526. Beautiful Arrangement -Medium

Question

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?

假设你有1…N的整数，我们定义“优美的安排”：对于第i个（1 <= i <= N）数字，它满足以下条件之一即可

1. 第i个位置上的元素可以被i整除

2. i可以被第i个位置上的元素整除

现在给出N，请问有多少个“优美的安排”

Example

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.

Solution

• 回溯解。对于每个位置，我们只需判断遍历数组的元素，只要有元素满足整除的条件就加入组合，然后回溯寻找下一位置即可。因为每次都需要遍历数组的元素，而同一元素不能重复使用，所以不要忘了标记已使用的元素。

  public class Solution {    public int countArrangement(int N) {        int[] nums = new int[N];        for(int i = 0;i < N; i++){            nums[i] = i + 1;        }        return backtracking(nums, 0, new boolean[N], 0);;    }    public int backtracking(int[] nums, int start, boolean[] isUsed, int temp){        int count = 0;        if(temp == nums.length){            count += 1;            return count;        }        // 在下标索引为start的位置分别搜索数组中的所有元素        for (int j = 0; j < nums.length; j++) {            if (isUsed[j]) continue;            // 只要能够整除就加入            if (nums[j] % (start + 1) == 0 || (start + 1) % nums[j] == 0) {                isUsed[j] = true;                count += backtracking(nums, start + 1, isUsed, temp + 1);                isUsed[j] = false;            }        }        return count;    }}