leetcode 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?

思路：两个相同数异或为0，结果中任一为1位，数分组；

注意：&按位与操作优先级低于==

参考：http://blog.csdn.net/sbitswc/article/details/48410781

class Solution {
public:
    vector<int> singleNumber(vector<int>& nums) {
        int tmp = 0;
        int len = nums.size();
        for(int i=0;i<len;i++)
        {
            tmp ^= nums[i];
        }
        tmp &= (-tmp);
        vector<int> ans(2,0);
        for(int i=0;i<len;i++)
        {
            if((tmp&nums[i])==0)
                ans[0] ^= nums[i];
            else
                ans[1] ^= nums[i];

        }
        return ans;
    }
};