Description:
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:

class Solution {
    public List<Integer> findDisappearedNumbers(int[] nums) {
        int len1 = nums.length;
        Set<Integer> set = new HashSet<Integer>();
        Set<Integer> set1 = new HashSet<Integer>();
        
 
for(int i = 0;i < len1;i++){
            set.add(Integer.valueOf(i + 1));
            set1.add(Integer.valueOf(nums[i]));
        }
        set.removeAll(set1);
        return new ArrayList<Integer>(set);
    }
}

Another Better Solution:

public class Solution {

    public List<Integer> findDisappearedNumbers(int
 
[] nums) {

        List<Integer> list = new ArrayList<>();

        boolean[] a = new boolean[nums.length + 1];

        for(int i = 0; i < nums.length; i++) {
            a[nums[i]] = true;
        }

        for(int i = 1; i < a.length; i++) {
            if(!a[i]){
                list.add(i);
            }
        }

        return list;
    }
}

總結：建立一個boolean數組,對應下標為true表示數值存在,false表示不存在,遍歷nums,將nums值作為下標，賦為true。遍歷boolean數組,為false值的下標即為不存在，添加到ArrayList即可(註意,boolean數組的長度為nums的長度加一，方便一一對應)。
重要的是要觀察到nums值與即為boolean數組下標。

leetCode-Find All Numbers Disappeared in an Array