[LeetCode] - 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?
一直感觉这道题的思路很是巧妙。3的情况也可以推广到n，这样的话就需要n个变量用来存储状态了。还有值得注意的是，twos，ones，threes这个顺序一定不能搞乱，先更新twos，再更新ones，要么twos就一直是0了。

public class Solution {
public int singleNumber(int[] A) {
int ones=0, twos=0, threes=0;

for(int i=0; i<=A.length-1; i++) {
twos |= ones & A[i];        // get 1s that have appeared two times
ones ^= A[i];               // get 1s that have appeared one time
threes = ~(twos & ones);    // get 1s that have appeared three times
ones &= threes;             // mod 3
twos &= threes;             // mod 3
}

return ones;
}
}