Given an array of integers where 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once.

Find all the elements of [1, n] inclusive that do not appear in this array.

Could you do it without extra space and in O(n) runtime? You may assume the returned list does not count as extra space.

Example:

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

Output:
[5,6]

My solution: straight forward

Sort the array and compare with the current index of (i+1), if not equal, check and add to list.

class Solution {
    public List<Integer> findDisappearedNumbers(int[] nums) {
        List<Integer> list = new ArrayList<>();
        Arrays.sort(nums);
        for(int i=0;i<nums.length;i++){
            if(nums[i]<i+1){
                int j=i+1;
                while(j<nums.length&&nums[j]<i+1) j++;
                if(j>=nums.length||nums[j]!=i+1) list.add(i+1);
            } else if(nums[i]>i+1){ 
                int j=i-1;
                while(j>-1&&nums[j]>i+1) j--;
                if(j<0||nums[j]!=i+1) list.add(i+1);
            }
        }
        return list;
    }
}

Another soultion: mark negative

The basic idea is that we iterate through the input array and mark elements as negative using nums[nums[i]-1] = -nums[nums[i]-1]. In this way all the numbers that we have seen will be marked as negative. In the second iteration, if a value is not marked as negative, it implies we have never seen that index before, so just add it to the return list.

public List<Integer> findDisappearedNumbers(int[] nums) {
    List<Integer> ret = new ArrayList<Integer>();

    for(int i = 0; i < nums.length; i++) {
        int val = Math.abs(nums[i]) - 1;
        if(nums[val] > 0) {
            nums[val] = -nums[val];
        }
    }

    for(int i = 0; i < nums.length; i++) {
        if(nums[i] > 0) {
            ret.add(i+1);
        }
    }
    return ret;
}