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;
}