[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!
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
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:
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:
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 }