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?

这道题仍然用位运算来做，不过要比那道只用位异或来做的题要复杂得多。还是从数的二进制表示看起，假设数组大小为n，每个数由32位组成，分别把这n个数的第i位相加，然后%3，结果非0即1，结果为0时single one对应的位为0，结果为1时single one对应的位为1，最后返回32位的结果。

class Solution {
public:
    int singleNumber(int A[],int n){
        int bit=0,temp,i,j;
        if(n<0)
        {
            return -1;
        }
        for(i=0;i<32;i++)
        {
            temp=0;
            for(j=0;j<n;j++)
            {
                temp+=((A[j]>>i)&1);
            }
            bit|=((temp%3)<<i);
        }
        return bit;
    }
};