1024. Video Stitching

//使用java dfs

    public int videoStitching(int[][] clips, int T) {
        //bfs
        Queue<Integer> queue = new LinkedList<>();
        Set<Integer> visited = new HashSet<>();
        for (int i = 0;i < clips.length; i ++)
            if (clips[i][0] == 0){
                queue.offer(clips[i][1]);
                queue.offer(1);
                visited.add(clips[i][0]* 1000 + clips[i][1]);
            }
        if (queue.isEmpty())
            return -1;
        
        int end = -1, step = -1;
        while ( !queue.isEmpty() ){
            end = queue.poll();
            step = queue.poll();
            if (end >= T)
                return step;
            for (int i = 0; i< clips.length; i++){
                if (!visited.contains(clips[i][0]* 1000 + clips[i][1]))
                    if (end >= clips[i][0]){
                        queue.offer(clips[i][1]);
                        queue.offer(step + 1);
                        visited.add(clips[i][0]* 1000 + clips[i][1]);
                    }
            }
        }
        return -1;
    }

python dp

def videoStitching(self, clips: List[List[int]], T: int) -> int:
        #dp[i] 代表 到i结尾时间的最小个数
        clips.sort()
        n = len(clips)
        dp = [float('inf')]* n
        if clips[0][0] != 0:
            return -1
        for i in range(n):
            if clips[i][0] == 0:
                dp[i] = 1
            else:
                break
        for i in range(n):
            for j in range(i):
                if dp[j] != float('inf') and clips[j][1] >= clips[i][0] :
                    dp[i] = min(dp[i], dp[j] + 1)
        res = float('inf')
        for i in range(n):
            if clips[i][1] >= T:
                res = min(res, dp[i])
        return res if res != float('inf') else - 1

标签：1024,Stitching,int,res,queue,clips,Video,return,dp
来源： https://www.cnblogs.com/whyaza/p/10668671.html