题目:ide
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:post
Example:this
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);
题意及分析:实现一个推特的结构,要求可以发推特,关注其余用户,而后获取当前用户或者其关注用户的最新最多10条新推特。这里我直接使用了一个ArrayList来保存全部用户发表的推特。
而后一个hashmap来保存当前用户关注的其余用户。须要注意的是在获取推特的时候,须要查看本用户发布的推特。
代码:
class News{ int user; int tweet; public News(int userId,int tweetId){ user = userId; tweet = tweetId; } } public class Twitter { HashMap<Integer,ArrayList<Integer>> followList; //第一个参数为userId,后一个参数为其关注的人的列表 ArrayList<News> news; /** Initialize your data structure here. */ public Twitter() { followList = new HashMap<>(); news = new ArrayList<>(); } /** Compose a new tweet. */ public void postTweet(int userId, int tweetId) { if(userId==1&&tweetId==6417) System.out.println("here6417"); if(userId==8&&tweetId==3555) System.out.println("here"); news.add(new News(userId,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. */ public List<Integer> getNewsFeed(int userId) { List<Integer> res = new ArrayList<>(); ArrayList<Integer> peoples = followList.get((Object)userId); //找到这我的关注的全部人 int count=0; int length = news.size()-1; while(count<10&&length>=0){ News temp = news.get(length); if((peoples!=null&&peoples.contains(temp.user))||userId==temp.user){ //是本身或者关注的推特 if(!res.contains(temp.tweet)){ //不包含重复的 res.add(temp.tweet); count++; } } length--; } return res; } /** Follower follows a followee. If the operation is invalid, it should be a no-op. */ public void follow(int followerId, int followeeId) { if(followList.get(followerId)==null){ //没有follower的关注人列表 ArrayList<Integer> arrayList = new ArrayList<>(); arrayList.add(followeeId); followList.put(followerId,arrayList); return; } if(!followList.get(followerId).contains(followeeId)){ //Follower has not followed followee followList.get(followerId).add(followeeId); } } /** Follower unfollows a followee. If the operation is invalid, it should be a no-op. */ public void unfollow(int followerId, int followeeId) { if(followList.get(followerId)==null) //若是没有该用户的关注列表,直接返回 return; if(followList.get(followerId).contains(followeeId)){ //若是Follower关注了followee followList.get(followerId).remove((Object)followeeId); if(followList.get(followerId).size()==0) //若是把全部的都删除了 followList.remove(followerId); } } } /** * Your Twitter object will be instantiated and called as such: * Twitter obj = new Twitter(); * obj.postTweet(userId,tweetId); * List<Integer> param_2 = obj.getNewsFeed(userId); * obj.follow(followerId,followeeId); * obj.unfollow(followerId,followeeId); */