645. Set Mismatch

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:

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

题目含义：1到n的等差递增数列，然而给的这个数列，有一个值重复了，也就是多了一个值，少了一个值。任务就是找出多了哪个值，少了哪个值

方法一：

 1     public int[] findErrorNums(int[] nums) {
 2         int[] res = new int[2];
 3         for (int i : nums) {
 4            if (nums[Math.abs(i) - 1] < 0) res[0] = Math.abs(i);
 5         　　else nums[Math.abs(i) - 1] *= -1;
 6         }
 7         for (int i=0;i<nums.length;i++) {
 8             if (nums[i] > 0) res[1] = i+1;
 9         }
10         return res;        
11     }

方法二：先求出1到n的总和，然后依次做减法

 1     public int[] findErrorNums(int[] nums) {
 2         Set<Integer> set = new HashSet<>();
 3         int duplicate = 0, n = nums.length;
 4         long sum = (n * (n+1)) / 2;
 5         for(int i : nums) {
 6             if(set.contains(i)) duplicate = i;
 7             sum -= i;
 8             set.add(i);
 9         }
10         return new int[] {duplicate, (int)sum + duplicate};
11     }