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]

//以下是两个参考代码，很好奇他们是怎么想出这种方法？，好好思考，学习一下

/*
      * 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;
    }

//5-line Java Easy-understanding

public List<Integer> findDisappearedNumbers(int[] nums) {
        List<Integer> res = new ArrayList<>();
        int n = nums.length;
        for (int i = 0; i < nums.length; i ++) nums[(nums[i]-1) % n] += n;  //a % b如果a<b,余数是它自己a
        for (int i = 0; i < nums.length; i ++) if (nums[i] <= n) res.add(i+1);
        return res;
    }