Add solution 196 [sql]

group by
min
not in

Author: yanglbme

README.md

@@ -40,6 +40,7 @@ Complete [solutions](https://github.com/doocs/leetcode/tree/master/solution) to
 | 182 | [Duplicate Emails](https://github.com/doocs/leetcode/tree/master/solution/182.Duplicate%20Emails) | `SQL` |
 | 183 | [Customers Who Never Order](https://github.com/doocs/leetcode/tree/master/solution/183.Customers%20Who%20Never%20Order) | `SQL` |
 | 189 | [Rotate Array](https://github.com/doocs/leetcode/tree/master/solution/189.Rotate%20Array) | `Array` |
+| 196 | [Delete Duplicate Emails](https://github.com/doocs/leetcode/tree/master/solution/196.Delete%20Duplicate%20Emails) | `SQL` |
 | 198 | [House Robber](https://github.com/doocs/leetcode/tree/master/solution/198.House%20Robber) | `Dynamic Programming` |
 | 203 | [Remove Linked List Elements](https://github.com/doocs/leetcode/tree/master/solution/203.Remove%20Linked%20List%20Elements) | `Linked List` |
 | 231 | [Power of Two](https://github.com/doocs/leetcode/tree/master/solution/231.Power%20of%20Two) | `Math`, `Bit Manipulation` |

solution/196.Delete Duplicate Emails/README.md

@@ -0,0 +1,59 @@
+## 删除重复的电子邮箱
+### 题目描述
+
+
+编写一个 SQL 查询，来删除 `Person` 表中所有重复的电子邮箱，重复的邮箱里只保留 **Id** 最小 的那个。
+```
++----+------------------+
+| Id | Email            |
++----+------------------+
+| 1  | [email protected] |
+| 2  | [email protected]  |
+| 3  | [email protected] |
++----+------------------+
+Id 是这个表的主键。
+```
+
+例如，在运行你的查询语句之后，上面的 `Person` 表应返回以下几行:
+```
++----+------------------+
+| Id | Email            |
++----+------------------+
+| 1  | [email protected] |
+| 2  | [email protected]  |
++----+------------------+
+```
+
+### 解法
+先根据 `Email` 分组，选出每个组中最小的 `Id`，作为一张临时表，再删除掉所有 Id 不在这张临时表的记录。
+
+```sql
+# Write your MySQL query statement below
+
+# 用这里注释的语句运行会报错，不能 select 之后再 update
+# You can't specify target table 'Person' for update in FROM clause
+# delete from Person
+# where Id not in(
+# select min(Id) as Id
+# from Person
+# group by Email);
+
+delete from Person
+where Id not in(
+    select Id from (
+        select min(id) as Id
+        from Person
+        group by Email
+    ) a
+);
+```
+
+#### Input
+```json
+{"headers": {"Person": ["Id", "Email"]}, "rows": {"Person": [[1, "[email protected]"], [2, "[email protected]"], [3, "[email protected]"]]}}
+```
+
+#### Output
+```json
+{"headers":["Id","Email"],"values":[[1,"[email protected]"],[2,"[email protected]"]]}
+```