leetcode 137

137. 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?

数组仅有一个数出现一次，其他的出现3次。找出那个出现一次的数。依然使用位运算来求解。

统计每一位上1出现的次数，1出现次数不为3的倍数的位所组成的数即为要找的数。

代码实现：

class Solution {
public:
    int singleNumber(vector<int>& nums) {
        int n = nums.size();
        int ones = 0;
        int twos = 0;
        int xthrees = 0;
        for(int i = 0; i < n; ++i)
        {
            twos |= (ones & nums[i]);
            ones ^= nums[i];
            xthrees = ~(ones & twos);
            ones &= xthrees;
            twos &= xthrees;
        }
        return ones;
    }
};