The set S originally contains numbers from 1 to n. But unfortunately, due to the data error, one of the numbers in the set got duplicated to another number in the set, which results in repetition of one number and loss of another number.

Given an array nums representing the data status of this set after the error. Your task is to firstly find the number occurs twice and then find the number that is missing. Return them in the form of an array.

Example 1:

Input: nums = [1,2,2,4]
Output: [2,3]

Note:

1. The given array size will in the range [2, 10000].
2. The given array’s numbers won’t have any order.

这道题看上去容易，但是写起来有点难度，用了取巧的方法，其实还有个高阶的办法（自遍历），之后题目在写。

class Solution {
    public int[] findErrorNums(int[] nums) {
        HashSet set = new HashSet();
        int[] res = new int[2];
        int sum = nums.length*(nums.length+1)/2;
        for(int i=1;i<=nums.length;i++){
            if(!set.add(nums[i-1])){
                res[0] = nums[i-1];
            }
            sum -= nums[i-1];
        }
        res[1] = sum+res[0];
        return res;
    }
}