👬 获取你好友已观看的视频

📝 题目描述

有n个人，每个人都有一个0到n-1的唯一id。

给你数组watchedVideos和friends，其中watchedVideos[i]和friends[i]分别表示id = i的人观看过的视频列表和他的好友列表。

Level 1的视频包含所有你好友观看过的视频，level 2的视频包含所有你好友的好友观看过的视频，以此类推。一般的，Level 为k的视频包含所有从你出发，最短距离为k的好友观看过的视频。

给定你的id和一个level值，请你找出所有指定level的视频，并将它们按观看频率升序返回。如果有频率相同的视频，请将它们按字母顺序从小到大排列。

📋 代码模板

javatypescript

java

class Solution {
    public List<String> watchedVideosByFriends(List<List<String>> watchedVideos, int[][] friends, int id, int level) {

    }
}

typescript

function watchedVideosByFriends(watchedVideos: string[][], friends: number[][], id: number, level: number): string[] {
    
}

💡 提示

$n == w a t c h e d V i d e o s . l e n g t h == f r i e n d s . l e n g t h$
$2 <= n <= 100$
$1 <= w a t c h e d V i d e o s [i] . l e n g t h <= 100$
$1 <= w a t c h e d V i d e o s [i] [j] . l e n g t h <= 8$
$0 <= f r i e n d s [i] . l e n g t h < n$
$0 <= f r i e n d s [i] [j] < n$
$0 <= i d < n$
$1 <= l e v e l < n$
如果 $f r i e n d s [i]$ 包含 $j$ ，那么 $f r i e n d s [j]$ 包含 $i$

🚀 示例

🖊️ 题解

可惜没有如果

javatypescript

java

class Solution {
    public List<String> watchedVideosByFriends(List<List<String>> watchedVideos, int[][] friends,
                                               int id, int level) {
        int n = friends.length;
        boolean[] visited = new boolean[n];
        Queue<Integer> queue = new LinkedList<>();
        queue.offer(id);
        visited[id] = true;
        while (!queue.isEmpty() && level > 0) {
            int size = queue.size();
            while (size > 0) {
                int i = queue.poll();
                for (int friend : friends[i]) {
                    if (!visited[friend]) {
                        queue.offer(friend);
                        visited[friend] = true;
                    }
                }
                size--;
            }
            level--;
        }
        List<String> videos = new ArrayList<>();
        Map<String, Integer> frequency = new HashMap<>();
        while (!queue.isEmpty()) {
            for (String video : watchedVideos.get(queue.poll())) {
                boolean isContains = frequency.containsKey(video);
                frequency.put(video, (isContains ? frequency.get(video) : 0) + 1);
                if (!isContains) {
                    videos.add(video);
                }
            }
        }
        videos.sort(Comparator.comparing((String video) -> frequency.get(video))
                .thenComparing(Comparator.naturalOrder()));
        return videos;
    }
}

typescript

function watchedVideosByFriends(watchedVideos: string[][], friends: number[][]
    , id: number, level: number): string[] {
    const n = friends.length;
    const visited: boolean[] = Array(n).fill(false);
    const queue: number[] = [];
    queue.push(id);
    visited[id] = true;
    while (queue.length > 0 && level > 0) {
        let size = queue.length;
        while (size > 0) {
            const i: number = queue.shift()!;
            for (const friend of friends[i]) {
                if (!visited[friend]) {
                    queue.push(friend);
                    visited[friend] = true;
                }
            }
            size--;
        }
        level--;
    }
    const frequency = new Map<string, number>();
    const videos: string[] = [];
    while (queue.length > 0) {
        for (const video of watchedVideos[queue.shift()!]) {
            const isContains = frequency.has(video);
            frequency.set(video, (isContains ? frequency.get(video)! : 0) + 1);
            if (!isContains) {
                videos.push(video);
            }
        }
    }
    videos.sort((v1, v2) => {
        const cnt1 = frequency.get(v1)!, cnt2 = frequency.get(v2)!
        if (cnt1 != cnt2) {
            return cnt1 - cnt2;
        }
        if (v1 != v2) {
            return v1 > v2 ? 1 : -1;
        }
        return 0;
    })
    return videos;
}

👬 获取你好友已观看的视频 ​

📝 题目描述 ​

📋 代码模板 ​

💡 提示 ​

🚀 示例 ​

🖊️ 题解 ​

💭 复杂度分析 ​