【Single Num II】cpp

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?

代码：

class Solution {
public:
    int singleNumber(vector<int>& nums) {
            int result = 0;
            const int W = sizeof(int)*8;
            int *count = new int[W]();
            for (size_t i = 0; i < nums.size(); ++i)
            {
                for (size_t j = 0; j < W; ++j)
                {
                    count[j] += (nums[i] >> j) & 1;
                }
            }
            for (size_t i = 0; i < W; ++i)
            {
                if ( count[i]%3!=0 )
                {
                    result += ( 1 << i);
                }
            }
            delete []count;
            return result;
    }
};

Tips:

1. 算法原理：由于都是int型的，只需要记录每个bit上1出现的次数；如果1出现的次数不是3的倍数，则将该位置置为1，并赋值到result中。

这道题主要是熟悉下位运算的各种操作，有时巧用位运算，可以大大提升代码效率。