LeetCode:Beautiful Arrangement

526. Beautiful Arrangement

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.

思路：这个题目的思路比较直接，生成全排列的过程中及时判断是够满足题目要求，及时剪枝就可以了。

 1 bool isv(vector<int>&path, int a)
 2 {
 3     if (path.size() == 0)
 4         return true;
 5     return a % (path.size()+1) == 0 || (path.size()+1) % a == 0;
 6 }
 7 void getalla(vector<vector<int>>&r,vector<int>&arr, int now,vector<int>path)
 8 {
 9     if (now >= arr.size())
10     {
11         r.push_back(path);
12         return;
13     }
14     for (int i = now; i <arr.size(); i++)
15     {
16         swap(arr[now], arr[i]);
17         if (isv(path, arr[now]))
18         {
19             path.push_back(arr[now]);
20             getalla(r, arr, now + 1, path);
21             path.pop_back();
22         }
23         swap(arr[now], arr[i]);
24     }
25 }
26 int countArrangement(int N) 
27 {
28     vector<vector<int>>r;
29     vector<int>arr,path;
30     for (int i = 0; i < N; i++)
31         arr.push_back(i + 1);
32     getalla(r, arr, 0, path);
33     return r.size();
34 }

 

转载于:https://www.cnblogs.com/maclearning/p/6559002.html