[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?

思路：int位32位，用32位的數組記錄每一bit的1的個數累加和，之後mod3，如果mod3！＝0，那麼。。。

注意：1. 可能有負數，所以多用一位記錄符號

2. 可能數組中元素個數很大，爲防止溢出用short型，

class Solution {
public:
    int singleNumber(int A[], int n) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        short ones[33]={0};
        
        for(int i=0;i<n;i++)
        {
            unsigned int number=A[i];
            int j=0;
            if(number < 0)
            {
                ones[32]+=1;
                number=0-number;
            }
            
            while(number > 0)
            {
                ones[j]+=number%2;
                number=number/2;
                j++;
            }
        }
        
        int result=0;
        for(int i=31;i>=0;i--)
        {
            ones[i]=ones[i]%3;
            if(ones[i]!=0)
                result=result*2+1;
            else
                result*=2;
        }
        ones[32]=ones[32]%3;
        if(ones[32]==1)//sign bit
            result=0-result;
            
        return result;
    }
};