第五周作业：Single Number III

题目:

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？
题解：求数组中两个唯一出现一次的两个数，由于知道用0遍历异或数组一遍可得到所求两个数的异或值，则由求所得值位为1则表示要求的两个数在那一位值不一样，则可用这不一样的位将原数组分为两个部分，两个不一样的数必然在不同部分中，各自异或遍历则可得到。
代码：

class Solution {
public:
vector<int> singleNumber(vector<int>& nums) {
vector<int>a(2,0);
int res=0;
int bit=0;
for(int i=0;i<nums.size();i++)
res^=nums[i];
bit=res&(~(res-1));
for(int j=0;j<nums.size();j++)
{
if((bit&nums[j])==0)
a[0]^=nums[j];
else
a[1]^=nums[j];
}
return a;
}
};