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?

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class Solution {
public:
    int singleNumber(int A[], int n) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        int ones = 0, twos = 0, not_three = 0;
        for (int i = 0; i < n; i++) {
            int x = A[i];
            twos |= ones & x;
            ones ^= x;
            not_three = ~(ones & twos);
            ones &= not_three;
            twos &= not_three;
        }
        
        return ones;
    }
};  

初始化 ： ones包含只出现1次bit信息， twos包含所有出现两次的bit信息，not_three包含所有出现少于三次的bit信息

对数组中每个整数(x)进行以下操作：

1. twos |= ones & x      --  获得当前整数x与现有ones的交集，也就是出现两次的bit信息,把这些得到的出现两次的bit信息 或 到 twos上

2. ones ^= x,   分三种情况 ： x出现1，2，3次， 除了出现两次外，ones将包含该数的bit信息

3. not_three = ~(ones & twos);  把当前非出现三次的bit信息赋值给 not_three

         4.  ones &= not_three;  从ones里边清除出现三次的bit信息
         5.  twos &= not_three;  从twos清除出现三次的bit信息