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_bytes() { ch_cnt[(c as u8 - b'a') as usize] += 1; } // 先将 s2 转成字符切片,方便后续遍历和引用 let s2 =...
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....
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...
个人博客:http://fuxuemingzhu.cn/ 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 ...
classSolution {publicbooleancheckInclusion(String s1, String s2) {int[] counts1 =newint[26];intlen =s1.length();for(inti = 0; i < len; i++){ counts1[s1.charAt(i)- 'a']++; }for(intj = 0; j <= s2.length() - len ; j++) { ...
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...
we add 1 to that character count. So once we see all zeros in the map, meaning equal numbers of every characters between s1 and the substring in the sliding window, we know the answer istrue.classSolution {publicbooleancheckInclusion(String s1, String s2) {//build a map of char and fr...
This might suggest a recursive solution: in order to find all permutations that begin with a given letter, call permute on the remainder of the string, thereby finding all permutations of the substring using the same algorithm.One way of coding this is as follows:#include <iostream> #include...
1: string getPermutation(int n, int k) { 2: vector<int> nums(n); 3: int permCount =1; 4: for(int i =0; i< n; i++) 5: { 6: nums[i] = i+1; 7: permCount *= (i+1); 8: } 9: // change K from (1,n) to (0, n-1) to accord to index 10: k--; 11: ...
A standard Sudoku puzzle has a 9 x 9 grid which is to be filled with numbers 1-9 in such a way that the numbers are not repeated in individual row, column or the 9 mini-grids of 3 x 3. A partially filled grid having a unique solution acts a puzzle to be completed. Sudoku ...