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 <iostream>#include <vector>using namespace std;int singleNumber(vector<int>& 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<int> nums;    int num;    while(cin>>num)    {        nums.push_back(num);    }    cout<<singleNumber(nums);}