时间复杂度:O(n),空间复杂度:O(n) funcfindDuplicates(nums []int)[]int{len:=len(nums);iflen<=1{returnnil} hashMap :=make(map[int]int) result := []int{}for_, v :=rangenums { _, exist := hashMap[v];if(exist) { result =append(result, v) }else{ hashMap[v] = v } }ret...
classSolution {public: vector<int> findDuplicates(vector<int>&nums) { vector<int>res;for(inti =0; i < nums.size(); ++i) {intidx = abs(nums[i]) -1;if(nums[idx] <0) res.push_back(idx +1); nums[idx]= -nums[idx]; }returnres; } }; 下面这种方法是将nums[i]置换到其对应的...
classSolution {publicList<Integer> findDuplicates(int[] nums) {//initializationList<Integer> result =newArrayList<Integer>();//for loop: get the new index, add to result if negative, change to negativefor(inti = 0; i < nums.length; i++) {//get the new indexintindex = Math.abs(nums...
442. Find All Duplicates in an Array Given an array of integers, 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once. Find all the elements that appear twice in this array. Could you do it without extra space and in O(n) runtime? Example: ...
const yourArray = [1, 1, 2, 3, 4, 5, 5] const yourArrayWithoutDuplicates = [...new Set(yourArray)] let duplicates = [...yourArray] yourArrayWithoutDuplicates.forEach((item) => { const i = duplicates.indexOf(item) duplicates = duplicates .slice(0, i) .concat(duplicates.slice(i...
LeetCode #442 - Find All Duplicates in an Array 题目描述: Given an array of integers, 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once. Find all the elements that appear twice&...[LeetCode]-442. Find All Duplicates in an Array 442. ...
Write a JavaScript program to find duplicate values in a JavaScript array. Sample Solution: JavaScript Code: // Function to find duplicates in an array function find_duplicate_in_array(arra1) { // Object to store the count of each element in the array ...
详见:https://leetcode.com/problems/find-all-duplicates-in-an-array/description/ C++: 方法一: AI检测代码解析 classSolution{public:vector<int>findDuplicates(vector<int>&nums){vector<int>res;for(inti=0;i<nums.size();++i){intidx=abs(nums[i])-1;if(nums[idx]<0){res.push_back(idx+1);...
442. Find All Duplicates in an Array 题目描述: Given an array of integers, 1 ≤ a[i] ≤ n (n = size of array), some elements appeartwice and others appear once. Find all the elements that appear twice in this array. Could you do it without extra space and in O(n) runtime?
Finally, the last method to find duplicates in an array is to use the for loop.Here is an example that compares each element of the array with all other elements of the array to check if two values are the same using nested for loop:...