526. Beautiful Arrangement (Medium)

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.

题意：第i个位置的数字可以被i整除，i可以被第i个位置的数字整除；
思路：
1.idx为当前数字的下标，nums为剩余待选数字；
2.初始令index = 1, nums = [1 .. N]；
3.遍历nums，记当前数字为n，除n以外的其余元素为nums[:i]+nums[i+1:]；
4.若n满足题设的整除条件，则将numSearch(index + 1, nums[:i]+nums[i+1:])累加至ans；

class Solution():
    def countArrangement(self, N):
        """
        :type N: int
        :rtype: int
        """
        
        temp = {}
        def numSearch(index, nums):
            if not nums:
                return 1
            key = index, tuple(nums)
            if key in temp:
                return temp[key]
            ans = 0
            for i, n in enumerate(nums):
                if n % index == 0 or index % n == 0:    
                    ans += numSearch(index + 1, nums[:i] + nums[i+1:])
            temp[key] = ans
            return ans
        return numSearch(1, range(1, N + 1))