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: 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:
- N is a positive integer and will not exceed 15.
1 public class Solution { 2 private int count = 0; 3 4 public int countArrangement(int N) { 5 helper(N, 1, new boolean[N]); 6 return count; 7 } 8 9 private void helper(int N, int level, boolean[] visited) { 10 if (level > N) { 11 count++; 12 return; 13 } 14 15 for (int i = 1; i <= N; i++) { 16 if ( !visited[i - 1] && (level % i == 0 || i % level == 0) ) { 17 visited[i - 1] = true; 18 helper(N, level + 1, visited); 19 visited[i - 1] = false; 20 } 21 } 22 } 23 }
时间: 2024-10-20 11:03:00