497 lines
9.1 KiB
Markdown
497 lines
9.1 KiB
Markdown
<!-- GFM-TOC -->
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* [175. Combine Two Tables](#175-combine-two-tables)
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* [181. Employees Earning More Than Their Managers](#181-employees-earning-more-than-their-managers)
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* [183. Customers Who Never Order](#183-customers-who-never-order)
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* [184. Department Highest Salary](#184-department-highest-salary)
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* [176. Second Highest Salary](#176-second-highest-salary)
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* [177. Nth Highest Salary](#177-nth-highest-salary)
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* [178. Rank Scores](#178-rank-scores)
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* [180. Consecutive Numbers](#180-consecutive-numbers)
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<!-- GFM-TOC -->
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# 175. Combine Two Tables
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https://leetcode.com/problems/combine-two-tables/description/
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## Description
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Person 表：
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```html
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+-------------+---------+
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| Column Name | Type |
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+-------------+---------+
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| PersonId | int |
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| FirstName | varchar |
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| LastName | varchar |
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+-------------+---------+
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PersonId is the primary key column for this table.
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```
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Address 表：
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```html
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+-------------+---------+
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| Column Name | Type |
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+-------------+---------+
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| AddressId | int |
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| PersonId | int |
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| City | varchar |
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| State | varchar |
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+-------------+---------+
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AddressId is the primary key column for this table.
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```
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查找 FirstName, LastName, City, State 数据，而不管一个用户有没有填地址信息。
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## SQL Schema
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```sql
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DROP TABLE
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IF
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EXISTS Person;
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CREATE TABLE Person ( PersonId INT, FirstName VARCHAR ( 255 ), LastName VARCHAR ( 255 ) );
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DROP TABLE
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IF
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EXISTS Address;
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CREATE TABLE Address ( AddressId INT, PersonId INT, City VARCHAR ( 255 ), State VARCHAR ( 255 ) );
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INSERT INTO Person ( PersonId, LastName, FirstName )
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VALUES
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( 1, 'Wang', 'Allen' );
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INSERT INTO Address ( AddressId, PersonId, City, State )
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VALUES
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( 1, 2, 'New York City', 'New York' );
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```
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## Solution
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使用左外连接。
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```sql
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SELECT
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FirstName,
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LastName,
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City,
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State
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FROM
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Person AS P
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LEFT JOIN Address AS A ON P.PersonId = A.PersonId;
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```
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# 181. Employees Earning More Than Their Managers
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https://leetcode.com/problems/employees-earning-more-than-their-managers/description/
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## Description
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Employee 表：
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```html
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+----+-------+--------+-----------+
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| Id | Name | Salary | ManagerId |
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+----+-------+--------+-----------+
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| 1 | Joe | 70000 | 3 |
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| 2 | Henry | 80000 | 4 |
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| 3 | Sam | 60000 | NULL |
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| 4 | Max | 90000 | NULL |
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+----+-------+--------+-----------+
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```
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查找所有员工，他们的薪资大于其经理薪资。
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## SQL Schema
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```sql
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DROP TABLE IF EXISTS Employee;
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CREATE TABLE Employee ( Id INT, NAME VARCHAR ( 255 ), Salary INT, ManagerId INT );
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INSERT INTO Employee ( Id, NAME, Salary, ManagerId )
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VALUES
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( '1', 'Joe', '70000', '3' ),
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( '2', 'Henry', '80000', '4' ),
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( '3', 'Sam', '60000', NULL ),
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( '4', 'Max', '90000', NULL );
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```
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## Solution
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```sql
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SELECT
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E1.NAME AS Employee
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FROM
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Employee AS E1
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INNER JOIN Employee AS E2 ON E1.ManagerId = E2.Id
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AND E1.Salary > E2.Salary;
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```
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# 183. Customers Who Never Order
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https://leetcode.com/problems/customers-who-never-order/description/
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## Description
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Curstomers 表：
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```html
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+----+-------+
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| Id | Name |
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+----+-------+
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| 1 | Joe |
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| 2 | Henry |
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| 3 | Sam |
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| 4 | Max |
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+----+-------+
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```
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Orders 表：
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```html
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+----+------------+
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| Id | CustomerId |
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+----+------------+
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| 1 | 3 |
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| 2 | 1 |
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+----+------------+
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```
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查找没有订单的顾客信息：
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```html
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+-----------+
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| Customers |
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+-----------+
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| Henry |
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| Max |
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+-----------+
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```
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## SQL Schema
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```sql
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DROP TABLE IF EXISTS Customers;
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CREATE TABLE Customers ( Id INT, NAME VARCHAR ( 255 ) );
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DROP TABLE IF EXISTS Orders;
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CREATE TABLE Orders ( Id INT, CustomerId INT );
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INSERT INTO Customers ( Id, NAME )
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VALUES
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( '1', 'Joe' ),
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( '2', 'Henry' ),
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( '3', 'Sam' ),
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( '4', 'Max' );
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INSERT INTO Orders ( Id, CustomerId )
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VALUES
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( '1', '3' ),
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( '2', '1' );
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```
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## Solution
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左外链接
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```sql
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SELECT
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C.NAME AS Customers
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FROM
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Customers AS C
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LEFT JOIN Orders AS O ON C.Id = O.CustomerId
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WHERE
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O.CustomerId IS NULL;
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```
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子查询
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```sql
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SELECT
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C.NAME AS Customers
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FROM
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Customers AS C
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WHERE
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C.Id NOT IN ( SELECT CustomerId FROM Orders );
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```
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# 184. Department Highest Salary
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https://leetcode.com/problems/department-highest-salary/description/
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## Description
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Employee 表：
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```html
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+----+-------+--------+--------------+
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| Id | Name | Salary | DepartmentId |
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+----+-------+--------+--------------+
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| 1 | Joe | 70000 | 1 |
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| 2 | Henry | 80000 | 2 |
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| 3 | Sam | 60000 | 2 |
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| 4 | Max | 90000 | 1 |
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+----+-------+--------+--------------+
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```
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Department 表：
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```html
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+----+----------+
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| Id | Name |
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+----+----------+
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| 1 | IT |
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| 2 | Sales |
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+----+----------+
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```
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查找一个 Department 中收入最高者的信息：
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```html
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+------------+----------+--------+
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| Department | Employee | Salary |
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+------------+----------+--------+
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| IT | Max | 90000 |
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| Sales | Henry | 80000 |
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+------------+----------+--------+
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```
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## SQL Schema
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```sql
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DROP TABLE IF EXISTS Employee;
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CREATE TABLE Employee ( Id INT, NAME VARCHAR ( 255 ), Salary INT, DepartmentId INT );
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DROP TABLE IF EXISTS Department;
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CREATE TABLE Department ( Id INT, NAME VARCHAR ( 255 ) );
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INSERT INTO Employee ( Id, NAME, Salary, DepartmentId )
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VALUES
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( 1, 'Joe', 70000, 1 ),
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( 2, 'Henry', 80000, 2 ),
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( 3, 'Sam', 60000, 2 ),
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( 4, 'Max', 90000, 1 );
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INSERT INTO Department ( Id, NAME )
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VALUES
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( 1, 'IT' ),
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( 2, 'Sales' );
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```
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## Solution
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创建一个临时表，包含了部门员工的最大薪资。可以对部门进行分组，然后使用 MAX() 汇总函数取得最大薪资。
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之后使用连接将找到一个部门中薪资等于临时表中最大薪资的员工。
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```sql
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SELECT
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D.NAME AS Department,
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E.NAME AS Employee,
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E.Salary
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FROM
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Employee AS E,
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Department AS D,
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( SELECT DepartmentId, MAX( Salary ) AS Salary FROM Employee GROUP BY DepartmentId ) AS M
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WHERE
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E.DepartmentId = D.Id
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AND E.DepartmentId = M.DepartmentId
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AND E.Salary = M.Salary;
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```
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# 176. Second Highest Salary
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https://leetcode.com/problems/second-highest-salary/description/
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## Description
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```html
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+----+--------+
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| Id | Salary |
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+----+--------+
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| 1 | 100 |
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| 2 | 200 |
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| 3 | 300 |
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+----+--------+
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```
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查找工资第二高的员工。
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```html
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+---------------------+
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| SecondHighestSalary |
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+---------------------+
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| 200 |
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+---------------------+
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```
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如果没有找到，那么就返回 null 而不是不返回数据。
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## SQL Schema
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```sql
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DROP TABLE
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IF
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EXISTS Employee;
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CREATE TABLE Employee ( Id INT, Salary INT );
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INSERT INTO Employee ( Id, Salary )
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VALUES
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( '1', '100' ),
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( '2', '200' ),
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( '3', '300' );
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```
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## Solution
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为了在没有查找到数据时返回 null，需要在查询结果外面再套一层 SELECT。
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```sql
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SELECT
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( SELECT DISTINCT Salary FROM Employee ORDER BY Salary DESC LIMIT 1, 1 ) AS SecondHighestSalary;
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```
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# 177. Nth Highest Salary
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## Description
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查找工资第 N 高的员工。
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## SQL Schema
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同 176。
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## Solution
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```sql
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CREATE FUNCTION getNthHighestSalary ( N INT ) RETURNS INT BEGIN
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SET N = N - 1;
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RETURN ( SELECT ( SELECT DISTINCT Salary FROM Employee ORDER BY Salary DESC LIMIT N, 1 ) );
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END
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```
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# 178. Rank Scores
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https://leetcode.com/problems/rank-scores/description/
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## Description
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得分表：
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```html
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+----+-------+
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| Id | Score |
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+----+-------+
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| 1 | 3.50 |
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| 2 | 3.65 |
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| 3 | 4.00 |
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| 4 | 3.85 |
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| 5 | 4.00 |
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| 6 | 3.65 |
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+----+-------+
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```
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将得分排序，并统计排名。
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```html
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+-------+------+
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| Score | Rank |
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+-------+------+
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| 4.00 | 1 |
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| 4.00 | 1 |
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| 3.85 | 2 |
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| 3.65 | 3 |
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| 3.65 | 3 |
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| 3.50 | 4 |
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+-------+------+
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```
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## SQL Schema
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```sql
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DROP TABLE
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IF
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EXISTS Scores;
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CREATE TABLE Scores ( Id INT, Score DECIMAL ( 3, 2 ) );
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INSERT INTO Scores ( Id, Score )
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VALUES
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( '1', '3.5' ),
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( '2', '3.65' ),
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( '3', '4.0' ),
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( '4', '3.85' ),
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( '5', '4.0' ),
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( '6', '3.65' );
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```
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## Solution
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```sql
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SELECT
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S1.score,
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COUNT( DISTINCT S2.score ) AS Rank
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FROM
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Scores AS S1
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INNER JOIN Scores AS S2 ON S1.score <= S2.score
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GROUP BY
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S1.id
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ORDER BY
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S1.score DESC;
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```
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# 180. Consecutive Numbers
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https://leetcode.com/problems/consecutive-numbers/description/
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## Description
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数字表：
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```html
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+----+-----+
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| Id | Num |
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+----+-----+
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| 1 | 1 |
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| 2 | 1 |
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| 3 | 1 |
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| 4 | 2 |
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| 5 | 1 |
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| 6 | 2 |
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| 7 | 2 |
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+----+-----+
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```
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查找连续出现三次的数字。
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```html
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+-----------------+
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| ConsecutiveNums |
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+-----------------+
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| 1 |
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+-----------------+
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```
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## SQL Schema
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```sql
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DROP TABLE
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IF
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EXISTS LOGS;
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CREATE TABLE LOGS ( Id INT, Num INT );
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INSERT INTO LOGS ( Id, Num )
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VALUES
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( '1', '1' ),
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( '2', '1' ),
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( '3', '1' ),
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( '4', '2' ),
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( '5', '1' ),
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( '6', '2' ),
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( '7', '2' );
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```
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## Solution
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```sql
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SELECT
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DISTINCT L1.num AS ConsecutiveNums
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FROM
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Logs AS L1,
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Logs AS L2,
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Logs AS L3
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WHERE L1.id = l2.id - 1
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AND L2.id = L3.id - 1
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AND L1.num = L2.num
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AND l2.num = l3.num;
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```
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