【Leetcode】196. Delete Duplicate Emails （Easy）
阿新 • • 發佈：2019-02-19
1.題目
Write a SQL query to delete all duplicate email entries in a table named Person, keeping only unique emails based on its smallest Id.
翻譯：寫一個SQL語句刪掉Person表中，所有email重複的實體（相同的保留一個），只保留I的最小的唯一email.
+----+------------------+ | Id | Email | +----+------------------+ | 1 | [email protected] | | 2 | [email protected] | | 3 | [email protected] | +----+------------------+ 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 | [email protected] | | 2 | [email protected] | +----+------------------+
2.思路
①將Person表按Email分組，篩選每組裡面的最小Id值作為minId。
②刪除所有Id不在minId中的實體。
這個sql語句的思路很簡單，但是要注意的細節很多，遇到的錯誤我將在第4點中進行總結。
3.演算法
delete from Person
where Person.Id not in (
select Person_derived.minId from(
select min(Id) as minId from Person group by Email
)Person_derived
)4.總結
①遇到的錯誤1：
delete from Person where Person.Id not in ( select min(Id) as minId from Person group by Email )
You can't specify target table 'Person' for update in FROM clauseSQL特有的一個錯誤，不能篩選表中一部分元素，然後直接更新它，需要引入一箇中間表。
具體原因我猜測可能是和sql的設計有關係。
②遇到的錯誤2：
delete from Person
where Person.Id not in (
select minId from(
select min(Id) as minId from Person group by Email
)
)Your answer
Every derived table must have its own aliasPS:Leetcode的Runtime很有意思，同一個演算法我提交兩次，一次打敗2%的使用者，一次打敗98%的使用者。
看來不能完全依賴這個指標，還是得學會計算時間複雜度呀。畢竟好幾次我都栽在時間複雜度上。