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: Input: [4,3,2,7,8,2,3,1] Output: [2...
遍历元素时, 当a[i] != i+1 时,就将位置 i 和 a[i]-1 上的两个值交换,反之则 i 移动到下一位。 当a[i] == a[a[i]-1]时,表示 a[i] 有重复,记录重复数字,并把下标 i 上的数字置为 0。 当i 上的数字是 0 时,i++ 遍历到数组末尾,结束🔚 时间复杂度:O(n),空间复杂度:O(1) f...
Arrays.sort(nums);intlast=nums[0];for(inti=1; i<nums.length; i++) {if(nums[i] == last) { ans.add(last); } last = nums[i]; }returnans; } Ref // when find a number i, flip the number at position i-1 to negative.// if the number at position i-1 is already negative,...
[2,3] 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(...
Since each value in a Set has to be unique, passing any duplicate item will be removed automatically:const numbers = [1, 2, 3, 2, 4, 5, 5, 6]; const unique = Array.from(new Set(numbers)); console.log(unique); // [ 1, 2, 3, 4, 5, 6 ] ...
442. Find All Duplicates in an ArrayMedium Topics Companies Given an integer array nums of length n where all the integers of nums are in the range [1, n] and each integer appears at most twice, return an array of all the integers that appears twice. You must write an algorithm that ...
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?
TheindexOf()method returns the first index at which a given element can be found in the array, or -1 if it is not present. Thefilter()method creates a shallow copy of a portion of a given array, filtered down to just the elements from the given array that pass the test implemented ...
I gets stuck in an endless loop. The problem is that this loop will see a duplicate...the number that was just entered. If the number 5 was entered into array[3], then when the loop runs, it's going to see that array[i] == array[j] if i = 3. I need something that will ...
You're dividing an int (arraycheck[j], by an int[]) in the if statement. Honestly, I would take a different approach and first fill the numbers array. Then sort it, and then iterate over the sorted array and just check if arrayNumbers=arrayNumbers[i+1] and then if it is iterate...