为有牺牲多壮志,敢教日月换新天。

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

★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤微信公众号:山青咏芝(shanqingyongzhi)
➤博客园地址:山青咏芝(https://www.cnblogs.com/strengthen/
➤GitHub地址:https://github.com/strengthen/LeetCode
➤原文地址:https://www.cnblogs.com/strengthen/p/9750997.html 
➤如果链接不是山青咏芝的博客园地址,则可能是爬取作者的文章。
➤原文已修改更新!强烈建议点击原文地址阅读!支持作者!支持原创!
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★

热烈欢迎,请直接点击!!!

进入博主App Store主页,下载使用各个作品!!!

注:博主将坚持每月上线一个新app!!!

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 首不同的歌,在旅途中,你的旅伴想要听 L 首歌(不一定不同,即,允许歌曲重复)。请你为她按如下规则创建一个播放列表:

  • 每首歌至少播放一次。
  • 一首歌只有在其他 K 首歌播放完之后才能再次播放。

返回可以满足要求的播放列表的数量。由于答案可能非常大,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 }
 
posted @ 2018-10-07 19:13  为敢技术  阅读(685)  评论(0编辑  收藏  举报