算法设计作业16

第十六周作业：

526. Beautiful Arrangement

解题思路：

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.

思路：在从1到N的一串数中，求解有多少种“漂亮的放法”，满足当前元素可以被当前位置数整除，或者当前位置数可以被当前元素整除。是一道排列组合的问题，但是如果用枚举法，复杂度将达到O(n!)，显然是不可行的。而考虑到可以采用递归来实现，从第一个位置开始判断，当第一个位置是可以的时候，我们从末尾向前逐一调用这个方法来判断。但是为了避免被重复判断，需要将判断过满足条件的放到末尾。当一个i值全部递归树完毕后，我们再将换位的值换回来以保证i+1后面的序列的正确性。

代码如下：

class Solution {
public:
    int countArrangement(int N) {
        vector<int> arrangement(N);
        iota(arrangement.begin(), arrangement.end(), 1);
        return countArrangementHelper(N, &arrangement);
    }

private:
    int countArrangementHelper(int n, vector<int> *arrangement) {
        if (n <= 0) {
            return 1;
        }
        int count = 0;
        for (int i = 0; i < n; ++i) {
            if ((*arrangement)[i] % n == 0 || n % (*arrangement)[i] == 0) {
                swap((*arrangement)[i], (*arrangement)[n - 1]);
                count += countArrangementHelper(n - 1, arrangement);
                swap((*arrangement)[i], (*arrangement)[n - 1]);
            }
        }
        return count;
    }
};