[LeetCode] 182.Duplicate Emails 重復(fù)的郵箱
Write a SQL query to find all duplicate emails in a table named Person.
+----+---------+
| Id | Email |
+----+---------+
| 1 | a@b.com |
| 2 | c@d.com |
| 3 | a@b.com |
+----+---------+
For example, your query should return the following for the above table:
+---------+
| Email |
+---------+
| a@b.com |
+---------+
Note: All emails are in lowercase.
這道題讓我們求重復(fù)的郵箱，那么最直接的方法就是用Group by...Having Count(*)...的固定搭配來(lái)做，代碼如下：
解法一：
SELECT Email FROM Person GROUP BY Email
HAVING COUNT(*) > 1;
我們還可以用內(nèi)交來(lái)做，用Email來(lái)內(nèi)交兩個(gè)表，然后返回Id不同的行，則說(shuō)明兩個(gè)不同的人使用了相同的郵箱：
解法二：
SELECT DISTINCT p1.Email FROM Person p1
JOIN Person p2 ON p1.Email = p2.Email
WHERE p1.Id > p2.Id;
參考資料：
https://leetcode.com/discuss/53206/a-solution-using-a-group-by-and-another-one-using-a-self-join
到此這篇關(guān)于SQL實(shí)現(xiàn)LeetCode(182.重復(fù)的郵箱)的文章就介紹到這了,更多相關(guān)SQL實(shí)現(xiàn)重復(fù)的郵箱內(nèi)容請(qǐng)搜索腳本之家以前的文章或繼續(xù)瀏覽下面的相關(guān)文章希望大家以后多多支持腳本之家！
您可能感興趣的文章:- SQL實(shí)現(xiàn)LeetCode(196.刪除重復(fù)郵箱)
- SQL實(shí)現(xiàn)LeetCode(185.系里前三高薪水)
- SQL實(shí)現(xiàn)LeetCode(184.系里最高薪水)
- SQL實(shí)現(xiàn)LeetCode(183.從未下單訂購(gòu)的顧客)
- SQL實(shí)現(xiàn)LeetCode(181.員工掙得比經(jīng)理多)
- SQL實(shí)現(xiàn)LeetCode(180.連續(xù)的數(shù)字)
- C++實(shí)現(xiàn)LeetCode(179.最大組合數(shù))
- SQL實(shí)現(xiàn)LeetCode(197.上升溫度)
