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  • [Swift]LeetCode1024. 视频拼接 | Video Stitching

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    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 }
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  • 原文地址:https://www.cnblogs.com/strengthen/p/10668090.html
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