LeetCode137：Single Number II

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?

解题思路：

这题比Single Number稍难些，不能用异或解决，但排序和bitmap还是可以的，只是时间复杂度和空间复杂度要多些

这里我用另一种方式实现，根据所给数组中元素的规律，可利用每一bit位上1的个数进行解决，直接看代码吧

实现代码：

#include <iostream>

using namespace std;
/*
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?
*/
class Solution {
public:
    int singleNumber(int A[], int n) {
        int once = 0;
        for(int i = 0; i < 32; i++)
        {
            int one_num = 0;//bit为第i位1的个数 
            for(int j = 0; j < n; j++)
                if((A[j] >> i) & 1 == 1)
                    one_num++;
            //因为数组中只有一个数出现一次，其他数都出现三次，
            //所以除非要找数的当前bit位为1，否则one_num为3的倍数 
            if(one_num % 3)
                once += (1 << i);

        }
        return once;
        
    }
};

int main(void)
{
    int arr[] = {2,4,5,5,4,1,2,4,2,5};
    int len = sizeof(arr) / sizeof(arr[0]);
    Solution solution;
    int once = solution.singleNumber(arr, len);
    cout<<once<<endl;
    return 0;
}