Solution: Similar toLeetcode-448 Mark the value of already apeared index to negative, then if the value appeared second time the index value should be negative, add it to the result list. publicclassSolution {publicIList<int> FindDuplicates(int[] nums) {intsum =0; IList<int> result =n...
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; } }; classSolution2{ public: vector<int> findDuplicates(vecto...
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...
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...
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: ...
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To check if there were duplicate items in the original array, just compare the length of both arrays:const numbers = [1, 2, 3, 2, 4, 5, 5, 6]; const unique = Array.from(new Set(numbers)); if(numbers.length === unique.length) { console.log(`Array doesn't contain duplicates.`...
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. ...
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I need to load an array with random integers, however, the numbers need to be unique. The size of the array can vary based on user input, so using a static "if array[0] == array[1]", etc., is not going to work. I'm assuming I will need to do this with a for loop, I'...