260. Single Number III

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

本题和single number1，2不同之处在于，本题有两个single number，其他都是成对存在的，我们首先可以考虑，a&(-a)为什么，他的结果是从右面数第一个1，而我们知道，singnumber1^singnumber2==a，说明1和2的位不同数是a，则右面数第一个不同位就是a&(-a)，这也是可以把它们区别开的办法，代码如下：

public class Solution {
    public int[] singleNumber(int[] nums) {
        int[] res = new int[2];
        int xor = 0;
        for(int i=0;i<nums.length;i++){
            xor= xor^nums[i];
        }
        xor = xor&(-xor);
        for(int i=0;i<nums.length;i++){
            if((xor&nums[i])==xor){
                res[0]=res[0]^nums[i];
            }else{
                res[1] = res[1]^nums[i];
            }
        }
        return res;
    }
}