[LeetCode] 526. Beautiful Arrangement

Suppose you have n integers labeled 1 through n. A permutation of those n integers perm (1-indexed) is considered a beautiful arrangement if for every i (1 <= i <= n), either of the following is true:

perm[i] is divisible by i.
i is divisible by perm[i].

Given an integer n, return the number of the beautiful arrangements that you can construct.

Example 1:

Input: n = 2
Output: 2
Explanation: 
The first beautiful arrangement is [1,2]:
    - perm[1] = 1 is divisible by i = 1
    - perm[2] = 2 is divisible by i = 2
The second beautiful arrangement is [2,1]:
    - perm[1] = 2 is divisible by i = 1
    - i = 2 is divisible by perm[2] = 1

Example 2:

Input: n = 1
Output: 1

Constraints:

1 <= n <= 15

优美的排列。

假设有从 1 到 N 的 N 个整数，如果从这 N 个数字中成功构造出一个数组，使得数组的第 i 位 (1 <= i <= N) 满足如下两个条件中的一个，我们就称这个数组为一个优美的排列。条件：

第 i 位的数字能被 i 整除
i 能被第 i 位上的数字整除
现在给定一个整数 N，请问可以构造多少个优美的排列？

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/beautiful-arrangement
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

思路是backtracking。既然这个所谓的优美的排列是需要试探所有的数字的，所以用backtracking是比较合适的。

时间O(k) - 一共有N个数字但是只有K种排列

空间O(n) - 回溯用到的栈空间

Java实现

 1 class Solution {
 2     int count = 0;
 3 
 4     public int countArrangement(int n) {
 5         boolean[] visited = new boolean[n + 1];
 6         helper(n, 1, visited);
 7         return count;
 8     }
 9 
10     private void helper(int n, int start, boolean[] visited) {
11         if (start > n) {
12             count++;
13             return;
14         }
15         for (int i = 1; i <= n; i++) {
16             if (!visited[i] && (start % i == 0 || i % start == 0)) {
17                 visited[i] = true;
18                 helper(n, start + 1, visited);
19                 visited[i] = false;
20             }
21         }
22     }
23 }

LeetCode 题目总结