[Algorithm] 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]

 

/**
 * @param {number[]} nums
 * @return {number[]}
 */
var findDisappearedNumbers = function(nums) {
    let a = new Set();
    for (let i = 1; i <= nums.length; i++) {
        a.add(i);
    }

    for (let num of nums) {
        if (a.has(num)) {
            a.delete(num)
        }
    }
   
    return [...a.values()];
};
         

 

---

 

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.

Because the elements in the Array are from 1 to n, so subtracting 1 will be 0 to n-1 which are the index of the array.
Take input [4.3.2.7.8.2.3.1] as an example, by subtracting 1 it becomes [3.2.1.6.7.1.2.0] which is an array of index.
For the first iteration
when i = 0 , it marks the nums[3] as negative, the array become [4.3.2.-7.8.2.3.1].
when i = 1, it marks the nums[2] as negative, the array become [4.3.-2.-7.8.2.3.1].
when i = 2, it marks the nums[1] as negative, the array become [4.-3.-2.-7.8.2.3.1].
when i = 3, it marks the nums[6] as negative, the array become [4.-3.-2.-7.8.2.-3.1].
...
...
when i = 6, it marks the nums[0] as negative, the array become [-4.-3.-2.-7.8.2.-3.-1].

 

For the second iteration
find nums[4] = 8 and nums[5] = 2 which > 0;
which means 4 and 5 are not in the index array[3.2.1.6.7.1.2.0].
by adding 1, 5 and 6 are not in the input[4.3.2.7.8.2.3.1]
return[5.6]

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

 

来源：https://www.cnblogs.com/Answer1215/p/12297660.html