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：
1.利用二进制中的位运算，有n*3个相同的数相加，那么其二进制的每一位必定是3的倍数
2.剩下的那个单独的数，就是每一位除以3的余数，之后在把每一位转换成十进制相加，总和就是单独的那个数。
1 int singleNumber(int* nums, int numsSize) {
2 int ary[32] = {0};
3 int res = 0;
4 int i;
5 int j;
6 for(i = 0; i < 32; i++)
7 {
8 for(j = 0; j < numsSize; j++)
9 {
10 ary[i] += ((nums[j] >> i) & 1);
11 ary[i] = ary[i] % 3;
12 }
13 res |= (ary[i] << i);
14 }
15 return res;
16 }
转：http://blog.csdn.net/feliciafay/article/details/19004479