LeetCode:Single NumberⅡ

Problems:

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?

 

解法一：按位统计0、1出现的次数。使用每个数的每位相加，取模运算，来获得结果。

class Solution {
public:

    int singleNumber(vector<int>& nums) {
        
        const int N=sizeof(int)*8;
        
        int count[N];
        fill_n(&count[0],N,0);
        
        for(int i=0;i<nums.size();i++)
            for(int j=0;j<N;j++)
                {
                    count[j]+=(nums[i]>>j)&1;
                    count[j]%=3;
                }
        //返回结果    
        int result=0;
        
        for(int j=0;j<N;j++)
            result+=(count[j]<<j);
        return result;
        
        
    }
};

The second solution is need update.