【leetcode】1024. Video Stitching

题目如下：

You are given a series of video clips from a sporting event that lasted Tseconds.  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

解题思路：由于选中的clip之间是允许有重叠的，因此尽量选择较长的clip，可以采用贪心算法。首先对clips按照start升序，end降序的方法排序。排序完成后的第一个元素是必选的，因为其start最小(并列)同时end最大。接下来再寻找和第一个元素有交集的区间，找出end最大的那个组成新的start,end区间。然后继续寻找end最大的区间直至clips遍历完成。最后判断start,end是否包含0,T区间即可。

代码如下：

class Solution(object):
    def videoStitching(self, clips, T):
        """
        :type clips: List[List[int]]
        :type T: int
        :rtype: int
        """
        def cmpf(v1,v2):
            if v1[0] != v2[0]:
                return v1[0] - v2[0]
            return v2[1] - v1[1]
        clips.sort(cmp=cmpf)
        #print clips

        start,end = clips[0][0],clips[0][1]
        clips.pop(0)
        res = 1
        flag = True
        while flag and len(clips) > 0 and end < T :
            maxEnd = end
            flag = False
            while len(clips) > 0:
                if end < clips[0][0]:
                    break
                flag = True
                maxEnd = max(maxEnd,clips.pop(0)[1])
            res += 1
            end = maxEnd
        return res if (start == 0 and end >= T) else -1