[Swift]LeetCode920. 播放列表的数量 | Number of Music Playlists

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Your music player contains N different songs and she wants to listen to L (not necessarily different) songs during your trip.  You create a playlist so that:

• Every song is played at least once
• A song can only be played again only if K other songs have been played

Return the number of possible playlists.  As the answer can be very large, return it modulo 10^9 + 7.

 Example 1:

Input: N = 3, L = 3, K = 1 Output: 6 Explanation: There are 6 possible playlists. [1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1], [3, 1, 2], [3, 2, 1]. 

Example 2:

Input: N = 2, L = 3, K = 0 Output: 6 Explanation: There are 6 possible playlists. [1, 1, 2], [1, 2, 1], [2, 1, 1], [2, 2, 1], [2, 1, 2], [1, 2, 2] 

Example 3:

Input: N = 2, L = 3, K = 1 Output: 2 Explanation: There are 2 possible playlists. [1, 2, 1], [2, 1, 2] 

 Note:

1. 0 <= K < N <= L <= 100

---

你的音乐播放器里有 N 首不一样的歌，在旅途中，你的旅伴想要听 L 首歌（不必定不一样，即，容许歌曲重复）。请你为她按以下规则建立一个播放列表：

• 每首歌至少播放一次。
• 一首歌只有在其余 K 首歌播放完以后才能再次播放。

返回能够知足要求的播放列表的数量。因为答案可能很是大，请返回它模 10^9 + 7 的结果。

 示例 1：

输入：N = 3, L = 3, K = 1
输出：6
解释：有 6 种可能的播放列表。[1, 2, 3]，[1, 3, 2]，[2, 1, 3]，[2, 3, 1]，[3, 1, 2]，[3, 2, 1].

示例 2：

输入：N = 2, L = 3, K = 0
输出：6
解释：有 6 种可能的播放列表。[1, 1, 2]，[1, 2, 1]，[2, 1, 1]，[2, 2, 1]，[2, 1, 2]，[1, 2, 2]

示例 3：

输入：N = 2, L = 3, K = 1
输出：2
解释：有 2 种可能的播放列表。[1, 2, 1]，[2, 1, 2]

 提示：

1. 0 <= K < N <= L <= 100

---

56 ms

 1 class Solution {
 2     func numMusicPlaylists(_ N: Int, _ L: Int, _ K: Int) -> Int {
 3         var mod:Int = 1000000007
 4         var dp = Array(repeating:0, count: N+1)
 5         dp[0] = 1
 6         for i in 0..<L
 7         {
 8             var ndp = Array(repeating:0, count: N+1)
 9             for j in 0..<N
10             {
11                 ndp[j + 1] += dp[j] * (N-j)
12                 ndp[j + 1] %= mod
13             }
14             for j in 0...N
15             {
16                 ndp[j] += dp[j] * max(j - K, 0)
17                 ndp[j] %= mod
18             }
19             dp = ndp
20         }
21         return Int(dp[N])
22     }
23 }