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.
Example:

Twitter twitter = new Twitter();

// User 1 posts a new tweet (id = 5).
twitter.postTweet(1, 5);

// User 1's news feed should return a list with 1 tweet id -> [5].
twitter.getNewsFeed(1);

// User 1 follows user 2.
twitter.follow(1, 2);

// User 2 posts a new tweet (id = 6).
twitter.postTweet(2, 6);

// User 1's news feed should return a list with 2 tweet ids -> [6, 5].
// Tweet id 6 should precede tweet id 5 because it is posted after tweet id 5.
twitter.getNewsFeed(1);

// User 1 unfollows user 2.
twitter.unfollow(1, 2);

// User 1's news feed should return a list with 1 tweet id -> [5],
// since user 1 is no longer following user 2.
twitter.getNewsFeed(1);

Solution：

思路：

1. 對于User的定義：
   {
   uid_num; //id for user
   tweet_list<Tweet>; //list for tweets the user has posted.
   follow_set<id_num>; //set for id_nums that the user follows

   方法：
   follow(id_num)：將要follow的id_num添加的此用戶的follow_set中
   unfollow(id_num)：將要unfollow的id_num從此用戶的follow_set中移除
   post(tid_num)：新建Tweet對象with id tid_num，插入到在此用戶tweet_list的head
   }

2. 對于Tweet的定義：
   {
   int timestamp; //for updates sorting
   tid_num; //id for tweet
   Tweet next; // Tweet defined for one user as a list
   }

3. 定義一個Map<id_num, User> 來通過id來訪問/修改User信息

4. 故要實現的public方法如下：
   (1)follow(followerId, followeeId): Follower follows a followee 定義：
   通過followerId從map中找到該user，執行user.follow(followeeId)：將要follow的id_num添加的此用戶的follow_set中

(2)unfollow(followerId, followeeId): Follower unfollows a followee 定義：
通過followerId從map中找到該user，執行user.unfollow(followeeId)：將要unfollow的id_num從此用戶的follow_set中移除

(3)postTweet(userId, tweetId): Compose a new tweet 定義：
通過userId從map中找到該user，執行user.post方法：新建Tweet對象with id tid_num，插入到在此用戶tweet_list的head

(4)getNewsFeed(userId): Retrieve the 10 most recent tweet ids in the user's news feed
(along with the updates from the users that he follows) 定義：
建立PriorityQueue(根據Tweet的timestamp排序)，通過userId從map中找到該user，將他follow的(默認用戶建立是follow自己)所有用戶的tweet_list
加入到此PriorityQueue中。后續以BFS的方式方式輸出10個NewsFeed.

Time Complexity: O(N) Space Complexity: O(N)

Solution Code：

public class Twitter {
    private static int timeStamp = 0;

    // easy to find if user exist
    private Map<Integer, User> userMap;

    // Tweet link to next Tweet so that we can save a lot of time
    // when we execute getNewsFeed(userId)
    private class Tweet {
        public int id;
        public int time;
        public Tweet next;

        public Tweet(int id) {
            this.id = id;
            time = timeStamp++;
            next = null;
        }
    }


    // OO design so User can follow, unfollow and post itself
    public class User {
        public int id;
        public Set<Integer> followed;
        public Tweet tweet_head;

        public User(int id) {
            this.id = id;
            followed = new HashSet<>();
            follow(id); // first follow itself
            tweet_head = null;
        }

        public void follow(int id) {
            followed.add(id);
        }

        public void unfollow(int id) {
            followed.remove(id);
        }


        // everytime user post a new tweet, add it to the head of tweet list.
        public void post(int id) {
            Tweet t = new Tweet(id);
            t.next = tweet_head;
            tweet_head = t;
        }
    }


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

    /**
     * Compose a new tweet.
     */
    public void postTweet(int userId, int tweetId) {
        if (!userMap.containsKey(userId)) {
            User u = new User(userId);
            userMap.put(userId, u);
        }
        userMap.get(userId).post(tweetId);

    }


    // Best part of this.
    // first get all tweets lists from one user including itself and all people it followed.
    // Second add all heads into a max heap. Every time we poll a tweet with
    // largest time stamp from the heap, then we add its next tweet into the heap.
    // So after adding all heads we only need to add 9 tweets at most into this
    // heap before we get the 10 most recent tweet.
    public List<Integer> getNewsFeed(int userId) {
        List<Integer> res = new LinkedList<>();

        if (!userMap.containsKey(userId)) return res;

        Set<Integer> users = userMap.get(userId).followed;
        PriorityQueue<Tweet> q = new PriorityQueue<Tweet>(users.size(), (a, b) -> (b.time - a.time));
        for (int user : users) {
            Tweet t = userMap.get(user).tweet_head;
            // very imporant! If we add null to the head we are screwed.
            if (t != null) {
                q.add(t);
            }
        }
        int n = 0;
        while (!q.isEmpty() && n < 10) {
            Tweet t = q.poll();
            res.add(t.id);
            n++;
            if (t.next != null)
                q.add(t.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 (!userMap.containsKey(followerId)) {
            User u = new User(followerId);
            userMap.put(followerId, u);
        }
        if (!userMap.containsKey(followeeId)) {
            User u = new User(followeeId);
            userMap.put(followeeId, u);
        }
        userMap.get(followerId).follow(followeeId);
    }

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