Leetcode: Single Number III

class Solution
{
public:
    vector<int> singleNumber(vector<int>& nums)
    {
    	size_t length = nums.size();
    
    	vector<int> results;
    	results.reserve(2);
    	if (0 == length) return results;
    
    	int axorb = 0;
    	for (size_t i = 0; i < length; i++)
    		axorb ^= nums[i];
    
    	int mask = axorb & (-axorb); // 取得axorb从低位向高位，第一个非0位所对应的数字
    	int a = 0;
    	int b = 0;
    	// axorb二进制位为1的位a和b一定不同，利用这个不为1的位将原数组元素分成两组异或即得结果
    	for (size_t i = 0; i < length; i++)
    	{
    		if (mask & nums[i]) a ^= nums[i];
    		else b ^= nums[i];
    	}
    	results.push_back(a);
    	results.push_back(b);
    	return results;
    }
};