Leetcode: 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:
Input:
[4,3,2,7,8,2,3,1]

Output:
[2,3]

my solution: put integer A[i] at index A[i]-1 unless A[i] already equals i +1 or A[i] == A[A[i]-1]

later scan again if at index i, A[i] != i+1, then A[i] is a duplicate

1 public class Solution {
2     public List<Integer> findDuplicates(int[] A) {
3         List<Integer> res = new ArrayList<Integer>();
4         for (int i=0; i<A.length; i++) {
5             if (A[i]!=i+1 && A[i]!=A[A[i]-1]) {
6                 int temp = A[A[i]-1];
7                 A[A[i]-1] = A[i];
8                 A[i] = temp;
9                 i--;
10             }
11         }
12         for (int i=0; i<A.length; i++) {
13             if (A[i] != i+1)
14                 res.add(A[i]);
15         }
16         return res;
17     }
18 }

Better solution: without destroying the input array

when find a number i, flip the number at position i-1 to negative.

if the number at position i-1 is already negative, i is the number that occurs twice.

1 public class Solution {
2     public List<Integer> findDuplicates(int[] nums) {
3         List<Integer> res = new ArrayList<Integer>();
4         for (int i=0; i<nums.length; i++) {
5             int index = Math.abs(nums[i])-1;
6             if (nums[index] < 0) {
7                 res.add(index+1); // index + 1 == Math.abs(nums[i])
8             }
9             else nums[index] = -nums[index];
10
11         }
12         return res;
13     }
14 }