LeetCode #355. Design Twitter

最新推荐文章于 2021-07-17 21:17:34 发布

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本文介绍了一个简化版Twitter的设计方案，包括发布推文、获取新闻动态、关注与取消关注等功能。设计支持用户发布推文、查看个人及关注者的最新10条推文，以及关注和取消关注其他用户。

题目描述：

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);

class Twitter {
public:
    /** Initialize your data structure here. */
    Twitter() {
        time=0;
    }
    
    /** Compose a new tweet. */
    void postTweet(int userId, int tweetId) {
        posts[userId].insert({time,tweetId});
        // 每个用户只需要保留最近的10个post即可
        if(posts[userId].size()>10) posts[userId].erase(posts[userId].begin());
        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. */
    vector<int> getNewsFeed(int userId) {
        vector<int> result;
        set<pair<int,int>> s;
        if(followees.count(userId))
        {   // 如果不加判断的话，下一个循环会给userId创建一个空的集合
            for(int followee:followees[userId])
            {
                if(posts.count(followee)==0) continue;
                for(auto post:posts[followee])
                {
                    s.insert(post);
                    if(s.size()>10) s.erase(s.begin());
                }
            }   
        }
        if(posts.count(userId))
        {
            for(auto post:posts[userId])
            {
                s.insert(post);
                if(s.size()>10) s.erase(s.begin());
            }      
        }
        for(auto post:s) result.push_back(post.second);
        reverse(result.begin(),result.end());
        return result;
    }
    
    /** Follower follows a followee. If the operation is invalid, it should be a no-op. */
    void follow(int followerId, int followeeId) { // 注意用户不能关注自己
        if(followerId!=followeeId) followees[followerId].insert(followeeId);
    }
    
    /** Follower unfollows a followee. If the operation is invalid, it should be a no-op. */
    void unfollow(int followerId, int followeeId) {
        if(followees.count(followerId)) followees[followerId].erase(followeeId);
    }

private:
    int time; // 用一个编号给每个post记录发布时间
    unordered_map<int,unordered_set<int>> followees;
    unordered_map<int,set<pair<int,int>>> posts;
};