Single Number III 两个不同的数出现一次，其余两次，异或

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]​​

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

Credits:
Special thanks to ​​​@jianchao.li.fighter​​ for adding this problem and creating all test cases.

class Solution {
public:
    //两个不同的数出现一次，其余两次，异或
    vector<int> singleNumber(vector<int>& nums) {
    
        int n,ans1,ans2;
        
        ans1=0;
        n=nums.size();
        
        for(int i=0;i<n;i++){
            ans1^=nums[i];
        }
        ans2=ans1;
        int k=0;
        while((ans2&1)==0)
        {
            ans2>>=1;
            k++;
        }
        
        ans2=ans1;
        for(int i=0;i<n;i++){
            
            if( (nums[i]>>k)&1)
                ans1^=nums[i];
        }
        ans2=ans1^ans2;
        vector<int> res;
        res.push_back(ans1);
        res.push_back(ans2);
        return res;
    }
};