[leetcode][ビット演算]Single Number III

タイトル:
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:

The order of the result is not important. So in the above example,  [5, 3]  is also correct.

Your algorithm should run in linear runtime complexity. Could you implement it using only constant space complexity?

class Solution {
public:
int single(vector &nums){
	if (nums.empty()) return -1;
	int res = nums[0];
	for (int i = 1; i < nums.size(); ++i) res ^= nums[i];
	return res;
}
vector singleNumber(vector& nums) {
	vector res;
	if (nums.empty()) return res;
	int xxor = nums[0];
	for (int i = 1; i < nums.size(); ++i){
		xxor ^= nums[i];
	}
	//xxor x1 x2     (  x1 x2   ，  xxor    0，    xxor   1  1)，   1           ，     nums    ，         ：           ，         
	int mask = 1;
	while ((mask & xxor) == 0) mask <<= 1;
	vector nums0, nums1;
	for (int i = 0; i < nums.size(); ++i){
		if ((mask & nums[i]) == 0) nums0.push_back(nums[i]);
		else nums1.push_back(nums[i]);
	}
	int x1 = single(nums0);
	int x2 = single(nums1);
	res.push_back(x1);
	res.push_back(x2);
	return res;
}
};