526. Beautiful Arrangement

1.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:

The number at the ith position is divisible by i.
i is divisible by the number at the ith 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:
N is a positive integer and will not exceed 15.

2.分析
  这道题，思路一般很容易想到就是使用回溯法，而我从常规思路来思考，结果想了好久，以下解法代码简介，思路比较巧妙。
3.解题

class Solution {int count = 0;public int countArrangement(int N) {    if(N==0){        return count;    }    helper(N,1,new int[N+1]);    return count;}public void helper(int n, int pos, int[]number){    if(pos>n){        count++;        return;    }    for(int i=1;i<=n;i++){        if(number[i]==0&&(i%pos==0||pos%i==0)){            number[i] = 1;            helper(n,pos+1,number);            number[i] = 0;        }    }}}

4.总结
  如果是一个正常的思路的话，会是一个全排列，然后进行1~n的%，这样时间复杂度比较高，而如果使用回溯法代码简洁，思路也比较清晰，可能有点烧脑。