LeetCode #526 - Beautiful Arrangement

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本文探讨了一个数学问题，即如何计算1到N的整数能够构成的美丽排列的数量。美丽排列是指在排列中，每个位置i上的数能被i整除，或者i能被该位置上的数整除。通过递归算法，我们展示了如何解决这个问题，并给出了具体的实现代码。

题目描述：

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:

N is a positive integer and will not exceed 15.

对于1~N个数字进行排序，如果第i位数字被i整除，或者i被第i位数字整除就说明是一个beautiful arrangment。那么只需要利用递归，尝试第i位上的所有可能。

class Solution {
public:
    int countArrangement(int N) {
        unordered_set<int> visited;
        int result=0;
        helper(N,visited,1,result);
        return result;
    }
    
    void helper(int& N, unordered_set<int>& visited, int pos, int& result)
    {
        if(pos==N+1)
        {
            result++;
            return;
        }
        for(int i=1;i<=N;i++)
        {
            if(visited.count(i)) continue;
            if(pos%i==0||i%pos==0)
            {
                visited.insert(i);
                helper(N,visited,pos+1,result);
                visited.erase(i);
            }
        }
    }
};