Single Number II

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。

int singleNumber(int* nums, int numsSize) {
    int ary[32] = {0};
    int res = 0;
    int i;
    int j;
    for(i = 0; i < 32; i++)
    {
        for(j = 0; j < numsSize; j++)
        {
            ary[i] += ((nums[j] >> i) & 1);
            ary[i] = ary[i] % 3;
        }
        res |= (ary[i] << i);
    }
    return res;
}

转：http://blog.csdn.net/feliciafay/article/details/19004479