Single Number II

题目：

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

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

解析：

其实这题好像把异或升级成三目运算，也就是说异或三个数，这三个数对应位都相同为0，都不同为1

可以用一个数组存储每位出现的个数，最后%3，余下的位组成的数字就是single num

class Solution {
public:
    int singleNumber(vector<int>& nums) {
        int n = sizeof(int) * 8;
        int count[n] = {0};
        vector<int>::size_type i = 0;
        for(i;i < nums.size();i++)
            for(int j = 0; j < n;j++)
                count[j] += ( nums[i] >> j ) & 1;
        for(int j = 0; j < n;j++)
            count[j] %= 3;

        int result = 0;
        for(int j = 0; j < n;j++)
            result +=( count[j] << j );
        return result;
    }
};