LeetCode:526. Beautiful Arrangement

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 i_th position (1 <= i <= N) in this array:

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

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

Example 1:

Input: 2Output: 2Explanation: 
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.

题意：将1-n个数放置到1-n的位置上，对于每个位置上的数必须满足a[i]可以被i整除或者被整除。
这是一个深度搜索的过程，和走迷宫问题类似，必须记录一个used数组，用来标记当前已经用过的数。主要难点是理解题意，并且转化成为一个便于理解的深度搜索过程。
代码如下：

class Solution {    int count=0;    public int countArrangement(int n) {        dfs(n,n,new boolean[n+1]);        return count;    }    public void dfs(int n,int k,boolean[] used){        if(k==1)        {            count++;            return ;        }            for(int i=1;i<=n;i++){            if(!used[i]&&(i%k==0||k%i==0))            {                used[i]=true;                dfs(n,k-1,used);                used[i]=false;            }        }    }}