LeetCode-1359. Count All Valid Pickup and Delivery Options [C++][Java]

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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

i = 1时， f(1) = 1;

...

i = n - 1时， f(i) = f(n-1);

i = n时，

收发相连，则一共分到的间隙有g=2(n-1)+1 = 2n-1,

手法分开，则将两个点分开放在g个位置上， 取

即f(i) = f(n) = f(n-1) (c+s) = (2n-1)nf(n-1);

class Solution {
public:
    int countOrders(int n) {
        if (n==0 || n==1) {return n;}
        vector dp(n+1);
        int m = 1e9+7;
        dp[0] = 0, dp[1] = 1 ;
        for(int i=2; i<=n; i++) {
            dp[i] = ((2*i-1) % m * dp[i-1]*i) % m;
        }
        return dp[n] ; 
    }
};

class Solution {
    public int countOrders(int n) {
        long[] dp = new long[n+1];
        dp[0] = 0; dp[1] = 1;
        long m = (long) 1e9 + 7;       
        for (int i=2; i<=n; i++) {
            dp[i] = (2*i-1) % m * dp[i-1] * i % m;
        }
        return (int)dp[n];
    }
}

也可以少用点辅助空间

class Solution {
    public int countOrders(int n) {
        if (n <= 0) {return 0;}
        long dp = 1;
        long m = (long) 1e9 + 7;       
        for (int i=2; i<=n; i++) {
            dp = (2*i-1) % m * dp * i % m; //注意%m的位置
        }
        return (int)dp;
    }
}