355. Design Twitter 设计推特
2022年3月7日
作者： 负雪明烛 id： fuxuemingzhu 个人博客： http://fuxuemingzhu.cn/
@TOC
题目地址：https://leetcode.com/problems/design-twitter/description/
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);
题目大意
产生一个推特系统，这个推特能关注、发推，解关注，获得信息流。
解题方法
这个题主要是考察自己的工程实践能力。这一块，使用了优先级队列。在Python中，优先级队列其实就是可以使用heapq模块实现。
我提交失误的地方主要在于，取消关注的时候，需要多进行判断，是否存在这个用户、被关注的用户，以及他们之间是否存在关注关系等等。
这些判断在其他的函数中并没有使用到，所以注意细节。
另外，下面的做法好像很复杂，足足有100行。其实可以使用defaultdict等数据结构优化代码，使用排序代替heapq的。按下不表。
代码如下：
class User(object):
"""
User structure
"""
def __init__(self, userId):
self.userId = userId
self.tweets = set()
self.following = set()
class Tweet(object):
"""
Tweet structure
"""
def __init__(self, tweetId, userId, ts):
self.tweetId = tweetId
self.userId = userId
self.ts = ts
def __cmp__(self, other):
#call global(builtin) function cmp for int
return cmp(other.ts, self.ts)
class Twitter(object):
def __init__(self):
"""
Initialize your data structure here.
"""
self.ts = 0
self.userMap = dict()
def postTweet(self, userId, tweetId):
"""
Compose a new tweet.
:type userId: int
:type tweetId: int
:rtype: void
"""
if userId not in self.userMap:
self.userMap[userId] = User(userId)
tweet = Tweet(tweetId, userId, self.ts)
self.userMap[userId].tweets.add(tweet)
self.ts += 1
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]
"""
res = list()
que = []
if userId not in self.userMap:
return res
mainUser = self.userMap[userId]
for t in mainUser.tweets:
heapq.heappush(que, t)
for u in mainUser.following:
for t in u.tweets:
heapq.heappush(que, t)
n = 0
while que and n < 10:
res.append(heapq.heappop(que).tweetId)
n += 1
return res
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 followeeId not in self.userMap:
self.userMap[followeeId] = User(followeeId)
if followerId not in self.userMap:
self.userMap[followerId] = User(followerId)
if followerId == followeeId:
return
followee = self.userMap[followeeId]
self.userMap[followerId].following.add(followee)
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
"""
if (followerId == followeeId) or (followerId not in self.userMap) or (followeeId not in self.userMap):
return
followee = self.userMap[followeeId]
if followee in self.userMap.get(followerId).following:
self.userMap.get(followerId).following.remove(followee)
# 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)
C++版本的代码如下：
class Twitter {
public:
/** Initialize your data structure here. */
Twitter() {
cnt = 0;
}
/** Compose a new tweet. */
void postTweet(int userId, int tweetId) {
follow(userId, userId);
tweets[userId].push_back({cnt++, 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) {
vector<int> res;
priority_queue<pair<int, int>, vector<pair<int, int>>, greater<>> q;
for (auto &it : friends[userId]) {
for (auto &a : tweets[it]) {
if (!q.empty() && q.top().first > a.first && q.size() > 10) break;
q.push(a);
if (q.size() > 10) q.pop();
}
}
while (!q.empty()) {
res.push_back(q.top().second);
q.pop();
}
reverse(res.begin(), res.end());
return res;
}
/** Follower follows a followee. If the operation is invalid, it should be a no-op. */
void follow(int followerId, int followeeId) {
friends[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 (followeeId != followerId)
friends[followerId].erase(followeeId);
}
private:
int cnt;
unordered_map<int, set<int>> friends;
unordered_map<int, vector<pair<int, int>>> tweets;
};
/**
* Your Twitter object will be instantiated and called as such:
* Twitter obj = new Twitter();
* obj.postTweet(userId,tweetId);
* vector<int> param_2 = obj.getNewsFeed(userId);
* obj.follow(followerId,followeeId);
* obj.unfollow(followerId,followeeId);
*/
参考资料：
- https://leetcode.com/problems/design-twitter/discuss/155493/Java-Solution-using-Two-classes-User-and-Tweet-and-using-just-one-Map
日期
2018 年 8 月 28 日 —— 雾霾天 2018 年 12 月 6 日 —— 周四啦！