Leetcode-Database 题解

技术2022-07-12 58

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 );

应该注意的是上述解法额外嵌套了一个 SELECT 语句，如果不这么做，会出现错误：You can’t specify target table ‘Person’ for update in FROM clause。以下演示了这种错误解法。

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

参考：pMySQL Error 1093 - Can’t specify target table for update in FROM clause

175. Combine Two Tables

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

Description

Person 表：

Address 表：

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

使用左外连接。

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 AS Employee 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

Customers 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

左外链接

SELECT C.Name AS Customers FROM Customers C LEFT JOIN Orders O ON C.Id = O.CustomerId WHERE O.CustomerId 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

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。

之后使用连接找到一个部门中薪资等于临时表中最大薪资的员工。

SELECT D.NAME Department, E.NAME Employee, E.Salary FROM Employee E, Department D, ( SELECT DepartmentId, MAX( Salary ) Salary FROM Employee GROUP BY DepartmentId ) M WHERE E.DepartmentId = D.Id AND E.DepartmentId = M.DepartmentId AND E.Salary = M.Salary;

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 DISTINCT 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 S1.score, COUNT( DISTINCT S2.score ) Rank FROM Scores S1 INNER JOIN Scores S2 ON S1.score <= S2.score GROUP BY S1.id, S1.score ORDER BY S1.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;

626. Exchange Seats

https://leetcode.com/problems/exchange-seats/description/

Description

seat 表存储着座位对应的学生。

+---------+---------+ | id | student | +---------+---------+ | 1 | Abbot | | 2 | Doris | | 3 | Emerson | | 4 | Green | | 5 | Jeames | +---------+---------+

要求交换相邻座位的两个学生，如果最后一个座位是奇数，那么不交换这个座位上的学生。

+---------+---------+ | id | student | +---------+---------+ | 1 | Doris | | 2 | Abbot | | 3 | Green | | 4 | Emerson | | 5 | Jeames | +---------+---------+

SQL Schema

DROP TABLE IF EXISTS seat; CREATE TABLE seat ( id INT, student VARCHAR ( 255 ) ); INSERT INTO seat ( id, student ) VALUES ( '1', 'Abbot' ), ( '2', 'Doris' ), ( '3', 'Emerson' ), ( '4', 'Green' ), ( '5', 'Jeames' );

Solution

使用多个 union。

SELECT s1.id - 1 AS id, s1.student FROM seat s1 WHERE s1.id MOD 2 = 0 UNION SELECT s2.id + 1 AS id, s2.student FROM seat s2 WHERE s2.id MOD 2 = 1 AND s2.id != ( SELECT max( s3.id ) FROM seat s3 ) UNION SELECT s4.id AS id, s4.student FROM seat s4 WHERE s4.id MOD 2 = 1 AND s4.id = ( SELECT max( s5.id ) FROM seat s5 ) ORDER BY id;

Processed: 0.010, SQL: 9