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?

思路是用一个32位的数result的每一位记录所有数在这一位出现的次数，如果是3的倍数则消去，最后留下来的数则是所要的结果。

注意，result需要定义为unsigned int类型，否则会出现负数的情况。

    int singleNumber(int A[], int n) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        unsigned int result = 0, MASK;
        int i, j, tmp;
        for(i = 0; i < 32; i++){
            MASK = 1<<i;
            tmp = 0;
            for(j = 0; j < n; j++){
                tmp += ((A[j]&MASK) > 0 ? 1:0);
            }
            result |= ((tmp%3)>0 ? MASK:0);
        }
        return result;
    }