+-------------+------------+
| followee | follower |
+-------------+------------+
| A | B |
| B | C |
| B | D |
| D | E |
+-------------+------------+
应该输出:
+-------------+------------+
| follower | num |
+-------------+------------+
| B | 2 |
| D | 1 |
+-------------+------------+
解释:
B 和 D 都在在 follower 字段中出现,作为被关注者,B 被 C 和 D 关注,D 被 E 关注。A 不在 follower 字段内,所以A不在输出列表中。
注意:
被关注者永远不会被他 / 她自己关注。 将结果按照字典序返回。
来源:力扣(LeetCode)
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Solution
自连接
select a.follower, count(distinct b.follower) num
from follow a inner join follow b on a.follower = b.followee
group by 1
order by 1
子查询
select followee follower, count(distinct follower) num
from follow
where followee in (select follower from follow)
group by 1
order by 1