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 = 0;  
        int twice = 0;  
  
        for (int i = 0; i < n; i++) {  
            twice |= once & A[i]; //将出现两次1的位，在twice中置1
            once ^= A[i];  //将出现一次1的位，在once中置1，出现两次则置0
            int not_three = ~(once & twice);  //将出现三次1的位，在not_tree中置0
            once = not_three & once;  //将出现三次1的位，在once中置0
            twice = not_three & twice;  //将出现三次1的位，在twice中置0
        }  
        return once;  
    }
};