zoukankan      html  css  js  c++  java
  • [LeetCode] 602. Friend Requests II: Who Has Most Friend? 朋友请求 II: 谁有最多的朋友?

    In social network like Facebook or Twitter, people send friend requests and accept others' requests as well.

    Table request_accepted holds the data of friend acceptance, while requester_id and accepter_id both are the id of a person.

    | requester_id | accepter_id | accept_date|
    |--------------|-------------|------------|
    | 1            | 2           | 2016_06-03 |
    | 1            | 3           | 2016-06-08 |
    | 2            | 3           | 2016-06-08 |
    | 3            | 4           | 2016-06-09 |
    

    Write a query to find the the people who has most friends and the most friends number. For the sample data above, the result is:

    | id | num |
    |----|-----|
    | 3  | 3   |
    

    Note:

      • It is guaranteed there is only 1 people having the most friends.
      • The friend request could only been accepted once, which mean there is no multiple records with the same requester_id and accepter_id value. Explanation:
        The person with id '3' is a friend of people '1', '2' and '4', so he has 3 friends in total, which is the most number than any others. Follow-up:
        In the real world, multiple people could have the same most number of friends, can you find all these people in this case?

    Algorithm

    Being friends is bidirectional, so if one person accepts a request from another person, both of them will have one more friend.

    Thus, we can union column requester_id and accepter_id, and then count the number of the occurrence of each person.

    select requester_id as ids from request_accepted
    union all
    select accepter_id from request_accepted;
    Note: Here we should use union all instead of union because union all will keep all the records even the 'duplicated' one.

    解法:

    select ids as id, cnt as num
    from
    (
    select ids, count(*) as cnt
       from
       (
            select requester_id as ids from request_accepted
            union all
            select accepter_id from request_accepted
        ) as tbl1
       group by ids
       ) as tbl2
    order by cnt desc
    limit 1
    ;  

    解法2:

    select a.id, count(*) as num
    from 
    (select requester_id as id from request_accepted
    union all 
    select accepter_id as id from request_accepted) a
    group by id
    order by num desc
    limit 1  

    解法3:

    select t2.Id as id, t2.num as num
    from (
    select t1.Id, sum(cnt) as num
    from(
    select accepter_id as Id, count(*) as cnt
    from request_accepted
    group by accepter_id
    
    union all
    
    select requester_id as Id, count(*) as cnt
    from request_accepted 
    group by requester_id) t1
    
    group by t1.Id ) t2 
    
    order by t2.num DESC
    limit 1
    

      

    All LeetCode Questions List 题目汇总

  • 相关阅读:
    看了关于全职女性的文字,我想到了一些事情
    通过一个大型项目来学习分布式算法(6)
    IO模式——同步(堵塞、非堵塞)、异步
    湖南省第九届大学生计算机程序设计竞赛 高桥和低桥
    为什么我的ECSHOP出现报错改正确了还是没有反应?
    wxWidgets刚開始学习的人导引(2)——下载、安装wxWidgets
    1096. Consecutive Factors (20)
    POJ 2955 Brackets
    (转载)单调栈题目总结
    20140708郑州培训第二题Impossible Game
  • 原文地址:https://www.cnblogs.com/lightwindy/p/9698975.html
Copyright © 2011-2022 走看看