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?

// Source : https://oj.leetcode.com/problems/single-number-ii/
// Author : Chao Zeng
// Date   : 2014-12-20

class Solution
{
public:
    int array[32];
    void countone(int n)
    {
        int k = 0;
        while (n)
        {
            array[k++] += n % 2;
            n /= 2;
        }
    }

    int singleNumber(int A[], int n)
    {
        for (int i = 0; i < 32; i++)
            array[i] = 0;
        int num = 0;
        //难点主要在于将负数转为正数然后求二进制数，然后确定所求数的符号
        for (int i = 0; i < n; i++){
            if (A[i] < 0){
                num ++;
                A[i] = -A[i];
            }
            countone (A[i]);
        }

        for (int i = 0; i < 32; i++)
            array[i] %= 3;

        int ans = 0;
        for (int i = 0; i < 32; i++){
            int bonus = 1;
            if (array[i])
            {
                for (int j = 0; j < i ; j++)
                    bonus = bonus * 2;
            }
            else
                bonus = 0;
            ans = bonus + ans;
        }
        if (num % 3 == 1)
            ans = -ans;
        return ans;
    }
};