Add solution 182 [sql]

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Author: yanglbme

README.md

@@ -37,6 +37,7 @@ Complete [solutions](https://github.com/doocs/leetcode/tree/master/solution) to
 | 175 | [Combine Two Tables](https://github.com/doocs/leetcode/tree/master/solution/175.Combine%20Two%20Tables) | `SQL` |
 | 176 | [Second Highest Salary](https://github.com/doocs/leetcode/tree/master/solution/176.Second%20Highest%20Salary) | `SQL` |
 | 181 | [Employees Earning More Than Their Managers](https://github.com/doocs/leetcode/tree/master/solution/181.Employees%20Earning%20More%20Than%20Their%20Managers) | `SQL` |
+| 182 | [Duplicate Emails](https://github.com/doocs/leetcode/tree/master/solution/182.Duplicate%20Emails) | `SQL` |
 | 189 | [Rotate Array](https://github.com/doocs/leetcode/tree/master/solution/189.Rotate%20Array) | `Array` |
 | 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` |

solution/182.Duplicate Emails/README.md

@@ -0,0 +1,47 @@
+## 查找重复的电子邮箱
+### 题目描述
+
+编写一个 SQL 查询，查找 `Person` 表中所有重复的电子邮箱。
+
+示例：
+```
++----+---------+
+| Id | Email   |
++----+---------+
+| 1  | [email protected] |
+| 2  | [email protected] |
+| 3  | [email protected] |
++----+---------+
+```
+
+根据以上输入，你的查询应返回以下结果：
+```
++---------+
+| Email   |
++---------+
+| [email protected] |
++---------+
+```
+
+**说明：**所有电子邮箱都是小写字母。
+
+### 解法
+对 `Email` 进行分组，选出数量大于 1 的 `Email` 即可。
+
+```sql
+# Write your MySQL query statement below
+select Email 
+from Person 
+group by Email 
+having count(Email) > 1;
+```
+
+#### Input
+```json
+{"headers": {"Person": ["Id", "Email"]}, "rows": {"Person": [[1, "[email protected]"], [2, "[email protected]"], [3, "[email protected]"]]}}
+```
+
+#### Output
+```json
+{"headers":["Email"],"values":[["[email protected]"]]}
+```