137. Single Number II (Bit)

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：

Step1：从Single Number我们知道，通过易或操作可以用一个int存储出现过单数次的

Step2：我们可以通过与单数出现的做与，来获得出现过双数次的数

Step3：想办法把Step1 int中出现三次的去掉。

class Solution {
public:
    int singleNumber(int A[], int n) {
        int once = 0;
        int twice = 0;  

        for (int i = 0; i < n; i++) {
            twice |= once & A[i]; //the num appeared 2 times
            once ^= A[i];  //the num appeared 1 times
            int not_three = ~(once & twice);  //the num not appeared 3 times
            once = not_three & once;  //remove num appeared 3 times from once
            twice = not_three & twice;  //remove num appeared 3 times from twice
        }
        return once;
    }
};