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&…
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.…
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.…
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…
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…
438. 找到字符串中所有字母异位词 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…
题目438. 找到字符串中所有字母异位词 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. 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. 题解 滑动窗口. 数据结构 准备两个HashMap,needs存模式串中各字符出现次数,window存待匹配串中(在needs中出现过的)各字符出现的次数. 用match来计数量匹配对的字符数. 具体方法 维护滑动窗口的左右指针. 右指针不断右移知道串包含了所有模式串中…
Given an input string, reverse the string word by word. A word is defined as a sequence of non-space characters.The input string does not contain leading or trailing spaces and the words are always separated by a single space.For example,Given s = "t…
Given a string, you need to reverse the order of characters in each word within a sentence while still preserving whitespace and initial word order. Example 1: Input: "Let's take LeetCode contest" Output: "s'teL ekat edoCteeL tsetnoc"…
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…
Given an input string, reverse the string word by word. A word is defined as a sequence of non-space characters.The input string does not contain leading or trailing spaces and the words are always separated by a single space.For example,Given s = "t…
Given a string, you need to reverse the order of characters in each word within a sentence while still preserving whitespace and initial word order. Example 1: Input: "Let's take LeetCode contest" Output: "s'teL ekat edoCteeL tsetnoc"…
今天,用Java读取配置文件占位符,使用String.Format(string format,object arg0)方法.以前只知“{0}”为索引占位符(即格式项),与参数列表中的第一个对象相对应,格式设置过程将每个格式项替换为对应对象的值的文本表示形式.但这次需将参数对象格式成一对大括号括起来的格式,即返回字符串“{对象arg0的文本表示形式}”. //直接加大括号 String.Format("{{0}}",1);//{0} 如果使用上面语句,并不能获得所需结果“{1}”,因为…
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★➤微信公众号:山青咏芝(shanqingyongzhi)➤博客园地址:山青咏芝(https://www.cnblogs.com/strengthen/)➤GitHub地址:https://github.com/strengthen/LeetCode➤原文地址:https://www.cnblogs.com/strengthen/p/10260347.html ➤如果链接不是山青咏芝的博客园地址,则可能是爬取作者的文章…
题目 Given an input string, reverse the string word by word. For example, Given s = "the sky is blue", return "blue is sky the". click to show clarification. Clarification: What constitutes a word? A sequence of non-space characters cons…
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.…
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…
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.…
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…
String.prototype.replaceCharAt = function(n,c){ return this.substr(0, n)+ c + this.substr(n+1,this.length-1-n); }…
In a string composed of 'L', 'R', and 'X' characters, like "RXXLRXRXL", a move consists of either replacing one occurrence of "XL" with "LX", or replacing one occurrence of "RX" with "XR". Given the starti…
[抄题]: You need to construct a binary tree from a string consisting of parenthesis and integers. The whole input represents a binary tree. It contains an integer followed by zero, one or two pairs of parenthesis. The integer represents the root's valu…
[抄题]: 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…
Given an input string, reverse the string word by word. Example: Input: "the sky is blue",Output: "blue is sky the".Note: A word is defined as a sequence of non-space characters.Input string may contain leading or trailing spaces. Howe…