LeetCode_Single Number II

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?

分析： 统计每一位1出现的个数，最后摸3求余数，1则置最终位为1

class Solution {
public:
    int singleNumber(int A[], int n) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        int res = ;
        int x, sum;
 
        for(int i = ; i < ; ++i){
 
            sum = ;
            x = (<<i);
 
            for(int j = ; j< n; ++j)
                if(A[j]&x)
                    ++sum;
 
            sum = sum%;
 
            if(sum == )
                res |= x;
        }
 
        return res;
    }
};