448. Find All Numbers Disappeared in an Array

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]

長度爲n的整形數組a中的全部數大於等於1，小於等於n，其中可能包含重複兩次的數字。

輸出[1, n]中不存在於數組a中的數字集合

private static List<Integer> findDisappearNUmber(int[] nums) {
         Set<Integer> set = new HashSet<>();
            List<Integer> list = new LinkedList<>();
            int index = 1;
            for(int i = 0; i < nums.length; i++) {
                set.add(nums[i]);
            }
            for(int i = 0; i < nums.length; i++, index++) {
                if(!set.contains(index)) {
                    list.add(index);
                }
            }
            return list;    
        
    }