# [LeetCode] 602. Friend Requests II: Who Has Most Friend? 朋友请求 II: 谁有最多的朋友？

2021年09月15日 阅读数：1

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

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

```| 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:post

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

Note:this

• 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?

Algorithmurl

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

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

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.blog

```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
;　　```

```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　　```

```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
```