题目链接：​​点击打开链接​​

题目大意：略。

解题思路：略。

AC 代码

-- 解决方案(1)
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;

-- 解决方案(2)
WITH t1 AS(SELECT requester_id id, COUNT(requester_id) cnt
FROM request_accepted
GROUP BY requester_id),

t2 AS(SELECT accepter_id id, COUNT(accepter_id) cnt
FROM request_accepted
GROUP BY accepter_id),

t3 AS(SELECT * FROM t1
UNION ALL
SELECT * FROM t2)

SELECT id, SUM(cnt) num
FROM t3
GROUP BY id
ORDER BY num DESC
LIMIT 1