Set Mismatch

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

Given an arraynumsrepresenting 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.

Tips:

1. 算法是先计算1-n的和，然后计算当前数列的总数，同时找出被复制的数，丢失的数为sum - curSum + du。
2. 注意sum得为long

Code：

public class Solution {
    public int[] findErrorNums(int[] nums) {
        Set<Integer> set = new HashSet<>();
        int du = 0, n = nums.length;
        long sum = (n * (n + 1) / 2);
        for (int i : nums) {
            if (set.contains(i)) du = i;
            sum -= i;
            set.add(i);
        }
        return new int[] {du, (int)sum + du};
    }
}