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   : 2014-12-20class 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;    }};