题目地址：https://leetcode.com/problems/video-stitching/
题目描述
Youare 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；
题目大意
给了一堆区间，要求选取最少的区间，这些区间能覆盖[0, T]区间。
解题方法
贪心
这个题还是很容易想到贪心的。贪心策略是在保证和前面的区间能连接的情况下，选择结尾最靠后的区间，这样的覆盖是最广的。举例来说，如果现在的区间是[0,3]，假设下一个区间要从[1,5]和[2,4]中选择，我们应该选择结尾最靠后的也就是[1,5]。
对于每个相同开头结尾的区间，我只保留结尾最大的那个。这个策略是相同开头的时候，覆盖越大越好。这样，总的区间数目不会超过100个。
然后，我看了Note里的提示，发现时间段的取值范围只有100，所以使用了一个暴力的方法：遍历。即对于区间[a,b]暴力遍历a+1到b中的每个元素，找出是否以该元素开头的区间：如果有，那么找出所有区间最靠后的结尾，则该区间是下一个应该选择的区间。如果对a+1和b之间的所有元素都遍历了，然而找不到存在的区间，那么一定断线了，所以返回-1.
这种贪心策略之下，则选取的区间个数是最少的。
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
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