# 1024. Video Stitching 视频拼接

@TOC

## # 题目描述

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`

## # 解题方法

### # 贪心

Python代码如下：

``````class Solution(object):
def videoStitching(self, clips, T):
"""
:type clips: List[List[int]]
:type T: int
:rtype: int
"""
count = collections.defaultdict(list)
for cl in clips:
if cl[0] in count:
if cl[1] - cl[0] > count[cl[0]][1] - count[cl[0]][0]:
count[cl[0]].pop()
count[cl[0]] = cl
else:
count[cl[0]] = cl
if 0 not in count: return -1
prev = 0
cur = count[0][1]
next = cur
res = 1
while cur < T:
hasFind = False
for c in range(cur, prev, -1):
if c in count:
if count[c][1] > next:
next = count[c][1]
prev = c
hasFind = True
if not hasFind:
return -1
cur = next
res += 1
return res
``````

## # 日期

2019 年 4 月 7 日 —— 周赛bug了3次。。