【leetcode】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?

class Solution {
public:
    vector<int> singleNumber(vector<int>& nums) {
        int AxorB = 0;
        for (vector<int>::iterator it = nums.begin(); it != nums.end(); ++it)
        {
            AxorB = AxorB ^ (*it);
        }
        
        int A = 0;
        int B = 0;
        int mask = AxorB & (-AxorB);
        for (vector<int>::iterator it = nums.begin(); it != nums.end(); ++it)
        {
            if ((mask & (*it)) == 0)
                A = A ^ (*it);
            else
                B = B ^ (*it);
        }
        
        vector<int> result;
        result.push_back(A);
        result.push_back(B);
        return result;
    }
};

Personal Note：

1. n & (-n)表示取的n中的最后一位二进制位
2. &的优先级小于==的优先级