[LeetCode] 355. Design Twitter 设计推特
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);
系统设计题,设计一个简单的推特,有发布消息,获得新鲜事,添加关注和取消关注等功能。
参考:linspiration
Java:
public class Twitter {
Map<Integer, Set<Integer>> userMap = new HashMap<>();
Map<Integer, LinkedList<Tweet>> tweets = new HashMap<>();
int timestamp = 0;
class Tweet {
int time;
int id;
Tweet(int time, int id) {
this.time = time;
this.id = id;
}
}
public void postTweet(int userId, int tweetId) {
if (!userMap.containsKey(userId)) userMap.put(userId, new HashSet<>());
userMap.get(userId).add(userId);
if (!tweets.containsKey(userId)) tweets.put(userId, new LinkedList<>());
tweets.get(userId).addFirst(new Tweet(timestamp++, tweetId));
}
public List<Integer> getNewsFeed(int userId) {
if (!userMap.containsKey(userId)) return new LinkedList<>();
PriorityQueue<Tweet> feed = new PriorityQueue<>((t1, t2) -> t2.time-t1.time);
userMap.get(userId).stream().filter(f -> tweets.containsKey(f))
.forEach(f -> tweets.get(f).forEach(feed::add));
List<Integer> res = new LinkedList<>();
while (feed.size() > 0 && res.size() < 10) res.add(feed.poll().id);
return res;
}
public void follow(int followerId, int followeeId) {
if (!userMap.containsKey(followerId)) userMap.put(followerId, new HashSet<>());
userMap.get(followerId).add(followeeId);
}
public void unfollow(int followerId, int followeeId) {
if (userMap.containsKey(followerId) && followeeId != followerId) userMap.get(followerId).remove(followeeId);
}
}
Python:
class Twitter(object): def __init__(self):
"""
Initialize your data structure here.
"""
self.__number_of_most_recent_tweets = 10
self.__followings = collections.defaultdict(set)
self.__messages = collections.defaultdict(list)
self.__time = 0 def postTweet(self, userId, tweetId):
"""
Compose a new tweet.
:type userId: int
:type tweetId: int
:rtype: void
"""
self.__time += 1
self.__messages[userId].append((self.__time, tweetId)) def getNewsFeed(self, 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.
:type userId: int
:rtype: List[int]
"""
max_heap = []
if self.__messages[userId]:
heapq.heappush(max_heap, (-self.__messages[userId][-1][0], userId, 0))
for uid in self.__followings[userId]:
if self.__messages[uid]:
heapq.heappush(max_heap, (-self.__messages[uid][-1][0], uid, 0)) result = []
while max_heap and len(result) < self.__number_of_most_recent_tweets:
t, uid, curr = heapq.heappop(max_heap)
nxt = curr + 1;
if nxt != len(self.__messages[uid]):
heapq.heappush(max_heap, (-self.__messages[uid][-(nxt+1)][0], uid, nxt))
result.append(self.__messages[uid][-(curr+1)][1]);
return result def follow(self, followerId, followeeId):
"""
Follower follows a followee. If the operation is invalid, it should be a no-op.
:type followerId: int
:type followeeId: int
:rtype: void
"""
if followerId != followeeId:
self.__followings[followerId].add(followeeId) def unfollow(self, followerId, followeeId):
"""
Follower unfollows a followee. If the operation is invalid, it should be a no-op.
:type followerId: int
:type followeeId: int
:rtype: void
"""
self.__followings[followerId].discard(followeeId)
C++:
class Twitter {
public:
/** Initialize your data structure here. */
Twitter() : time_(0) { } /** Compose a new tweet. */
void postTweet(int userId, int tweetId) {
messages_[userId].emplace_back(make_pair(++time_, 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. */
vector<int> getNewsFeed(int userId) {
using RIT = deque<pair<size_t, int>>::reverse_iterator;
priority_queue<tuple<size_t, RIT, RIT>> heap; if (messages_[userId].size()) {
heap.emplace(make_tuple(messages_[userId].rbegin()->first,
messages_[userId].rbegin(),
messages_[userId].rend()));
}
for (const auto& id : followings_[userId]) {
if (messages_[id].size()) {
heap.emplace(make_tuple(messages_[id].rbegin()->first,
messages_[id].rbegin(),
messages_[id].rend()));
}
}
vector<int> res;
while (!heap.empty() && res.size() < number_of_most_recent_tweets_) {
const auto& top = heap.top();
size_t t;
RIT begin, end;
tie(t, begin, end) = top;
heap.pop(); auto next = begin + 1;
if (next != end) {
heap.emplace(make_tuple(next->first, next, end));
} res.emplace_back(begin->second);
}
return res;
} /** Follower follows a followee. If the operation is invalid, it should be a no-op. */
void follow(int followerId, int followeeId) {
if (followerId != followeeId && !followings_[followerId].count(followeeId)) {
followings_[followerId].emplace(followeeId);
}
} /** Follower unfollows a followee. If the operation is invalid, it should be a no-op. */
void unfollow(int followerId, int followeeId) {
if (followings_[followerId].count(followeeId)) {
followings_[followerId].erase(followeeId);
}
} private:
const size_t number_of_most_recent_tweets_ = 10;
unordered_map<int, unordered_set<int>> followings_;
unordered_map<int, deque<pair<size_t, int>>> messages_;
size_t time_;
};
All LeetCode Questions List 题目汇总
[LeetCode] 355. Design Twitter 设计推特的更多相关文章
- 355 Design Twitter 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)的最近十条推文.你的设计需要支持以下的几个功能: postTweet(userI ...
- [leetcode]355. Design Twitter设计实现一个微博系统
//先定义一个数据结构,代表一条微博,有两个内容:发布者id,微博id(代表微博内容) class TwitterData { int userId; int twitterId; public Tw ...
- [LeetCode] Design Twitter 设计推特
Design a simplified version of Twitter where users can post tweets, follow/unfollow another user and ...
- leetcode@ [355] Design Twitter (Object Oriented Programming)
https://leetcode.com/problems/design-twitter/ Design a simplified version of Twitter where users can ...
- 【LeetCode】355. Design Twitter 解题报告(Python & C++)
作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 日期 题目地址:https://leetcode.c ...
- 【Leetcode】355. Design Twitter
题目描述: Design a simplified version of Twitter where users can post tweets, follow/unfollow another us ...
- 355. Design Twitter
二刷尝试了别的办法,用MAP代表关注列表. 然后不初始化,但是只要有用户被使用,而他又不在MAP里,就把他加进去,然后让他关注自己.. 但是这样做超时了. 问题在于这个题解法太多,有很多不同的情况. ...
- leetcode 355 Design Twitte
题目连接 https://leetcode.com/problems/design-twitter Design Twitte Description Design a simplified vers ...
- [LeetCode] 534. Design TinyURL 设计短网址
Note: For the coding companion problem, please see: Encode and Decode TinyURL. How would you design ...
随机推荐
- HTTP get post 请求实例
package wzh.Http; import java.io.BufferedReader; import java.io.IOException; import java.io.InputStr ...
- jq function return value
所有 JS 函数 都会返回值 假如 没有 return 则返回 undefined
- Sql 数据库 用户密码MD5加密
直接给代码先 DECLARE @TAB TABLE( NAEM VARCHAR(50) ) DECLARE @PA VARCHAR(50) DECLARE @A VARCHAR(10) SET @A= ...
- 【python】Requests 库支持RESTFUL的几种方式
Get: 查看资源 POST: 增加资源 PUT:修改资源,更新全部资源 PATCH:修改资源,更新局部资源 DELETE: 删除资源 HEAD: 查看响应头 OPTIONS: 查看请求方式
- Gym-100648B: Hie with the Pie(状态DP)
题意:外卖员开始在0号节点,有N个人点了外卖,(N<=10),现在告诉两两间距离,问怎么配送,使得每个人的外卖都送外,然后回到0号点的总时间最短,注意,同一个点可以多次经过. 思路:TSP问题( ...
- 【python爬虫】动态html
一.反爬策略 1.请求头 ——user-agent ——referer ——cookie 2.访问频率限制 ——代理池 ——再用户访问高峰期进行爬取,冲散日志.12-13 7-10 ——设置等待时长. ...
- nginx 配置文件正确性测试
今日思语:每天都要不一样,那么每天就应该多学习 在安装完nginx之后,我们可以使用nginx的测试命令来验证下nginx.conf的配置是否正确: 方式一:不指定文件 nginx -t 如上可知/e ...
- JPA EnableJpaAuditing 审计功能
关于自动填充或更新实体中的 CreateDate.CreatedBy 等在之前有一篇 jeecg 默认为空的字段值是如何被填充的? 有提到通过拦截器的方式实现,但是今天带大家了解一下如果使用 JPA ...
- c++ 去掉所有空格及换行符
string get_string(string res){ //删除换行符 int r = res.find('\r\n'); while (r != string::npos) { if (r ! ...
- C++ EH Exception(0xe06d7363)----抛出过程
C++ EH Exception是Windows系统VC++里对c++语言的throw的分类和定义,它的代码就是0xe06d7363.在VC++里其本质也是SEH结构化异常机制.在我们分析用户崩溃的例 ...