182. 查找重復的電子郵箱 - 力扣（LeetCode） (leetcode-cn.com)
# Write your MySQL query statement below
//解法1
select email from person group by email having count(email)>1
//解法2
select email from (select count(1) as t,email from person group by email) r where r.t>1;
//解法3
select distinct(p1.email) from person p1 join person p2 on p1.email = p2.email and p1.id!=p2.id;
