645. Set Mismatch

Description

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.

Solution

挺妙的一道题。最开始的想法是使用异或，但后来发现对于这道题，只能算出x ^ y，得不到x的值，所以放弃。
后来发现对于值为1到n这类型的题，都可以试试交换元素使其归位的做法，这道题中所有元素归位之后，唯一的一个mismatch的元素和位置即为所求。

class Solution {
    public int[] findErrorNums(int[] nums) {
        int targetIndex = -1;
        for (int i = 0; i < nums.length; ++i) {
            while (nums[i] != i + 1 && nums[i] != nums[nums[i] - 1]) {
                swap(nums, i, nums[i] - 1);
            }
            if (nums[i] != i + 1) {
                targetIndex = i;
            }
        }
        
        return new int[]{nums[targetIndex], targetIndex + 1};
    }
    
    public void swap(int[] nums, int i, int j) {
        int tmp = nums[i];
        nums[i] = nums[j];
        nums[j] = tmp;
    }
}