leetcode第14题（single-number-ii）

题目：

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：

如果能定义三进制底下的某种位操作，也可以达到相同的效果，Single Number II中想要记录每个bit出现的次数，一个数搞不定就加两个数，用ones来记录只出现过一次的bits，用twos来记录只出现过两次的bits，ones&twos实际上就记录了出现过三次的bits，这时候我们来模拟进行出现3次就抵消为0的操作，抹去ones和twos中都为1的bits

代码：

public class Solution
{
    public int singleNumber(int []nums)
    {
        int ones = 0;
        int twos = 0;
        int threes = 0;
        for(int num : nums)
        {
            twos = twos | (num&ones);
            ones = ones ^ num;
            threes = ones & twos;
            ones = ones & (~threes);
            twos = twos & (~threes);
        }
        int res = ones | twos;
        return res;
    }
}