SQL實(shí)現(xiàn)LeetCode(196.刪除重復(fù)郵箱)

[LeetCode] 196.Delete Duplicate Emails 刪除重復(fù)郵箱

Write a SQL query to delete all duplicate email entries in a table named Person, keeping only unique emails based on its smallest Id.

+----+------------------+
| Id | Email            |
+----+------------------+
| 1  | john@example.com |
| 2  | bob@example.com  |
| 3  | john@example.com |
+----+------------------+
Id is the primary key column for this table.

For example, after running your query, the above Person table should have the following rows:

+----+------------------+
| Id | Email            |
+----+------------------+
| 1  | john@example.com |
| 2  | bob@example.com  |
+----+------------------+

這道題讓我們刪除重復(fù)郵箱，那我們可以首先找出所有不重復(fù)的郵箱，然后取個(gè)反就是重復(fù)的郵箱，都刪掉即可，那么我們?nèi)绾握页鏊胁恢貜?fù)的郵箱呢，我們可以按照郵箱群組起來(lái)，然后用Min關(guān)鍵字挑出較小的，然后取補(bǔ)集刪除即可：

解法一：

DELETE FROM Person WHERE Id NOT IN
(SELECT Id FROM (SELECT MIN(Id) Id FROM Person GROUP BY Email) p);

我們也可以使用內(nèi)交讓兩個(gè)表以郵箱關(guān)聯(lián)起來(lái)，然后把相同郵箱且Id大的刪除掉，參見(jiàn)代碼如下：

解法二：

DELETE p2 FROM Person p1 JOIN Person p2 
ON p2.Email = p1.Email WHERE p2.Id > p1.Id;

我們也可以不用Join，而直接用where將兩表關(guān)聯(lián)起來(lái)也行：

解法三：

DELETE p2 FROM Person p1, Person p2
WHERE p1.Email = p2.Email AND p2.Id > p1.Id;

類(lèi)似題目：

Duplicate Emails

參考資料：

https://leetcode.com/discuss/61176/simple-solution-using-a-self-join

https://leetcode.com/discuss/48403/my-answer-delete-duplicate-emails-with-double-nested-query

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