为有牺牲多壮志,敢教日月换新天。

[Swift]LeetCode1024. 视频拼接 | Video Stitching

★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤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/10668090.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!!!

You are given a series of video clips from a sporting event that lasted T seconds.  These video clips can be overlapping with each other and have varied lengths.

Each video clip clips[i] is an interval: it starts at time clips[i][0] and ends at time clips[i][1].  We can cut these clips into segments freely: for example, a clip [0, 7] can be cut into segments [0, 1] + [1, 3] + [3, 7].

Return the minimum number of clips needed so that we can cut the clips into segments that cover the entire sporting event ([0, T]).  If the task is impossible, return -1.

Example 1:

Input: clips = [[0,2],[4,6],[8,10],[1,9],[1,5],[5,9]], T = 10
Output: 3
Explanation: 
We take the clips [0,2], [8,10], [1,9]; a total of 3 clips.
Then, we can reconstruct the sporting event as follows:
We cut [1,9] into segments [1,2] + [2,8] + [8,9].
Now we have segments [0,2] + [2,8] + [8,10] which cover the sporting event [0, 10].

Example 2:

Input: clips = [[0,1],[1,2]], T = 5
Output: -1
Explanation: 
We can't cover [0,5] with only [0,1] and [0,2].

Example 3:

Input: clips = [[0,1],[6,8],[0,2],[5,6],[0,4],[0,3],[6,7],[1,3],[4,7],[1,4],[2,5],[2,6],[3,4],[4,5],[5,7],[6,9]], T = 9
Output: 3
Explanation: 
We can take clips [0,4], [4,7], and [6,9].

Example 4:

Input: clips = [[0,4],[2,8]], T = 5
Output: 2
Explanation: 
Notice you can have extra video after the event ends.

Note:

  1. 1 <= clips.length <= 100
  2. 0 <= clips[i][0], clips[i][1] <= 100
  3. 0 <= T <= 100

你将会获得一系列视频片段,这些片段来自于一项持续时长为 T 秒的体育赛事。这些片段可能有所重叠,也可能长度不一。

视频片段 clips[i] 都用区间进行表示:开始于 clips[i][0] 并于 clips[i][1] 结束。我们甚至可以对这些片段自由地再剪辑,例如片段 [0, 7] 可以剪切成 [0, 1] + [1, 3] + [3, 7] 三部分。

我们需要将这些片段进行再剪辑,并将剪辑后的内容拼接成覆盖整个运动过程的片段([0, T])。返回所需片段的最小数目,如果无法完成该任务,则返回 -1 。

示例 1:

输入:clips = [[0,2],[4,6],[8,10],[1,9],[1,5],[5,9]], T = 10
输出:3
解释:
我们选中 [0,2], [8,10], [1,9] 这三个片段。
然后,按下面的方案重制比赛片段:
将 [1,9] 再剪辑为 [1,2] + [2,8] + [8,9] 。
现在我们手上有 [0,2] + [2,8] + [8,10],而这些涵盖了整场比赛 [0, 10]。

示例 2:

输入:clips = [[0,1],[1,2]], T = 5
输出:-1
解释:
我们无法只用 [0,1] 和 [0,2] 覆盖 [0,5] 的整个过程。

示例 3:

输入:clips = [[0,1],[6,8],[0,2],[5,6],[0,4],[0,3],[6,7],[1,3],[4,7],[1,4],[2,5],[2,6],[3,4],[4,5],[5,7],[6,9]], T = 9
输出:3
解释: 
我们选取片段 [0,4], [4,7] 和 [6,9] 。

示例 4:

输入:clips = [[0,4],[2,8]], T = 5
输出:2
解释:
注意,你可能录制超过比赛结束时间的视频。

提示:

  1. 1 <= clips.length <= 100
  2. 0 <= clips[i][0], clips[i][1] <= 100
  3. 0 <= T <= 100

8ms
 1 class Solution {
 2     func videoStitching(_ clips: [[Int]], _ T: Int) -> Int {
 3         let clips = clips.sorted { first, second in
 4                                   return first[0] < second[0]
 5                                  }
 6         var result = 0
 7         var curEnd = -1
 8         var nextEnd = 0   
 9         for clip in clips {
10             if nextEnd >= T || clip[0] > nextEnd {
11                 break
12             }
13             if curEnd < clip[0] {
14                 result += 1
15                 curEnd = nextEnd
16             }
17             nextEnd = max(clip[1], nextEnd)
18         }
19         return nextEnd >= T ? result : -1
20     }
21 }

Runtime: 16 ms

Memory Usage: 18.8 MB
 1 class Solution {
 2     func videoStitching(_ clips: [[Int]], _ T: Int) -> Int {
 3         var last:Int = 0
 4         var cnt:Int = 0
 5         while(true)
 6         {
 7             if last >= T {break}
 8             var found:Bool = false
 9             var mx:Int = -1
10             for i in 0..<clips.count
11             {
12                 if clips[i][0] <= last
13                 {
14                     mx = max(mx, clips[i][1])
15                 }
16             }
17             if mx > last
18             {
19                 last = mx
20                 cnt += 1
21                 found = true
22             }
23             if !found {break}
24         }
25         if last >= T {return cnt}
26         return -1
27     }
28 }

 

posted @ 2019-04-07 23:32  为敢技术  阅读(971)  评论(0编辑  收藏  举报