原题链接在这里：https://leetcode.com/problems/video-stitching/
题目：
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 <= 1000 <= clips[i][0], clips[i][1] <= 1000 <= T <= 100
题解：
Could sort the clips based on starting time.
For each step, go through all the clips having starting time <= current start time, get to the furtherest.
if furtherest doesn't change, start == end, that means there is a gap. return -1.
Otherwise, update start to end and increment res.
End as soon as current furtherest reaches T.
Time Complexity: O(nlogn). n = clips.length.
Space: O(1).
AC Java:
1 class Solution {
2 public int videoStitching(int[][] clips, int T) {
3 if(clips == null || clips.length == 0 || clips[0].length != 2 || T <= 0){
4 return -1;
5 }
6
7 Arrays.sort(clips, (a, b) -> a[0] - b[0]);
8
9 int start = 0;
10 int end = 0;
11 int res = 0;
12 int i = 0;
13 while(end < T){
14 while(i<clips.length && clips[i][0] <= start){
15 end = Math.max(end, clips[i][1]);
16 i++;
17 }
18
19 if(start == end){
20 return -1;
21 }
22
23 start = end;
24 res++;
25 }
26
27 return res;
28 }
29 }
Let dp[i] denotes minimum number of clips needed to cover up to i.
For each clip, if clip[0] <= i <= clip[1], then i could be covered using minimum clips covered up to clip[0] plus this clip.
Update dp[i].
Check if dp[T] is never updated. If it is never updated, then return -1.
Time Complexity: O(nT). n = clips.length.
Space: O(T).
AC Java:
1 class Solution {
2 public int videoStitching(int[][] clips, int T) {
3 int [] dp = new int[T+1];
4 dp[0] = 0;
5
6 for(int i = 1; i<=T; i++){
7 dp[i] = T+1;
8 for(int [] clip : clips){
9 if(clip[0]<=i && clip[1]>=i){
10 dp[i] = Math.min(dp[i], dp[clip[0]]+1);
11 }
12 }
13 }
14
15 return dp[T] == T+1 ? -1 : dp[T];
16 }
17 }