标签：str positive pos hat amp and osi can app
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:
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:
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;
}
};
标签：str positive pos hat amp and osi can app
原文地址：https://www.cnblogs.com/ruruozhenhao/p/10354265.html
