You are given a series of video clips from a sporting event that lasted time seconds. These video clips can be overlapping with each other and have varying lengths.
Each video clip is described by an array clips where clips[i] = [starti, endi] indicates that the ith clip started at starti and ended at endi.
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, time]. If the task is impossible, return -1.
Example 1:
Input: clips = [[0,2],[4,6],[8,10],[1,9],[1,5],[5,9]], time = 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]], time = 5
Output: -1
Explanation: We cannot cover [0,5] with only [0,1] and [1,2].
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]], time = 9
Output: 3
Explanation: We can take clips [0,4], [4,7], and [6,9].
Constraints:
1 <= clips.length <= 100
0 <= starti <= endi <= 100
1 <= time <= 100
感觉更像是贪心算法
1. 先对clips排序，因为对一批相同起始点的clip，一定是选end最大的clip
2. 选好第一个片之后[s,e], 下一个片的起始点必须要在区间[s+1,e]中而且选择，而且还是选end最大的clip
3. 循环知道选到end>time的分片。
func videoStitching(clips [][]int, time int) int {
sort.Slice(clips, func(i, j int) bool {
return clips[i][0] < clips[j][0]
})
end, res, curMax := 0, 0, 0
for i := 0; i < len(clips); i++ {
if clips[i][0] <= end {
if clips[i][1] >= curMax {
curMax = clips[i][1]
}
}
if curMax >= time {
return res + 1
}
if i+1 < len(clips) && clips[i+1][0] > end {
res++
end = curMax
}
}
return -1
}