英文原文
Table: Person
+-------------+---------+ | Column Name | Type | +-------------+---------+ | id | int | | email | varchar | +-------------+---------+ id is the primary key column for this table. Each row of this table contains an email. The emails will not contain uppercase letters.
Write an SQL query to report all the duplicate emails.
Return the result table in any order.
The query result format is in the following example.
Example 1:
Input: Person table: +----+---------+ | id | email | +----+---------+ | 1 | a@b.com | | 2 | c@d.com | | 3 | a@b.com | +----+---------+ Output: +---------+ | Email | +---------+ | a@b.com | +---------+ Explanation: a@b.com is repeated two times.
中文题目
编写一个 SQL 查询,查找 Person
表中所有重复的电子邮箱。
示例:
+----+---------+ | Id | Email | +----+---------+ | 1 | a@b.com | | 2 | c@d.com | | 3 | a@b.com | +----+---------+
根据以上输入,你的查询应返回以下结果:
+---------+ | Email | +---------+ | a@b.com | +---------+
说明:所有电子邮箱都是小写字母。
通过代码
官方题解
方法一:使用 GROUP BY
和临时表
算法
重复的电子邮箱存在多次。要计算每封电子邮件的存在次数,我们可以使用以下代码。
select Email, count(Email) as num
from Person
group by Email;
| Email | num |
|---------|-----|
| a@b.com | 2 |
| c@d.com | 1 |
以此作为临时表,我们可以得到下面的解决方案。
select Email from
(
select Email, count(Email) as num
from Person
group by Email
) as statistic
where num > 1
;
方法二:使用 GROUP BY
和 HAVING
条件
向 GROUP BY
添加条件的一种更常用的方法是使用 HAVING
子句,该子句更为简单高效。
所以我们可以将上面的解决方案重写为:
select Email
from Person
group by Email
having count(Email) > 1;
统计信息
通过次数 | 提交次数 | AC比率 |
---|---|---|
192529 | 242181 | 79.5% |
提交历史
提交时间 | 提交结果 | 执行时间 | 内存消耗 | 语言 |
---|