【leetcode】1359. Count All Valid Pickup and Delivery Options

题目如下：

Given n orders, each order consist in pickup and delivery services. 

Count all valid pickup/delivery possible sequences such that delivery(i) is always after of pickup(i). 

Since the answer may be too large, return it modulo 10^9 + 7.

Example 1:

Input: n = 1
Output: 1
Explanation: Unique order (P1, D1), Delivery 1 always is after of Pickup 1.

Example 2:

Input: n = 2
Output: 6
Explanation: All possible orders: 
(P1,P2,D1,D2), (P1,P2,D2,D1), (P1,D1,P2,D2), (P2,P1,D1,D2), (P2,P1,D2,D1) and (P2,D2,P1,D1).
This is an invalid order (P1,D2,P2,D1) because Pickup 2 is after of Delivery 2.

Example 3:

Input: n = 3
Output: 90

Constraints:

• 1 <= n <= 500

解题思路：记dp[i]为i个订单时可以组成的配送顺序的总数。当有i+1个订单的，对于最后一个配送的订单，一共有i+1种可能，因为任何一个订单都可以最后配送。而第i+1个订单的收取则有 (2*i-1)种可能。确定了最后一个收取以及配送的订单的后， 剩下的问题就转换成i个订单的收取和配送了，所以有 dp[i+1]  = dp[i] * i * (2*i-1)。

代码如下：

class Solution(object):
    def countOrders(self, n):
        """
        :type n: int
        :rtype: int
        """
        dp = [0] * (n+1)
        dp[1] = 1 
        for i in range(2,n+1):
            dp[i] = i * (2*i-1) * dp[i-1]
        return dp[-1] % (10 ** 9 + 7)