[LeetCode]Single Number II

题目描述：

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?

解题方案：

该题目的一般方法是开辟一个空间为32的数组bitsum[32]，然后遍历题目中给定的数组，将数组中每个整数的每一bit位给分出来，加到bitsum的对应位置。最后再遍历bitsum数组，同时求bitsum[i]%3，将结果做移位操作！下面是该题的代码：

class Solution {
public:
    int singleNumber(int A[], int n) {
        int bitsum[32] = {0};
        int result = 0;
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < 32; ++j) {
                bitsum[j] += A[i]>>j & 1;
            }
        }
        for (int i = 0; i < 32; ++i) {
            result |=bitsum[i] % 3 << i;
        }
        return result;
    }
};