182 查找重复的电子邮箱

关键词：重复

解法一：group by + having

231 ms

select Email 
from Person
group by Email
having count(Email) > 1;

解法二：distinct + 自连接

237 ms

select distinct p1.Email 
from Person as p1, Person as p2 
where p1.Email = p2.Email and p1.Id != p2.Id;

239 ms

select distinct p1.Email
from Person as p1
inner join Person as p2
on p1.Email = p2.Email and p1.Id < p2.Id;

解法三：子查询 + where

232 ms

select Email 
from (
      select Email, count(Email) as num
      from Person
      group by Email
     ) as statistic
where num > 1;