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.…

problem 438. Find All Anagrams in a String solution1: class Solution { public: vector<int> findAnagrams(string s, string p) { if(s.empty()) return {}; vector<, ); for(auto a:p) pv[a]++; int sn = s.size(); ; while(i<sn) { vector<int> tmp…

原题: 438. Find All Anagrams in a String 解题: 两个步骤 1)就是从s中逐步截取p长度的字符串 2)将截取出的字符串和p进行比较,比较可以用排序,或者字典比较(这两种方法提交后都超时了) 代码如下(提交超时): class Solution { public: vector<int> findAnagrams(string s, string p) { int lenp = p.length(); int lens = s.length(); int i=…

Question 438. Find All Anagrams in a String Solution 题目大意:给两个字符串,s和p,求p在s中出现的位置,p串中的字符无序,ab=ba 思路:起初想的是求p的全排列,保存到set中,遍历s,如果在set中出现,s中的第一个字符位置保存到结果中,最后返回结果.这种思路执行超时.可能是求全排列超时的. 思路2:先把p中的字符及字符出现的次数统计出来保存到map中,再遍历s,这个思路和169. Majority Element - LeetCode…

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.…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 滑动窗口 双指针 日期 题目地址:https://leetcode.com/problems/find-all-anagrams-in-a-string/description/ 题目描述 Given a string s and a non-empty string p, find all the start indices of p's anag…

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);…

class Solution(object):    def findAnagrams(self, s, p):        """        :type s: str        :type p: str        :rtype: List[int]        """        reslist=[];sdic={};pdic={}        ls=len(s)        lp=len(p)              …

1. 原始题目 给定一个字符串 s 和一个非空字符串 p,找到 s 中所有是 p 的字母异位词的子串,返回这些子串的起始索引. 字符串只包含小写英文字母,并且字符串 s 和 p 的长度都不超过 20100. 说明: 字母异位词指字母相同,但排列不同的字符串. 不考虑答案输出的顺序. 示例 1: 输入: s: "cbaebabacd" p: "abc" 输出: [0, 6] 解释: 起始索引等于 0 的子串是 "cba", 它是 "abc…

438. Find All Anagrams in a String DescriptionHintsSubmissionsDiscussSolution   Pick One 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…

[leetcode]438. Find All Anagrams in a String 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…

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.…

原题链接在这里:https://leetcode.com/problems/find-all-anagrams-in-a-string/ 题目: 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 a…

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.Ex…

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 pwill not be larger than 20,100. The order of output does not matter.…

这是悦乐书的第228次更新,第240篇原创 01 看题和准备 今天介绍的是LeetCode算法题中Easy级别的第95题(顺位题号是438).给定一个字符串s和一个非空字符串p,找到s中p的字谜的所有起始索引.字符串仅由小写英文字母组成,字符串s和p的长度不会大于20,100.输出顺序无关紧要.例如: 输入:s:"cbaebabacd" p:"abc" 输出:[0,6] 说明: 起始索引等于0的子字符串是"cba",它是"abc&quo…

Level:   Easy 题目描述: 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 sand p will not be larger than 20,100. The order of outp…

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.…

1.题目描述 2.题目分析 直接使用 哈希表记录子串的信息,然后对比两个哈希表信息即可 3.代码 vector<int> findAnagrams(string s, string p) { vector<int> b; if( s.size() < p.size() ) return b; vector<int> ans; map<char,int> m; for(auto c: p){ m[c]++; } map<char,int> s…

题目描述 Given a string, find the first non-repeating character in it and return it's index. If it doesn't exist, return -1. Examples: s = "leetcode" return 0. s = "loveleetcode", return 2. Note: You may assume the string contain only lowe…

给定一个字符串 s 和一个非空字符串 p,找到 s 中所有是 p 的字母异位词的子串,返回这些子串的起始索引. 字符串只包含小写英文字母,并且字符串 s 和 p 的长度都不超过 20100. 说明: 字母异位词指字母相同,但排列不同的字符串. 不考虑答案输出的顺序. 示例 1: 输入: s: "cbaebabacd" p: "abc" 输出: [0, 6] 解释: 起始索引等于 0 的子串是 "cba", 它是 "abc" 的…

Count the number of segments in a string, where a segment is defined to be a contiguous sequence of non-space characters. Please note that the string does not contain any non-printablecharacters. Example: Input: "Hello, my name is John" Output:…

Given two strings s1 and s2, write a function to return true if s2 contains the permutation of s1. In other words, one of the first string's permutations is the substring of the second string. Example 1: Input:s1 = "ab" s2 = "eidbaooo"…

终于将LeetCode的免费题刷完了,真是漫长的第一遍啊,估计很多题都忘的差不多了,这次开个题目汇总贴,并附上每道题目的解题连接,方便之后查阅吧~ 477 Total Hamming Distance 44.10% Meidum 475 Heaters  30.20% Easy 474 Ones and Zeroes  34.90% Meidum 473 Matchsticks to Square  31.80% Medium 472 Concatenated Words 29.20% Hard…

463. Island Perimeterhttps://leetcode.com/problems/island-perimeter/就是逐一遍历所有的cell,用分离的cell总的的边数减去重叠的边的数目即可.在查找重叠的边的数目的时候有一点小技巧,就是沿着其中两个方向就好,这种题目都有类似的规律,就是可以沿着上三角或者下三角形的方向来做.一刷一次ac,但是还没开始注意codestyle的问题,需要再刷一遍. class Solution { public: int islandPerime…

299. Bulls and Cows 思路:抽屉法,放进secrets,拿出guess,最终cows = cows - bulls public String getHint(String secret, String guess) { int bulls = 0; int cows = 0; int[] numbers = new int[10]; for(int i = 0; i < secret.length(); i++){ if(secret.charAt(i) == guess.c…

滑动窗口基础 滑动窗口常用来解决求字符串子串问题,借助map和计数器,其能在O(n)时间复杂度求子串问题.滑动窗口和双指针(Two pointers)有些类似,可以理解为往同一个方向走的双指针.常用滑动窗口代码框架如下: //3. Longest Substring Without Repeating Characters int lengthOfLongestSubstring(string s) { vector<,); //用于对窗口内的各个字符计数 ,end=,res=; //窗口计数器…