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问题：

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?

利用运算，所有位加，能除尽3说明多余的一个数该位为0，否则为1.

参考：

代码：http://blog.csdn.net/kenden23/article/details/13625297

http://www.cnblogs.com/daijinqiao/p/3352893.html

class Solution {public:    int singleNumber(int A[], int n) {        int sum[32] = {0};		int i,j,temp,result;		result = 0;        for(i = 0; i < n; i++)        {            for(j = 0; j<32; j++)			{				sum[j] += (A[i]>>j)&0x01;			}        }		for(j = 0; j < 32; ++j)		{			if(sum[j]%3 == 0)				temp = 0;			else				temp = 1;			result += (temp<