https://leetcode.com/problems/design-twitter
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:
1. postTweet(userId, tweetId): Compose a new tweet.
2. 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.
3. follow(followerId, followeeId): Follower follows a followee.
4. 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);
根据题意直接模拟即可，
我用了一个固定大小的set来保存10个最近的id
如果tweets少于10个直接插入即可，否则每插入一个就与set的最后一个元素判断
(已重载比较函数，set里面的元素是以时间戳从大到小排序的)
PS：好久都没有更新blogs了，太懒了(⊙﹏⊙)b
class Twitter {
private:
struct P {
int id, ref;
P(int i = 0, int j = 0) :id(i), ref(j) {}
inline bool operator<(const P &a) const { return ref > a.ref; }
};
public:
Twitter() { time = 1; }
void postTweet(int userId, int tweetId) {
userPost[userId].insert(P(tweetId, time++));
userFollow[userId].insert(userId);
}
vector<int> getNewsFeed(int userId) {
q.clear();
vector<int> res;
for(auto &r: userFollow[userId]) {
int n = userPost[r].size();
auto it = userPost[r].begin();
n = min(n, 10);
while(n--) {
if(q.size() < 10) {
q.insert(*it++);
} else {
auto c = q.end();
if(*it < *--c) {
q.erase(c);
q.insert(*it++);
}
}
}
}
for(auto &r: q) res.push_back(r.id);
return res;
}
void follow(int followerId, int followeeId) {
userFollow[followerId].insert(followeeId);
}
void unfollow(int followerId, int followeeId) {
if(followerId == followeeId) return;
userFollow[followerId].erase(followeeId);
}
private:
int time;
set<P> q;
unordered_map<int, set<P>> userPost;
unordered_map<int, set<int>> userFollow;
};
原文：http://www.cnblogs.com/GadyPu/p/5578594.html