137. 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?

class Solution {
public:
    int singleNumber(vector<int>& nums) {
        int w = sizeof(int)*8;  //一个整数的字长
        int *count = new int[w];
        memset(count,0,sizeof(int)*w); //数组一定要初始化，否则会出现意外的值
        int n = nums.size();
        for(int i=0;i<w;i++)
          {
              for(int j=0;j<n;j++)
              {
                 count[i] += (nums[j]>>i)&0x01;
              }
              count[i] = count[i]%3;
          }
        int result=0;
        for(int i=0;i<w;i++)
        {
            result += (count[i]<<i);
        }
        return result;
    }
};