★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤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!
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
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 <= clips.length <= 100
0 <= clips[i][0], clips[i][1] <= 100
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 <= clips.length <= 100
0 <= clips[i][0], clips[i][1] <= 100
0 <= T <= 100
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
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 }