448. Find All Numbers Disappeared in an Array

QUESTION:

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]

EXPLANATION:

这个地方采用的是正负位置标记法的解决办法，首先创建一个都是负数的数组，然后将原数组中的元素都填充到这个负数数组中，最后再判断其中的负数位置就是缺少的元素了。

SOLUTION:

    public List<Integer> findDisappearedNumbers(int[] nums) {
        ArrayList<Integer> result = new ArrayList<>();
        int n = nums.length;
        int[] num = new int[n];
        Arrays.fill(num,-1);
        for(int i = 0;i < n;i++)
            num[nums[i] - 1] = nums[i];
        for(int i = 0;i<n;i++){
            if(num[i]==-1){
                result.add(i+1);
            }
        }
        return result;
    }
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