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设计推特

设计一个简化版的推特(Twitter),可以让用户实现发送推文,关注/取消关注其他用户,能够看见关注人(包括自己)的最近 10 条推文。 实现 Twitter 类: Twitter() 初始化简易版推特对象 void postTweet(int userId, int tweetId) 根据给定的 …

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中等 · 链表指针操作

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答案摘要

class Twitter: def __init__(self):

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题目描述

设计一个简化版的推特(Twitter),可以让用户实现发送推文,关注/取消关注其他用户,能够看见关注人(包括自己)的最近 10 条推文。

实现 Twitter 类:

  • Twitter() 初始化简易版推特对象
  • void postTweet(int userId, int tweetId) 根据给定的 tweetIduserId 创建一条新推文。每次调用此函数都会使用一个不同的 tweetId
  • List<Integer> getNewsFeed(int userId) 检索当前用户新闻推送中最近  10 条推文的 ID 。新闻推送中的每一项都必须是由用户关注的人或者是用户自己发布的推文。推文必须 按照时间顺序由最近到最远排序
  • void follow(int followerId, int followeeId) ID 为 followerId 的用户开始关注 ID 为 followeeId 的用户。
  • void unfollow(int followerId, int followeeId) ID 为 followerId 的用户不再关注 ID 为 followeeId 的用户。

 

示例:

输入
["Twitter", "postTweet", "getNewsFeed", "follow", "postTweet", "getNewsFeed", "unfollow", "getNewsFeed"]
[[], [1, 5], [1], [1, 2], [2, 6], [1], [1, 2], [1]]
输出
[null, null, [5], null, null, [6, 5], null, [5]]

解释
Twitter twitter = new Twitter();
twitter.postTweet(1, 5); // 用户 1 发送了一条新推文 (用户 id = 1, 推文 id = 5)
twitter.getNewsFeed(1);  // 用户 1 的获取推文应当返回一个列表,其中包含一个 id 为 5 的推文
twitter.follow(1, 2);    // 用户 1 关注了用户 2
twitter.postTweet(2, 6); // 用户 2 发送了一个新推文 (推文 id = 6)
twitter.getNewsFeed(1);  // 用户 1 的获取推文应当返回一个列表,其中包含两个推文,id 分别为 -> [6, 5] 。推文 id 6 应当在推文 id 5 之前,因为它是在 5 之后发送的
twitter.unfollow(1, 2);  // 用户 1 取消关注了用户 2
twitter.getNewsFeed(1);  // 用户 1 获取推文应当返回一个列表,其中包含一个 id 为 5 的推文。因为用户 1 已经不再关注用户 2

 

提示:

  • 1 <= userId, followerId, followeeId <= 500
  • 0 <= tweetId <= 104
  • 所有推特的 ID 都互不相同
  • postTweetgetNewsFeedfollowunfollow 方法最多调用 3 * 104
  • 用户不能关注自己
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解题思路

方法一

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class Twitter:
    def __init__(self):
        """
        Initialize your data structure here.
        """
        self.user_tweets = defaultdict(list)
        self.user_following = defaultdict(set)
        self.tweets = defaultdict()
        self.time = 0

    def postTweet(self, userId: int, tweetId: int) -> None:
        """
        Compose a new tweet.
        """
        self.time += 1
        self.user_tweets[userId].append(tweetId)
        self.tweets[tweetId] = self.time

    def getNewsFeed(self, userId: int) -> List[int]:
        """
        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.
        """
        following = self.user_following[userId]
        users = set(following)
        users.add(userId)
        tweets = [self.user_tweets[u][::-1][:10] for u in users]
        tweets = sum(tweets, [])
        return nlargest(10, tweets, key=lambda tweet: self.tweets[tweet])

    def follow(self, followerId: int, followeeId: int) -> None:
        """
        Follower follows a followee. If the operation is invalid, it should be a no-op.
        """
        self.user_following[followerId].add(followeeId)

    def unfollow(self, followerId: int, followeeId: int) -> None:
        """
        Follower unfollows a followee. If the operation is invalid, it should be a no-op.
        """
        following = self.user_following[followerId]
        if followeeId in following:
            following.remove(followeeId)


# Your Twitter object will be instantiated and called as such:
# obj = Twitter()
# obj.postTweet(userId,tweetId)
# param_2 = obj.getNewsFeed(userId)
# obj.follow(followerId,followeeId)
# obj.unfollow(followerId,followeeId)
speed

复杂度分析

指标
时间complexity varies by operation: posting is O(1), follow/unfollow is O(1), and getting the news feed is O(N log k) where N is total followed tweets and k is number of tweets to retrieve. Space complexity depends on the number of users and tweets stored, roughly O(U + T).
空间Depends on the final approach
psychology

面试官常问的追问

外企场景
  • question_mark

    Focus on correctly merging multiple user feeds while preserving tweet order.

  • question_mark

    Ensure follow/unfollow updates maintain user relationships in the hash map.

  • question_mark

    Handle edge cases like unfollowing a non-followed user or self-follow attempts.

warning

常见陷阱

外企场景
  • error

    Forgetting to update pointers when inserting new tweets can break feed order.

  • error

    Not limiting news feed to 10 most recent tweets leads to performance issues.

  • error

    Incorrectly handling self-follow or unfollow scenarios can cause unexpected results.

swap_horiz

进阶变体

外企场景
  • arrow_right_alt

    Retrieve news feed with N most recent tweets instead of fixed 10.

  • arrow_right_alt

    Support deleting a tweet and updating feeds dynamically.

  • arrow_right_alt

    Implement trending tweets aggregation based on user likes or retweets.

help

常见问题

外企场景

设计推特题解:链表指针操作 | LeetCode #355 中等