[Leetcode] 526. Beautiful Arrangement 解题报告

题目：

Suppose you have N integers from 1 to N. We define a beautiful arrangement as an array that is constructed by these N numbers successfully if one of the following is true for the ith position (1 <= i <= N) in this array:

1. The number at the ith position is divisible by i.
2. i is divisible by the number at the ith position.

Now given N, how many beautiful arrangements can you construct?

Example 1:

Input: 2
Output: 2
Explanation: 

The first beautiful arrangement is [1, 2]:

Number at the 1st position (i=1) is 1, and 1 is divisible by i (i=1).

Number at the 2nd position (i=2) is 2, and 2 is divisible by i (i=2).

The second beautiful arrangement is [2, 1]:

Number at the 1st position (i=1) is 2, and 2 is divisible by i (i=1).

Number at the 2nd position (i=2) is 1, and i (i=2) is divisible by 1.

Note:

1. N is a positive integer and will not exceed 15.

思路：

一道典型的DFS+BackTracking题目：我们用一个哈希表来记录当前还剩余的可以填充的数字（初始化为从1到N，共N个数字）。然后用DFS分别填充第1到N个位置。在填充第i个位置的时候，我们就遍历hash表中当前还剩余的可供选择的数字，如果该数字符合条件填充在该位置，那么就试着填充，然后用DFS搜索下一个位置，当然在DFS之后还需要回溯。最后返回符合条件的填充个数即可。

代码：

class Solution {
public:
    int countArrangement(int N) {
        unordered_set hash;
        int ret = 0;
        for (int i = 1; i <= N; ++i) {
            hash.insert(i);
        }
        DFS(N, 1, hash, ret);
        return ret;
    }
private:
    void DFS(int N, int n, unordered_set &hash, int &ret) {
        if (n > N) {
            ++ret;
            return;
        }
        vector left_nums(hash.begin(), hash.end());
        for (auto num : left_nums) {
            if (num % n == 0 || n % num == 0) {
                hash.erase(num);
                DFS(N, n + 1, hash, ret);
                hash.insert(num);
            }
        }
    }
};