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 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.

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

看清楚dfs里面谁在变谁不变：变得是pos，不变的是for循环的界限

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

换一种写法居然能变快