MaRui

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SQL 题目进阶

1686字约6分钟

2024-08-10

595. Big Countries

https://leetcode.com/problems/big-countries/description/

Description

+-----------------+------------+------------+--------------+---------------+
| name            | continent  | area       | population   | gdp           |
+-----------------+------------+------------+--------------+---------------+
| Afghanistan     | Asia       | 652230     | 25500100     | 20343000      |
| Albania         | Europe     | 28748      | 2831741      | 12960000      |
| Algeria         | Africa     | 2381741    | 37100000     | 188681000     |
| Andorra         | Europe     | 468        | 78115        | 3712000       |
| Angola          | Africa     | 1246700    | 20609294     | 100990000     |
+-----------------+------------+------------+--------------+---------------+

查找面积超过 3,000,000 或者人口数超过 25,000,000 的国家。

+--------------+-------------+--------------+
| name         | population  | area         |
+--------------+-------------+--------------+
| Afghanistan  | 25500100    | 652230       |
| Algeria      | 37100000    | 2381741      |
+--------------+-------------+--------------+

SQL Schema

DROP TABLE
IF
    EXISTS World;
CREATE TABLE World ( NAME VARCHAR ( 255 ), continent VARCHAR ( 255 ), area INT, population INT, gdp INT );
INSERT INTO World ( NAME, continent, area, population, gdp )
VALUES
    ( 'Afghanistan', 'Asia', '652230', '25500100', '203430000' ),
    ( 'Albania', 'Europe', '28748', '2831741', '129600000' ),
    ( 'Algeria', 'Africa', '2381741', '37100000', '1886810000' ),
    ( 'Andorra', 'Europe', '468', '78115', '37120000' ),
    ( 'Angola', 'Africa', '1246700', '20609294', '1009900000' );

Solution

SELECT name,
population,
area
FROM world
WHERE area > 3000000 
      or population > 25000000

627. Swap Salary

https://leetcode.com/problems/swap-salary/description/

Description

| id | name | sex | salary |
|----|------|-----|--------|
| 1  | A    | m   | 2500   |
| 2  | B    | f   | 1500   |
| 3  | C    | m   | 5500   |
| 4  | D    | f   | 500    |

只用一个 SQL 查询，将 sex 字段反转。

| id | name | sex | salary |
|----|------|-----|--------|
| 1  | A    | f   | 2500   |
| 2  | B    | m   | 1500   |
| 3  | C    | f   | 5500   |
| 4  | D    | m   | 500    |

SQL Schema

DROP TABLE
IF
    EXISTS salary;
CREATE TABLE salary ( id INT, NAME VARCHAR ( 100 ), sex CHAR ( 1 ), salary INT );
INSERT INTO salary ( id, NAME, sex, salary )
VALUES
    ( '1', 'A', 'm', '2500' ),
    ( '2', 'B', 'f', '1500' ),
    ( '3', 'C', 'm', '5500' ),
    ( '4', 'D', 'f', '500' );

Solution

UPDATE salary
SET sex = CHAR ( ASCII(sex) ^ ASCII( 'm' ) ^ ASCII( 'f' ) );

620. Not Boring Movies

https://leetcode.com/problems/not-boring-movies/description/

Description

+---------+-----------+--------------+-----------+
|   id    | movie     |  description |  rating   |
+---------+-----------+--------------+-----------+
|   1     | War       |   great 3D   |   8.9     |
|   2     | Science   |   fiction    |   8.5     |
|   3     | irish     |   boring     |   6.2     |
|   4     | Ice song  |   Fantacy    |   8.6     |
|   5     | House card|   Interesting|   9.1     |
+---------+-----------+--------------+-----------+

查找 id 为奇数，并且 description 不是 boring 的电影，按 rating 降序。

+---------+-----------+--------------+-----------+
|   id    | movie     |  description |  rating   |
+---------+-----------+--------------+-----------+
|   5     | House card|   Interesting|   9.1     |
|   1     | War       |   great 3D   |   8.9     |
+---------+-----------+--------------+-----------+

SQL Schema

DROP TABLE
IF
    EXISTS cinema;
CREATE TABLE cinema ( id INT, movie VARCHAR ( 255 ), description VARCHAR ( 255 ), rating FLOAT ( 2, 1 ) );
INSERT INTO cinema ( id, movie, description, rating )
VALUES
    ( 1, 'War', 'great 3D', 8.9 ),
    ( 2, 'Science', 'fiction', 8.5 ),
    ( 3, 'irish', 'boring', 6.2 ),
    ( 4, 'Ice song', 'Fantacy', 8.6 ),
    ( 5, 'House card', 'Interesting', 9.1 );

Solution

SELECT
	*
FROM 
	cinema
WHERE 
	id % 2 = 1 
	and description != 'boring'
ORDER BY 
	rating desc;

596. Classes More Than 5 Students

https://leetcode.com/problems/classes-more-than-5-students/description/

Description

+---------+------------+
| student | class      |
+---------+------------+
| A       | Math       |
| B       | English    |
| C       | Math       |
| D       | Biology    |
| E       | Math       |
| F       | Computer   |
| G       | Math       |
| H       | Math       |
| I       | Math       |
+---------+------------+

查找有五名及以上 student 的 class。

+---------+
| class   |
+---------+
| Math    |
+---------+

SQL Schema

DROP TABLE
IF
    EXISTS courses;
CREATE TABLE courses ( student VARCHAR ( 255 ), class VARCHAR ( 255 ) );
INSERT INTO courses ( student, class )
VALUES
    ( 'A', 'Math' ),
    ( 'B', 'English' ),
    ( 'C', 'Math' ),
    ( 'D', 'Biology' ),
    ( 'E', 'Math' ),
    ( 'F', 'Computer' ),
    ( 'G', 'Math' ),
    ( 'H', 'Math' ),
    ( 'I', 'Math' );

Solution

SELECT
	class
FROM 
    courses
GROUP BY 
    class
HAVING 
    count(DISTINCT student) >= 5;

182. Duplicate Emails

https://leetcode.com/problems/duplicate-emails/description/

Description 邮件地址表:

+----+---------+
| Id | Email   |
+----+---------+
| 1  | a@b.com |
| 2  | c@d.com |
| 3  | a@b.com |
+----+---------+

查找重复的邮件地址:

+---------+
| Email   |
+---------+
| a@b.com |
+---------+

SQL Schema

DROP TABLE
IF
    EXISTS Person;
CREATE TABLE Person ( Id INT, Email VARCHAR ( 255 ) );
INSERT INTO Person ( Id, Email )
VALUES
    ( 1, 'a@b.com' ),
    ( 2, 'c@d.com' ),
    ( 3, 'a@b.com' );

Solution

SELECT
	Email
FROM
	person
GROUP BY Email
HAVING count(*) >= 2;

196. Delete Duplicate Emails

https://leetcode.com/problems/delete-duplicate-emails/description/

Description 邮件地址表:

+----+---------+
| Id | Email   |
+----+---------+
| 1  | a@b.com |
| 2  | c@d.com |
| 3  | a@b.com |
+----+---------+

删除重复的邮件地址:

+----+------------------+
| Id | Email            |
+----+------------------+
| 1  | john@example.com |
| 2  | bob@example.com  |
+----+------------------+

SQL Schema

与 182 相同。

Solution

连接:

DELETE p1
FROM
    Person p1,
    Person p2
WHERE
    p1.Email = p2.Email
    AND p1.Id > p2.Id

子查询:

DELETE
FROM
    Person
WHERE
    id NOT IN ( SELECT id FROM ( SELECT min( id ) AS id FROM Person GROUP BY email ) AS m );

175. Combine Two Tables

https://leetcode.com/problems/combine-two-tables/description/

Description Person 表:

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| PersonId    | int     |
| FirstName   | varchar |
| LastName    | varchar |
+-------------+---------+
PersonId is the primary key column for this table.

Address 表:

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| AddressId   | int     |
| PersonId    | int     |
| City        | varchar |
| State       | varchar |
+-------------+---------+
AddressId is the primary key column for this table.

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

DROP TABLE
IF
    EXISTS Person;
CREATE TABLE Person ( PersonId INT, FirstName VARCHAR ( 255 ), LastName VARCHAR ( 255 ) );
DROP TABLE
IF
    EXISTS Address;
CREATE TABLE Address ( AddressId INT, PersonId INT, City VARCHAR ( 255 ), State VARCHAR ( 255 ) );
INSERT INTO Person ( PersonId, LastName, FirstName )
VALUES
    ( 1, 'Wang', 'Allen' );
INSERT INTO Address ( AddressId, PersonId, City, State )
VALUES
    ( 1, 2, 'New York City', 'New York' );

Solution

SELECT
	FirstName, 
	LastName, 
	City, 
	State
FROM person p
LEFT JOIN address 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:

+----+-------+--------+-----------+
| Id | Name  | Salary | ManagerId |
+----+-------+--------+-----------+
| 1  | Joe   | 70000  | 3         |
| 2  | Henry | 80000  | 4         |
| 3  | Sam   | 60000  | NULL      |
| 4  | Max   | 90000  | NULL      |
+----+-------+--------+-----------+

查找薪资大于其经理薪资的员工信息。

SQL Schema

DROP TABLE
IF
    EXISTS 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

SELECT
 e1.name
FROM 
	employee e1
inner join employee e2 on 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:

+----+-------+
| Id | Name  |
+----+-------+
| 1  | Joe   |
| 2  | Henry |
| 3  | Sam   |
| 4  | Max   |
+----+-------+

Orders 表:

+----+------------+
| Id | CustomerId |
+----+------------+
| 1  | 3          |
| 2  | 1          |
+----+------------+

查找没有订单的顾客信息:

+-----------+
| Customers |
+-----------+
| Henry     |
| Max       |
+-----------+

SQL Schema

DROP TABLE
IF
    EXISTS Customers;
CREATE TABLE Customers ( Id INT, NAME VARCHAR ( 255 ) );
DROP TABLE
IF
    EXISTS 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

左外连接

SELECT
	c.name as customers
FROM customers c
LEFT JOIN orders o on c.Id = o.CustomerId
WHERE o.id is null;

子查询

SELECT
	name as customers
FROM customers
WHERE id not in (SELECT CustomerId FROM orders);

184. Department Highest Salary

https://leetcode.com/problems/department-highest-salary/description/

Description Employee 表:

+----+-------+--------+--------------+
| Id | Name  | Salary | DepartmentId |
+----+-------+--------+--------------+
| 1  | Joe   | 70000  | 1            |
| 2  | Henry | 80000  | 2            |
| 3  | Sam   | 60000  | 2            |
| 4  | Max   | 90000  | 1            |
+----+-------+--------+--------------+

Department 表:

+----+----------+
| Id | Name     |
+----+----------+
| 1  | IT       |
| 2  | Sales    |
+----+----------+

查找一个 Department 中收入最高者的信息:

+------------+----------+--------+
| Department | Employee | Salary |
+------------+----------+--------+
| IT         | Max      | 90000  |
| Sales      | Henry    | 80000  |
+------------+----------+--------+

SQL Schema

DROP TABLE IF EXISTS Employee;
CREATE TABLE Employee ( Id INT, NAME VARCHAR ( 255 ), Salary INT, DepartmentId INT );
DROP TABLE IF EXISTS 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

SELECT
 d.name as Department,
 e1.name as Employee,
 e1.Salary as Salary
FROM
	employee e1
inner join (
	SELECT departmentId,max(salary) salary
	FROM employee GROUP BY departmentId) e2 on e1.departmentId = e2.departmentId and e1.Salary = e2.salary
inner join department d on e1.departmentId = d.id;

176. Second Highest Salary

https://leetcode.com/problems/second-highest-salary/description/

Description

+----+--------+
| Id | Salary |
+----+--------+
| 1  | 100    |
| 2  | 200    |
| 3  | 300    |
+----+--------+

查找工资第二高的员工。

+---------------------+
| SecondHighestSalary |
+---------------------+
| 200                 |
+---------------------+

没有找到返回 null 而不是不返回数据。

SQL Schema

DROP TABLE
IF
    EXISTS Employee;
CREATE TABLE Employee ( Id INT, Salary INT );
INSERT INTO Employee ( Id, Salary )
VALUES
    ( 1, 100 ),
    ( 2, 200 ),
    ( 3, 300 );

Solution

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。

SELECT
	(SELECT salary FROM employee ORDER BY Salary desc limit 1,1) secondHighestSalary;

177. Nth Highest Salary

Description

查找工资第 N 高的员工。

SQL Schema

同 176。

Solution

CREATE FUNCTION getNthHighestSalary ( N INT ) RETURNS INT BEGIN

SET N = N - 1;
RETURN ( SELECT ( SELECT DISTINCT Salary FROM Employee ORDER BY Salary DESC LIMIT N, 1 ) );

END

178. Rank Scores

https://leetcode.com/problems/rank-scores/description/

Description 得分表:

+----+-------+
| Id | Score |
+----+-------+
| 1  | 3.50  |
| 2  | 3.65  |
| 3  | 4.00  |
| 4  | 3.85  |
| 5  | 4.00  |
| 6  | 3.65  |
+----+-------+

将得分排序，并统计排名。

+-------+------+
| Score | Rank |
+-------+------+
| 4.00  | 1    |
| 4.00  | 1    |
| 3.85  | 2    |
| 3.65  | 3    |
| 3.65  | 3    |
| 3.50  | 4    |
+-------+------+

SQL Schema

DROP TABLE
IF
    EXISTS Scores;
CREATE TABLE Scores ( Id INT, Score DECIMAL ( 3, 2 ) );
INSERT INTO Scores ( Id, Score )
VALUES
    ( 1, 3.5 ),
    ( 2, 3.65 ),
    ( 3, 4.0 ),
    ( 4, 3.85 ),
    ( 5, 4.0 ),
    ( 6, 3.65 );

Solution

SELECT
	Score,
	@rank:=@rank+1 as rank
FROM
	scores
ORDER BY score desc

180. Consecutive Numbers

https://leetcode.com/problems/consecutive-numbers/description/

Description 数字表:

+----+-----+
| Id | Num |
+----+-----+
| 1  |  1  |
| 2  |  1  |
| 3  |  1  |
| 4  |  2  |
| 5  |  1  |
| 6  |  2  |
| 7  |  2  |
+----+-----+

查找连续出现三次的数字。

+-----------------+
| ConsecutiveNums |
+-----------------+
| 1               |
+-----------------+

SQL Schema

DROP TABLE
IF
    EXISTS LOGS;
CREATE TABLE LOGS ( Id INT, Num INT );
INSERT INTO LOGS ( Id, Num )
VALUES
    ( 1, 1 ),
    ( 2, 1 ),
    ( 3, 1 ),
    ( 4, 2 ),
    ( 5, 1 ),
    ( 6, 2 ),
    ( 7, 2 );

Solution

SELECT
    DISTINCT L1.num ConsecutiveNums
FROM
    Logs L1,
    Logs L2,
    Logs L3
WHERE L1.id = l2.id - 1
    AND L2.id = L3.id - 1
    AND L1.num = L2.num
    AND l2.num = l3.num;