567. Permutation in String Given two stringss1ands2, returntrueifs2contains apermutationofs1, orfalseotherwise. In other words, returntrueif one ofs1's permutations is the substring ofs2. Example 1: Input:s1 = "ab", s2 = "eidbaooo"Output:trueExplanation:s2 contains one permutation of s1 ...
进阶:找到所有这种子串的起始下标,这就是前几天做过的 LeetCode 438 - 找到字符串中所有字母异位词 (Rust) 这道题了。 代码 impl Solution { pub fn check_inclusion(s1: String, s2: String) -> bool { // 统计 s1 中每个字符出现的次数 let mut ch_cnt = vec![0; 26]; for &c in s1.as_...
题目地址:https://leetcode.com/problems/permutation-in-string/description/ 题目描述: Given two stringss1ands2, 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: Inp...
classSolution {public:boolcheckInclusion(strings1,strings2) {intn1 = s1.size(), n2 = s2.size(), cnt = n1, left =0; vector<int> m(128);for(charc : s1) ++m[c];for(intright =0; right < n2; ++right) {if(m[s2[right]]-- >0) --cnt;while(cnt ==0) {if(right - left ...
题目地址:https://leetcode.com/problems/permutation-in-string/description/ 题目描述: Given two stringss1ands2, 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. ...
Given a strings, you can transform every letter individually to be lowercase or uppercase to create another string. Returna list of all possible strings we could create. Return the output inany order. Example 1: Input:s = "a1b2"Output:["a1b2","a1B2","A1b2","A1B2"] ...
Given a string, determine if a permutation of the string could form a palindrome. Example 1: Input: "code" Output: false 1. 2. Example 2: Input: "aab" Output: true 1. 2. Example 3: Input: "carerac" Output: true 1. 2.
classSolution{funccheckInclusion(_s1:String,_s2:String)->Bool{ifs1.length>s2.length||s2.length==0{returnfalse}ifs1.length==0{returntrue}lets1CharList=Array(s1)lets2CharList=Array(s2)lets1Count=s1CharList.countlets2Count=s2CharList.countletmap1:[Character:Int]=generateMap(s1CharList)foriin0....
代码func checkInclusion(s1 string, s2 string) bool { need,windows:=make(map[byte]int,0),make(map[byte]int,0) for _,v:=range s1{ need[byte(v)]++ } l,r,valid:=0,0,0 for r<len(s2){ tmp:=s2[r] r++ if _,ok:=need[tmp];ok{ windows[tmp]++ ...
leetcode里面应该有很多个与permutation相关的问题,那么首先就先写出一个全排列把。 code: #include <iostream> #include <vector> using namespace std; void helper(vector<int>& nums, vector<int> rec, vector<vector<int>>& ans, vector<int> visited){ if( rec.size() == nums.size() ){ ans.pus...