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.

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

Example 1:

  1. Input: n = 2
  2. Output: 2
  3. Explanation: 
  4. The first beautiful arrangement is [1, 2]:
  5. Number at the 1st position (i=1) is 1, and 1 is divisible by i (i=1).
  6. Number at the 2nd position (i=2) is 2, and 2 is divisible by i (i=2).
  7. The second beautiful arrangement is [2, 1]:
  8. Number at the 1st position (i=1) is 2, and 2 is divisible by i (i=1).
  9. Number at the 2nd position (i=2) is 1, and i (i=2) is divisible by 1.

Example 2:

  1. Input: n = 1
  2. Output: 1

Constraints:

  • 1 <= n <= 15

题意

对1-n这n个数进行排列，使得对于序列中第i个数字x满足i是x的倍数或者x是i的倍数。

思路

回溯法，对1-n每个位置挑选一个满足的数字放上去，判断最终得到的序列是否有效。

代码实现

Java

  1. class Solution {
  2.     public int countArrangement(int n) {
  3.         return dfs(1, n, new boolean[n + 1]);
  4.     }
  6.     private int dfs(int index, int n, boolean[] used) {
  7.         if (index == n + 1) {
  8.             return 1;
  9.         }
  11.         int count = 0;
  12.         for (int i = 1; i <= n; i++) {
  13.             if (!used[i] && (index % i == 0 || i % index == 0)) {
  14.                 used[i] = true;
  15.                 count += dfs(index + 1, n, used);
  16.                 used[i] = false;
  17.             }
  18.         }
  20.         return count;
  21.     }
  22. }