Add solution 181 [sql]

inner join

Author: yanglbme

README.md

@@ -36,6 +36,7 @@ Complete [solutions](https://github.com/doocs/leetcode/tree/master/solution) to
 | 155 | [Min Stack](https://github.com/doocs/leetcode/tree/master/solution/155.Min%20Stack) | `Stack`, `Design` |
 | 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` |
 | 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/181.Employees Earning More Than Their Managers/README.md

@@ -0,0 +1,52 @@
+## 连续出现的数字
+### 题目描述
+
+`Employee` 表包含所有员工，他们的经理也属于员工。每个员工都有一个 Id，此外还有一列对应员工的经理的 Id。
+```
++----+-------+--------+-----------+
+| Id | Name  | Salary | ManagerId |
++----+-------+--------+-----------+
+| 1  | Joe   | 70000  | 3         |
+| 2  | Henry | 80000  | 4         |
+| 3  | Sam   | 60000  | NULL      |
+| 4  | Max   | 90000  | NULL      |
++----+-------+--------+-----------+
+```
+
+给定 `Employee` 表，编写一个 SQL 查询，该查询可以获取收入超过他们经理的员工的姓名。在上面的表格中，Joe 是唯一一个收入超过他的经理的员工。
+```
++----------+
+| Employee |
++----------+
+| Joe      |
++----------+
+```
+
+### 解法
+联表查询。
+
+```sql
+# Write your MySQL query statement below
+
+# select s.Name as Employee
+# from Employee s 
+# where s.ManagerId is not null and s.Salary > (select Salary from Employee where Id = s.ManagerId);
+
+
+select s1.Name as Employee
+from Employee s1
+inner join Employee s2
+on s1.ManagerId = s2.Id
+where s1.Salary > s2.Salary;
+
+```
+
+#### Input
+```json
+{"headers": {"Employee": ["Id", "Name", "Salary", "ManagerId"]}, "rows": {"Employee": [[1, "Joe", 70000, 3], [2, "Henry", 80000, 4], [3, "Sam", 60000, null], [4, "Max", 90000, null]]}}
+```
+
+#### Output
+```json
+{"headers":["Employee"],"values":[["Joe"]]}
+```