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：

只有一个元素出现一次，其他元素均出现三次，故计算所有元素的对应位的1的个数，如：计算所有元素的最低位的1的个数，再模3，得到的结果即为single number的对应位。

代码：

int singleNumber(int A[], int n) {
        int count=0;
int res=0;
int size=sizeof(int)*8;


for(int i=0; i<size; ++i)
{
for(int j=0; j<n; ++j)
{
if((A[j]&0x0001)!=0)
{
++count;
}
A[j]=A[j]>>1;
}
count=count%3;
int tmp=0;
if(count == 1)
{
tmp=1;
for(int k=0; k<i; ++k)
{
tmp=2*tmp;
}
}
res=res+tmp;
tmp=0;
count=0;
}
return res;
    }