MySQL高阶1949-坚定地友谊

目录

题目

准备数据

分析数据

实现

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题目

如果 x  和 y 为 朋友 且他们 至少 有三个共同的朋友 ，那么 x 和 y 之间的友谊就是 坚定的。

写一个解决方案来找到所有的 坚定的友谊。

注意，结果表不应该包含重复的行，并且 user1_id < user2_id。

以 任何顺序 返回结果表。

准备数据

Create table If Not Exists Friendship (user1_id int, user2_id int)
Truncate table Friendship
insert into Friendship (user1_id, user2_id) values ('1', '2')
insert into Friendship (user1_id, user2_id) values ('1', '3')
insert into Friendship (user1_id, user2_id) values ('2', '3')
insert into Friendship (user1_id, user2_id) values ('1', '4')
insert into Friendship (user1_id, user2_id) values ('2', '4')
insert into Friendship (user1_id, user2_id) values ('1', '5')
insert into Friendship (user1_id, user2_id) values ('2', '5')
insert into Friendship (user1_id, user2_id) values ('1', '7')
insert into Friendship (user1_id, user2_id) values ('3', '7')
insert into Friendship (user1_id, user2_id) values ('1', '6')
insert into Friendship (user1_id, user2_id) values ('3', '6')
insert into Friendship (user1_id, user2_id) values ('2', '6')

分析数据

第一步:

用UNION ALL合并，UNION ALL不会消除重复的行，意味着如果user1_id和user2_id互为好友，那么这两行都会被包含。并且按照user_id和friend_id的顺序。

select user1_id user_id,user2_id friend_id
from Friendship
union all
select user2_id,user1_id
from Friendship
order by 1,2;

第二步:

• where (t1.user_id, t2.user_id) in (select * from t)：确保t1和t2中的user_id对在CTE t中存在。
• and t1.friend_id = t2.friend_id：确保t1和t2有相同的friend_id，即它们有共同的好友。
• and t1.user_id < t2.user_id：确保user1_id小于user2_id，避免重复的组合（例如，用户1和用户2与用户2和用户1被视为相同的组合）。
• group by 1, 2：按user1_id和user2_id分组。
• having count(distinct t1.friend_id) >= 3：筛选出至少有3个不同共同好友的用户对。

with t as (
    (select user1_id user_id,user2_id friend_id
     from Friendship
     union all
     select user2_id,user1_id
     from Friendship)
        order by 1,2
)
select *
from t t1,t t2
where (t1.user_id,t2.user_id) in (select * from t)
    and t1.friend_id = t2.friend_id
  and t1.user_id < t2.user_id
order by 1,2;

第三步:

• select t1.user_id user1_id, t2.user_id user2_id, count(distinct t1.friend_id) common_friend：选择两列user1_id和user2_id，以及它们之间不同的共同好友数量common_friend。

with t as (
    (select user1_id user_id,user2_id friend_id
     from Friendship
     union all
     select user2_id,user1_id
     from Friendship)
        order by 1,2
)
select t1.user_id user1_id
     ,t2.user_id user2_id
     ,count(distinct t1.friend_id) common_friend
from t t1,t t2
where
    (t1.user_id,t2.user_id) in (select * from t)
  and t1.friend_id = t2.friend_id
  and t1.user_id < t2.user_id
group by 1,2
having count(distinct t1.friend_id) >=3;

实现

with t as (
    (select user1_id user_id,user2_id friend_id
     from Friendship
     union all
     select user2_id,user1_id
     from Friendship)
        order by 1,2
)
select t1.user_id user1_id
     ,t2.user_id user2_id
     ,count(distinct t1.friend_id) common_friend
from t t1,t t2
where
    (t1.user_id,t2.user_id) in (select * from t)
  and t1.friend_id = t2.friend_id
  and t1.user_id < t2.user_id
group by 1,2
having count(distinct t1.friend_id) >=3;

原文地址：https://blog.csdn.net/weixin_58305115/article/details/142531778

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