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]

长度为n的整形数组a中的所有数大于等于1，小于等于n，其中可能包含重复两次的数字。

输出[1, n]中不存在于数组a中的数字集合

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;    
        
    }