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?

Solution:

 看到大神的解法： http://www.mitbbs.com/article_t/JobHunting/32547143.html

整数的32个bits，出现次数mod 3后必余0, 1, 2，其中余1的就是答案

public int singleNumber(int[] A) {        int n1=0,n2=0;            for(int i=0;i<A.length;++i){            int n0=~(n1|n2);         // 若非余1也非余2，就是余0了            n2=(n1&A[i])|(n2&~A[i]); // 若「原本就余2且bit为0」或「原本余1且bit为1」，则该bit更新后余2            n1=(n1&~A[i])|(n0&A[i]); // 若「原本就余1且bit为0」或「原本余0且bit为1」，则该bit更新后余1        }        return n1;    }

[Leetcode] Single Number II