[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位的数组记录每一bit的1的个数累加和，之后mod3，如果mod3！＝0，那么。。。

注意：1. 可能有负数，所以多用一位记录符号

2. 可能数组中元素个数很大，为防止溢出用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;    }};