LeetCode (Single number 2)

题目要求：

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?

代码：

class Solution {
public:
    int singleNumber(int A[], int n) {
        int res = 0;
        for (size_t i =0; i < 32; ++i) {
            int d = 1 << i;
            int cnt = 0;
            for (size_t j = 0; j < n; ++j) {//统计第i位是1的个数， 如果出现统计结果不能被3整除，那么single number 这位上肯定是1
                if(A[j] & d)
                    cnt++;
            }
            if((cnt % 3) != 0)
                res |= d;

        }
        return res;
    }
};