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：用变量分别存储出现一次、两次、三次的number

出现一次的number: 抑或操作去除了出现两次的可能，与未出现三次的number作“与操作”去除了出现三次的可能

出现两次的number:与出现一次的number作“与操作”

出现三次的number: 出现一次的number“与操作”出现两次的number

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

        for (int i = ; 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;
    }
};