Leetcode: 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?
分析：

两个相等的数异或结果为0。因此，首次扫描数组，得到两个单独的数a、b的异或结果axorb。因为a和b不相等，因此axorb一定不为0，且二进制位为1的位a和b一定不同。任取axorb中的一个不为0的二进制位，可以将原数组元素分成两组异或即得结果。

注：n & (-n)为n从低位向高位，第一个非0位所对应的数字

参考代码：

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;
}
};