auto commit

notes/Leetcode-Database 题解.md

 <!-- GFM-TOC -->
 * [175. Combine Two Tables](#175-combine-two-tables)
+* [181. Employees Earning More Than Their Managers](#181-employees-earning-more-than-their-managers)
+* [183. Customers Who Never Order](#183-customers-who-never-order)
+* [184. Department Highest Salary](#184-department-highest-salary)
 * [未完待续...](#未完待续)
 <!-- GFM-TOC -->
 
@@ -8,9 +11,9 @@
 
 https://leetcode.com/problems/combine-two-tables/description/
 
-## Describe
+## Description
 
-Table: Person
+Person 表：
 
 ```html
 +-------------+---------+
@@ -23,7 +26,7 @@ Table: Person
 PersonId is the primary key column for this table.
 ```
 
-Table: Address
+Address 表：
 
 ```html
 +-------------+---------+
@@ -37,18 +40,18 @@ Table: Address
 AddressId is the primary key column for this table.
 ```
 
-取出 FirstName, LastName, City, State 数据，而不管一个用户有没有填地址信息。
+查找 FirstName, LastName, City, State 数据，而不管一个用户有没有填地址信息。
 
 ## SQL Schema
 
 ```sql
+DROP TABLE Person;
 CREATE TABLE Person ( PersonId INT, FirstName VARCHAR ( 255 ), LastName VARCHAR ( 255 ) );
+DROP TABLE Address;
 CREATE TABLE Address ( AddressId INT, PersonId INT, City VARCHAR ( 255 ), State VARCHAR ( 255 ) );
-TRUNCATE TABLE Person;
 INSERT INTO Person ( PersonId, LastName, FirstName )
 VALUES
     ( 1, 'Wang', 'Allen' );
-TRUNCATE TABLE Address;
 INSERT INTO Address ( AddressId, PersonId, City, State )
 VALUES
     ( 1, 2, 'New York City', 'New York' );
@@ -60,8 +63,205 @@ VALUES
 
 ```sql
 SELECT FirstName, LastName, City, State
-FROM Person AS p LEFT JOIN address AS a
-ON p.PersonId = a.PersonId;
+FROM Person AS P LEFT JOIN Address AS A
+ON P.PersonId = A.PersonId;
+```
+
+# 181. Employees Earning More Than Their Managers
+
+https://leetcode.com/problems/employees-earning-more-than-their-managers/description/
+
+## Description
+
+Employee 表：
+
+```html
++----+-------+--------+-----------+
+| Id | Name  | Salary | ManagerId |
++----+-------+--------+-----------+
+| 1  | Joe   | 70000  | 3         |
+| 2  | Henry | 80000  | 4         |
+| 3  | Sam   | 60000  | NULL      |
+| 4  | Max   | 90000  | NULL      |
++----+-------+--------+-----------+
+```
+
+查找所有员工，它们的薪资大于其经理薪资。
+
+## SQL Schema
+
+```sql
+DROP TABLE Employee;
+CREATE TABLE Employee ( Id INT, NAME VARCHAR ( 255 ), Salary INT, ManagerId INT );
+INSERT INTO Employee ( Id, NAME, Salary, ManagerId )
+VALUES
+    ( '1', 'Joe', '70000', '3' ),
+    ( '2', 'Henry', '80000', '4' ),
+    ( '3', 'Sam', '60000', NULL ),
+    ( '4', 'Max', '90000', NULL );
+```
+
+## Solution
+
+```sql
+SELECT E1.Name AS Employee
+FROM Employee AS E1, Employee AS E2
+WHERE E1.ManagerId = E2.Id AND E1.Salary > E2.Salary;
+```
+
+# 183. Customers Who Never Order
+
+https://leetcode.com/problems/customers-who-never-order/description/
+
+## Description
+
+Curstomers 表：
+
+```html
++----+-------+
+| Id | Name  |
++----+-------+
+| 1  | Joe   |
+| 2  | Henry |
+| 3  | Sam   |
+| 4  | Max   |
++----+-------+
+```
+
+Orders 表：
+
+```html
++----+------------+
+| Id | CustomerId |
++----+------------+
+| 1  | 3          |
+| 2  | 1          |
++----+------------+
+```
+
+查找没有订单的顾客信息：
+
+```html
++-----------+
+| Customers |
++-----------+
+| Henry     |
+| Max       |
++-----------+
+```
+
+## SQL Schema
+
+```sql
+DROP TABLE Customers;
+CREATE TABLE Customers ( Id INT, NAME VARCHAR ( 255 ) );
+DROP TABLE Orders;
+CREATE TABLE Orders ( Id INT, CustomerId INT );
+INSERT INTO Customers ( Id, NAME )
+VALUES
+    ( '1', 'Joe' ),
+    ( '2', 'Henry' ),
+    ( '3', 'Sam' ),
+    ( '4', 'Max' );
+INSERT INTO Orders ( Id, CustomerId )
+VALUES
+    ( '1', '3' ),
+    ( '2', '1' );
+```
+
+## Solution
+
+左外链接
+
+```sql
+SELECT C.Name AS Customers
+FROM Customers AS C LEFT JOIN Orders AS O
+ON C.Id = O.CustomerId
+WHERE O.CustomerId IS NULL;
+```
+
+子查询
+
+```sql
+SELECT C.Name AS Customers
+FROM Customers AS C
+WHERE C.Id NOT IN (
+    SELECT CustomerId
+    FROM Orders
+);
+```
+
+# 184. Department Highest Salary
+
+https://leetcode.com/problems/department-highest-salary/description/
+
+## Description
+
+Employee 表：
+
+```html
++----+-------+--------+--------------+
+| Id | Name  | Salary | DepartmentId |
++----+-------+--------+--------------+
+| 1  | Joe   | 70000  | 1            |
+| 2  | Henry | 80000  | 2            |
+| 3  | Sam   | 60000  | 2            |
+| 4  | Max   | 90000  | 1            |
++----+-------+--------+--------------+
+```
+
+Department 表：
+
+```html
++----+----------+
+| Id | Name     |
++----+----------+
+| 1  | IT       |
+| 2  | Sales    |
++----+----------+
+```
+
+查找一个 Department 中收入最高者的信息：
+
+```html
++------------+----------+--------+
+| Department | Employee | Salary |
++------------+----------+--------+
+| IT         | Max      | 90000  |
+| Sales      | Henry    | 80000  |
++------------+----------+--------+
+```
+
+## SQL Schema
+
+```sql
+DROP TABLE Employee;
+CREATE TABLE Employee ( Id INT, NAME VARCHAR ( 255 ), Salary INT, DepartmentId INT );
+DROP TABLE Department;
+CREATE TABLE Department ( Id INT, NAME VARCHAR ( 255 ) );
+INSERT INTO Employee ( Id, NAME, Salary, DepartmentId )
+VALUES
+    ( 1, 'Joe', 70000, 1 ),
+    ( 2, 'Henry', 80000, 2 ),
+    ( 3, 'Sam', 60000, 2 ),
+    ( 4, 'Max', 90000, 1 );
+INSERT INTO Department ( Id, NAME )
+VALUES
+    ( 1, 'IT' ),
+    ( 2, 'Sales' );
+```
+
+## Solution
+
+```sql
+SELECT D.Name AS Department, E.Name AS Employee, E.Salary
+FROM Employee AS E, Department AS D,
+    (SELECT DepartmentId, MAX(Salary) AS Salary
+    FROM Employee
+    GROUP BY DepartmentId) AS M
+WHERE E.DepartmentId = D.Id
+    AND E.DepartmentId = M.DepartmentId
+    AND E.Salary = M.Salary;
 ```
 
 # 未完待续...