctdvc"); --> t function firstUniqueChar(str){ var str = str || "", i = 0, k = "", _char = "", charMap = {}, result = {name: "&quo…

已知字符串"aabbbcddddeeffffghijklmnopqrst"编程找出出现最多的字符和次数,要求时间复杂度小于O(n^2) /******************************************************** Copyright (C), 2016-2017, FileName: main9 Author: woniu201 Email: wangpengfei.201@163.com Created: 2017/10/31 Descripti…

一.循环obj let testStr = 'asdasddsfdsfadsfdghdadsdfdgdasd'; function getMax(str) { let obj = {}; for(let i in str) { if(obj[str[i]]) { obj[str[i]]++; }else{ obj[str[i]] = 1; } } let keys = Object.keys(obj); // 获取对象中所有key的值返回数组 let values = Object.values…