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?

class Solution {
public:
    int singleNumber(int A[], int n) {
#if 0
		int ans[32] = {0};
		int res = 0;
		for (int i = 0; i < 32; i++)
		{
			for (int j = 0; j < n; j++)
			{
				ans[i] += (A[j] >> i) & 1;
			}
			res |= (ans[i] % 3) << i;
		}
		return res;
#endif // 1
		//一般解法，可以统计每个数字的每位bit的1的个数，最后的mod 3得到结果
#if 1
		int one = 0, two = 0, three = 0;
		for (int i = 0; i < n; i++)
		{
			two |= one & A[i];
			one ^= A[i];
			three = one & two;
			one &= ~three;
			two &= ~three;
		}
		return one;
#endif // 1
		//one标记为当前数组成员为止，二进制中1出现1次，two标记二进制中1出现2次，
		//three标记二进制中1出现3次，那么，当three = one & two,如果，某一个位
		//one和two都有，那么，肯定是出现了三次，然后 ~three，则清除那些出现三次的数字
    }
};