Description
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.
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 。
题目链接：https://leetcode.com/problems/video-stitching/
Difficulty: medium
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 <= clips.length <= 100
- 0 <= clips[i][0], clips[i][1] <= 100
- 0 <= T <= 100
分析
- 找到一个视频之后，下一个视频应该跟上一个视频可以连上，同时保证延续的时间最长；
- 将clips排序，第一个视频应从零开始，最后一个视频的结束时间应该大于T，否则返回-1；
- index记录遍历的位置，end记录已选择视频的结束时间，s记录选择的视频数，若index位置的clip的开始时间小于end，继续后移，知道找到满足条件的最完结束的视频，修改end；
- 若index没有超出范围，修改start，s；
- 反之返回-1；
- 结束返回s。
参考代码
class Solution(object):
def videoStitching(self, clips, T):
clips.sort()
start=0
end=0
if(clips[0][0] > 0 or clips[-1][1] < T):
return -1
s=0
index=-1
while(end < T):
while(index+1 < len(clips) and clips[index+1][0] <= start):
index+=1
end=max(end,clips[index][1])
if(index < len(clips)):
start=end
s+=1
else:
return -1
return s