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.
Approach #1: backtracking. [C++]
class Solution { public: int countArrangement(int N) { vector<int> path; for (int i = 1; i <= N; ++i) path.push_back(i); return helper(N, path); } private: int helper(int n, vector<int> path) { if (n <= 0) return 1; int ans = 0; for (int i = 0; i < n; ++i) { if (path[i] % n == 0 || n % path[i] == 0) { swap(path[i], path[n-1]); ans += helper(n-1, path); swap(path[i], path[n-1]); } } return ans; } };
原文地址:https://www.cnblogs.com/ruruozhenhao/p/10354265.html
时间: 2024-10-31 13:07:01