QUESTION:

Given an array of numbers nums, in which exactly two elements appear only once and all the other elements appear exactly twice. Find the two elements that appear only once.

For example:

Given nums = [1, 2, 1, 3, 2, 5], return [3, 5].

Note:

1. The order of the result is not important. So in the above example, [5, 3] is also correct.
2. Your algorithm should run in linear runtime complexity. Could you implement it using only constant space complexity?

EXPLANATION:

这次的情况是有两个数只出现了一次，那么就需要将这两个数区分出来就可以，区分的关键就在于找到两个数的不同。这里用到的方法就是

1.异或所有数。这样得到的就是3^5=6

2.然后再得出补位，再&，就能得到最右边一位不同的

3.然后再将所有数进行异或

如 3的二进制 是 011， 5的二进制是 101 所以他们异或出来是110也就是6；

6的补位就是010 那么他们再&一下，那么就是010，就可以得到最右边一位不同的数了。就可以区分出对应的两个数了。

SOLUTION:

public class Solution {
    public int[] singleNumber(int[] nums) {
        int[] ans = new int[2];
        int diff = 0;
        for(int num : nums) {
            diff ^= num;
        }
        diff &= -diff;//解决方法的关键，key point
        for(int num : nums) {
            if((num & diff) == 0) {
                ans[0] ^= num;
            }else {
                ans[1] ^= num;
            }
        }
        return ans;
    }
}