Design Twitter

https://leetcode.com/problems/design-twitter/description/

Design a simplified version of Twitter where users can post tweets, follow/unfollow another user and is able to see the 10 most recent tweets in the user's news feed. Your design should support the following methods:

postTweet(userId, tweetId): Compose a new tweet.

getNewsFeed(userId): Retrieve the 10 most recent tweet ids in the user's news feed. Each item in the news feed must be posted by users who the user followed or by the user herself. Tweets must be ordered from most recent to least recent.

follow(followerId, followeeId): Follower follows a followee.

unfollow(followerId, followeeId): Follower unfollows a followee.

Thoughts

这道题的难点在于如何快速的获取一个人的news. 一个user的news不光包含本人的所有post, 还包括他所follow所有人的post。返回还必须按照时间排序。刚开始想的是每个user都维持一个news list，但这样当任何一个followee发出新post，就要往news加入这个这个post；当unfollow时要把followee的所有post全部从list中删除，耗时O(mn). 当新follow一个人时，要把新followee所有post加入这个list中，如果用list则同样耗时O(mn). 因此我们可以用pq来维持news, 这样每次加入新元素只需要logn时间，总耗时mlgn. 不过总体来说这么写不光时间慢，空间大，还容易出Bug, 因为post, follow和unfollow时都要更新news。

看到大神写的lazy动态get news，只有当要get news时才计算。由于要按时间排序，一种是所有元素放一起调用排序算法，但这肯定不合实际，因为所有follow tweets会很多。还有就是用pq了，用pq把每个followee（以及user本身）最新的post放到pq里排序，当一个最新的tweet抽出来，对应用户的下一个post要放到pq里。因此每个用户的posts可以用一个linked list存起来。当t表示follower数目时，这样get news所花时间就是O(tlgt)

Code

class Twitter {
    private int time = 0;
    private class Tweet {
        private int uid, tid, time;
        public Tweet next;
        Tweet(int uid, int tid, int time) {
            this.uid = uid;
            this.tid = tid;
            this.time = time;
        }

        public int getUid() {
            return uid;
        }
        public int getTid() {
            return tid;
        }
        public int getTime() {
            return time;
        }

        @Override
        public String toString() {
            return "[tweet: " + uid + "," + tid + "]";
        }
    }

    private class User {
        private int uid;
        private Tweet head;
        Set<User> followees;
        public User(int uid) {
            this.uid = uid;
            followees = new HashSet<>();
        }

        public void post(Tweet tw) {
            tw.next = head;
            head = tw;
        }

        public void follow(User user) {
            addFollowee(user);
        }

        public Tweet getHead() {
            return head;
        }

        public boolean hasFollowees(User user) {
            return followees.contains(user);
        }

        public Set<User> getFollowees() {
            return followees;
        }

        public void addFollowee(User user) {
            followees.add(user);
        }

        public void rmFollowee(User user) {
            followees.remove(user);
        }

        @Override
        public String toString() {
            return "[user:" + uid + "]";
        }

    }

    class TweetComparator implements Comparator<Tweet> {
        @Override
        public int compare(Tweet a, Tweet b) {
            return b.getTime() - a.getTime();
        }
    }

    Map<Integer, User> users;

    /** Initialize your data structure here. */
    public Twitter() {
        users = new HashMap<>();
    }

    /** Compose a new tweet. */
    public void postTweet(int userId, int tweetId) {
        if (!users.containsKey(userId)) {
            users.put(userId, new User(userId));
        }
        users.get(userId).post(new Tweet(userId, tweetId, time));
        time++;
    }

    /** Retrieve the 10 most recent tweet ids in the user's news feed. Each item in the news feed must be posted by users who the user followed or by the user herself. Tweets must be ordered from most recent to least recent. */
    public List<Integer> getNewsFeed(int userId) {
        List<Integer> res = new ArrayList<>();
        if (!users.containsKey(userId)) {
            return res;
        }

        PriorityQueue<Tweet> pq = new PriorityQueue(new TweetComparator());
        User user = users.get(userId);
        if (user.getHead() != null) pq.add(user.getHead());
        for (User followee : user.getFollowees()) {
            if (followee.getHead() != null) pq.add(followee.getHead());
        }

        for (int i = 0; i < 10 && !pq.isEmpty(); i++) {
            Tweet tw = pq.poll();
            res.add(tw.getTid());
            if (tw.next != null) pq.offer(tw.next);
        }

        return res;
    }

    /** Follower follows a followee. If the operation is invalid, it should be a no-op. */
    public void follow(int followerId, int followeeId) {
        if (followerId == followeeId) {
            return;
        }
        if (!users.containsKey(followerId)) {
            users.put(followerId, new User(followerId));
        }
        if (!users.containsKey(followeeId)) {
            users.put(followeeId, new User(followeeId));
        }
        users.get(followerId).follow(users.get(followeeId));
    }

    /** Follower unfollows a followee. If the operation is invalid, it should be a no-op. */
    public void unfollow(int followerId, int followeeId) {
        if (followerId == followeeId || !users.containsKey(followerId) || !users.containsKey(followeeId)) {
            return;
        }
        users.get(followerId).rmFollowee(users.get(followeeId));
    }
}

/**
 * Your Twitter object will be instantiated and called as such:
 * Twitter obj = new Twitter();
 * obj.postTweet(userId,tweetId);
 * List<Integer> param_2 = obj.getNewsFeed(userId);
 * obj.follow(followerId,followeeId);
 * obj.unfollow(followerId,followeeId);
 */

Analysis

每次get news时间复杂度是O(tlgt).