LeetCode每日一题(1024. Video Stitching)

You are given a series of video clips from a sporting event that lasted time seconds. These video clips can be overlapping with each other and have varying lengths.

Each video clip is described by an array clips where clips[i] = [starti, endi] indicates that the ith clip started at starti and ended at endi.

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, time]. If the task is impossible, return -1.

Example 1:

Input: clips = [[0,2],[4,6],[8,10],[1,9],[1,5],[5,9]], time = 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]], time = 5
Output: -1

Explanation: We cannot cover [0,5] with only [0,1] and [1,2].

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]], time = 9
Output: 3

Explanation: We can take clips [0,4], [4,7], and [6,9].

Constraints:

• 1 <= clips.length <= 100
• 0 <= starti <= endi <= 100
• 1 <= time <= 100

---

先按开始时间对片段进行排序， 如果开始时间相同则按结束时间倒序排序， 然后维护一个栈，保存需要进行拼接的片段， 遍历已排好序的片段， 考虑如下几种情况:

1. 当前片段结束时间小于等于栈顶的结束时间， 当前片段不入栈
2. 当前片段结束时间大于栈顶的结束时间，开始时间大于栈顶-1 的结束时间， 当前片段正常入栈
3. 当前片段结束时间大于栈顶的结束时间，开始时间小于等于栈顶-1 的结束时间, 弹栈， 直到不符合此条件的时候，当前片段入栈

其他细节方面要注意的是片段必须有开始时间为 0 并且有一定时间跨度的片段，否则不可能完成拼接， 遍历过程中， 如果栈顶的结束时间大于等于 time 了那就不需要继续遍历了

---

impl Solution {
    
    pub fn video_stitching(mut clips: Vec<Vec<i32>>, time: i32) -> i32 {
    
        clips.sort_by(|a, b| {
    
            if a[0] == b[0] {
    
                return b[1].cmp(&a[1]);
            }
            a[0].cmp(&b[0])
        });
        if clips.first().unwrap()[0] != 0 {
    
            return -1;
        }
        let mut stack: Vec<Vec<i32>> = Vec::new();
        'outer: for c in clips {
    
            while let Some(last) = stack.pop() {
    
                if last[1] >= time {
    
                    stack.push(last);
                    break 'outer;
                }
                if c[0] > last[1] || c[1] <= last[1] {
    
                    stack.push(last);
                    continue 'outer;
                }
                if stack.is_empty() {
    
                    stack.push(last);
                    stack.push(c);
                    continue 'outer;
                }
                if stack.last().unwrap()[1] >= c[0] {
    
                    continue;
                }
                stack.push(last);
                break;
            }
            stack.push(c);
            continue;
        }
        if stack.last().unwrap()[1] < time {
    
            return -1;
        }
        stack.len() as i32
    }
}