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. 我是这样想的，将N看成N个格子，把1~N个数一个个填进去，使得i%n===0或者n%i==0。在方法的参数中thisNum指当前要填入的数字，nums是当前形成的填入数组（index1对应1，N对应N），leftNums指还剩下多少个数字没有填。 public class Beautiful_Arrangement_526 { int count=0; public void getNextPosition(int thisNum,int N,int[] nums,int leftNums){ if(thisNum>N){ return; } for(int i=1;i<=N;i++){ if(nums[i]==0&&(i%thisNum==0||thisNum%i==0)){ nums[i]=thisNum; leftNums--; if(leftNums==0){ count++; } getNextPosition(thisNum+1, N, nums,leftNums); nums[i]=0; leftNums++; } } } public int countArrangement(int N) { if(N==1) return 1; for(int i=1;i<=N;i++){ int[] nums=new int[N+1]; nums[i]=1; getNextPosition(2,N,nums,N-1); } return count; } public static void main(String[] args) { // TODO Auto-generated method stub Beautiful_Arrangement_526 b=new Beautiful_Arrangement_526(); System.out.println(b.countArrangement(5)); } }大神的解法跟我差不多，但是很明显人家就用的DFS的思想，不像我是瞎编出来的。 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; } } }而且人家是从后往前遍历，显然后面数字大，所要试的格子数会减少，这样也能增快速度。The back tracking start from the back so that each search won't go too deep before it fails because smaller numbers have higher chance to be divisible among themselves. Also I don't use "visited" boolean array but use swap of an array of 1~N to avoid duplication.
