Single Number II

Given an array of integers, every element appears three times except for one, which appears exactly once. Find that single one.

Note:

Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?

解题技巧：

这道题类似于 Single Number，不同的是，除了一个数字出现过一次，其他数字都出现了三次，因此，我们可以统计所有数字在某一位上1的个数，如果是3的整数倍，则说明只出现1次的那个数在该位为0，否则，为1。

代码：

#include 
#include 
using namespace std;

int singleNumber(vector& nums)
{
    int length = nums.size();
    int result = 0;
    for(int i = 0; i < 32; i ++)
    {
        int count = 0;
        int mask = 1 << i;
        for(int j = 0; j < length; j ++)
        {
            if(nums[j] & mask)  count ++;
        }
        if(count % 3)  result |= mask;
    }
    return result;
}

int main()
{
    vector nums;
    int num;
    while(cin>>num)
    {
        nums.push_back(num);
    }
    cout<