题目要求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);设计一个迷你推特，要求能够支持以下几个方法：发布推特，关注用户，取关用户，查看最近的十条关注用户发送的推特。思路和代码这道题目本质上是考察是否能将数据结构的知识灵活的运用于现实生活中。从最直观的想法来看，我们会有一个用户实体，每个用户会记录自己关注的用户的id，以及记录自己发表的所有tweet。这里唯一的难点在于我们如何按照时间顺序获取tweet流。这么一想，这题其实就转换为如何将N个有序排列的数组汇合成一个有序的数组。这题等价于我们每次都会比较当前所有被关注者发布的最新未读tweet，选出结果后将其插入结果集。这里我们可以利用等价队列帮助我们更快的完成选择的过程。public class Twitter { public Twitter() { users = new HashMap<>(); } public static int timestamp = 0; private Map<Integer, User> users; /** Compose a new tweet. / public void postTweet(int userId, int tweetId) { if(!users.containsKey(userId)) { User user = new User(userId); users.put(userId, user); } User user = users.get(userId); user.tweet(tweetId); } /* 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> result = new ArrayList<Integer>(); if(!users.containsKey(userId)) { return result; } User user = users.get(userId); PriorityQueue<Tweet> queue = new PriorityQueue<>(user.followed.size()); for(int followee : user.followed) { User tmp = users.get(followee); if(tmp != null && tmp.headTweet != null) { queue.offer(tmp.headTweet); } } while(!queue.isEmpty() && result.size() < 10) { Tweet t = queue.poll(); result.add(t.tweetId); if(t.next != null) { queue.offer(t.next); } } return result; } /* Follower follows a followee. If the operation is invalid, it should be a no-op. / public void follow(int followerId, int followeeId) { if(!users.containsKey(followerId)) { User user = new User(followerId); users.put(followerId, user); } if(!users.containsKey(followeeId)) { User user = new User(followeeId); users.put(followeeId, user); } User user = users.get(followerId); user.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(!users.containsKey(followerId) || followerId == followeeId) { return; } User user = users.get(followerId); user.unfollow(followeeId); } public static class User{ int userId; Set<Integer> followed; Tweet headTweet; public User(int userId) { this.userId = userId; this.followed = new HashSet<>(); follow(userId); } public void follow(int userId) { followed.add(userId); } public void unfollow(int userId) { followed.remove(userId); } public void tweet(int tweetId) { Tweet tweet = new Tweet(tweetId); tweet.next = headTweet; headTweet = tweet; } } public static class Tweet implements Comparable<Tweet>{ int tweetId; Tweet next; int time; public Tweet(int tweetId) { this.tweetId = tweetId; this.time = timestamp++; } @Override public int compareTo(Tweet o) { return o.time - this.time; } }}
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