LeetCode | Single Number II【转】

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：
一个比较好的方法是从位运算来考虑。每一位是0或1，由于除了一个数其他数都出现三次，我们可以检测出每一位出现三次的位运算结果再加上最后一位即可。

class Solution {
public:
    int singleNumber(int A[], int n) {
        int ones = , twos = , threes = ;
        for(int i = ; i < n; i++)
        {
            threes = twos & A[i]; //已经出现两次并且再次出现
            twos = twos | ones & A[i]; //曾经出现两次的或者曾经出现一次但是再次出现的
            ones = ones | A[i]; //出现一次的
 
            twos = twos & ~threes; //当某一位出现三次后，我们就从出现两次中消除该位
            ones = ones & ~threes; //当某一位出现三次后，我们就从出现一次中消除该位
        }
        return ones; //twos, threes最终都为0.ones是只出现一次的数
    }
};