LeetCode Beautiful Arrangement

Posted 2020-10-09 Dylan_Java_NYC

原题链接在这里：https://leetcode.com/problems/beautiful-arrangement/description/

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:

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?

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:

1. N is a positive integer and will not exceed 15.

题解：

典型的backtracking. 用visited记录用过的位置, 挨个position往后permute. 遇到不合规矩的直接返回, 看能不能走到N.

Time Complexity: exponential. 每次走到最后才遇到不和规矩的backtrack回来.

Space: O(N).

AC Java:

 1 class Solution {
 2     int res = 0;
 3     public int countArrangement(int N) {
 4         if(N <= 0){
 5             return 0;
 6         }
 7         
 8         boolean [] visited = new boolean[N+1];
 9         dfs(visited, 1, N);
10         return res;
11     }
12     
13     private void dfs(boolean [] visited, int pos, int N){
14         if(pos > N){
15             res++;
16             return;
17         }
18         
19         for(int i = 1; i<=N; i++){
20             if(!visited[i] && (i%pos==0 || pos%i==0)){
21                 visited[i] = true;
22                 dfs(visited, pos+1, N);
23                 visited[i] = false;
24             }
25         }
26     }
27 }