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notes/Leetcode-Database 题解.md
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@ -1,4 +1,10 @@
<!-- GFM-TOC --> <!-- GFM-TOC -->
* [595. Big Countries](#595-big-countries)
* [627. Swap Salary](#627-swap-salary)
* [620. Not Boring Movies](#620-not-boring-movies)
* [596. Classes More Than 5 Students](#596-classes-more-than-5-students)
* [182. Duplicate Emails](#182-duplicate-emails)
* [196. Delete Duplicate Emails](#196-delete-duplicate-emails)
* [175. Combine Two Tables](#175-combine-two-tables) * [175. Combine Two Tables](#175-combine-two-tables)
* [181. Employees Earning More Than Their Managers](#181-employees-earning-more-than-their-managers) * [181. Employees Earning More Than Their Managers](#181-employees-earning-more-than-their-managers)
* [183. Customers Who Never Order](#183-customers-who-never-order) * [183. Customers Who Never Order](#183-customers-who-never-order)
@ -10,6 +16,360 @@
<!-- GFM-TOC --> <!-- GFM-TOC -->
# 595. Big Countries
https://leetcode.com/problems/big-countries/description/
## Description
```html
+-----------------+------------+------------+--------------+---------------+
| 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。
```html
+--------------+-------------+--------------+
| name | population | area |
+--------------+-------------+--------------+
| Afghanistan | 25500100 | 652230 |
| Algeria | 37100000 | 2381741 |
+--------------+-------------+--------------+
```
## SQL Schema
```sql
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
```sql
SELECT name,
population,
area
FROM
World
WHERE
area > 3000000
OR population > 25000000;
```
# 627. Swap Salary
https://leetcode.com/problems/swap-salary/description/
## Description
```html
| 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。
```html
| id | name | sex | salary |
|----|------|-----|--------|
| 1 | A | m | 2500 |
| 2 | B | f | 1500 |
| 3 | C | m | 5500 |
| 4 | D | f | 500 |
```
## SQL Schema
```sql
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
```sql
UPDATE salary
SET sex = CHAR ( ASCII(sex) ^ ASCII( 'm' ) ^ ASCII( 'f' ) );
```
# 620. Not Boring Movies
https://leetcode.com/problems/not-boring-movies/description/
## Description
邮件地址表：
```html
+---------+-----------+--------------+-----------+
| 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。
```html
+---------+-----------+--------------+-----------+
| id | movie | description | rating |
+---------+-----------+--------------+-----------+
| 5 | House card| Interesting| 9.1 |
| 1 | War | great 3D | 8.9 |
+---------+-----------+--------------+-----------+
```
## SQL Schema
```sql
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
```sql
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
```html
+---------+------------+
| 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。
```html
+---------+
| Email |
+---------+
| a@b.com |
+---------+
```
## SQL Schema
```sql
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
```sql
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：
```html
+----+---------+
| Id | Email |
+----+---------+
| 1 | a@b.com |
| 2 | c@d.com |
| 3 | a@b.com |
+----+---------+
```
查找重复的邮件地址：
```html
+---------+
| Email |
+---------+
| a@b.com |
+---------+
```
## SQL Schema
```sql
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
```sql
SELECT
Email
FROM
Person
GROUP BY
Email
HAVING
COUNT( * ) >= 2;
```
# 196. Delete Duplicate Emails
## Description
邮件地址表：
```html
+----+---------+
| Id | Email |
+----+---------+
| 1 | a@b.com |
| 2 | c@d.com |
| 3 | a@b.com |
+----+---------+
```
查找重复的邮件地址：
```html
+---------+
| Email |
+---------+
| a@b.com |
+---------+
```
## SQL Schema
与 182 相同。
## Solution
连接：
```sql
DELETE p1
FROM
Person p1,
Person p2
WHERE
p1.Email = p2.Email
AND p1.Id > p2.Id
```
子查询：
```sql
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。
```sql
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](https://stackoverflow.com/questions/45494/mysql-error-1093-cant-specify-target-table-for-update-in-from-clause)
# 175. Combine Two Tables # 175. Combine Two Tables
https://leetcode.com/problems/combine-two-tables/description/ https://leetcode.com/problems/combine-two-tables/description/
@ -75,7 +435,7 @@ SELECT
City, City,
State State
FROM FROM
Person AS P Person P
LEFT JOIN Address AS A ON P.PersonId = A.PersonId; LEFT JOIN Address AS A ON P.PersonId = A.PersonId;
``` ```
@ -103,14 +463,16 @@ Employee 表：
## SQL Schema ## SQL Schema
```sql ```sql
DROP TABLE IF EXISTS Employee; DROP TABLE
IF
EXISTS Employee;
CREATE TABLE Employee ( Id INT, NAME VARCHAR ( 255 ), Salary INT, ManagerId INT ); CREATE TABLE Employee ( Id INT, NAME VARCHAR ( 255 ), Salary INT, ManagerId INT );
INSERT INTO Employee ( Id, NAME, Salary, ManagerId ) INSERT INTO Employee ( Id, NAME, Salary, ManagerId )
VALUES VALUES
( '1', 'Joe', '70000', '3' ), ( 1, 'Joe', 70000, 3 ),
( '2', 'Henry', '80000', '4' ), ( 2, 'Henry', 80000, 4 ),
( '3', 'Sam', '60000', NULL ), ( 3, 'Sam', 60000, NULL ),
( '4', 'Max', '90000', NULL ); ( 4, 'Max', 90000, NULL );
``` ```
## Solution ## Solution
@ -119,8 +481,8 @@ VALUES
SELECT SELECT
E1.NAME AS Employee E1.NAME AS Employee
FROM FROM
Employee AS E1 Employee E1
INNER JOIN Employee AS E2 ON E1.ManagerId = E2.Id INNER JOIN Employee E2 ON E1.ManagerId = E2.Id
AND E1.Salary > E2.Salary; AND E1.Salary > E2.Salary;
``` ```
@ -168,20 +530,24 @@ Orders 表：
## SQL Schema ## SQL Schema
```sql ```sql
DROP TABLE IF EXISTS Customers; DROP TABLE
IF
EXISTS Customers;
CREATE TABLE Customers ( Id INT, NAME VARCHAR ( 255 ) ); CREATE TABLE Customers ( Id INT, NAME VARCHAR ( 255 ) );
DROP TABLE IF EXISTS Orders; DROP TABLE
IF
EXISTS Orders;
CREATE TABLE Orders ( Id INT, CustomerId INT ); CREATE TABLE Orders ( Id INT, CustomerId INT );
INSERT INTO Customers ( Id, NAME ) INSERT INTO Customers ( Id, NAME )
VALUES VALUES
( '1', 'Joe' ), ( 1, 'Joe' ),
( '2', 'Henry' ), ( 2, 'Henry' ),
( '3', 'Sam' ), ( 3, 'Sam' ),
( '4', 'Max' ); ( 4, 'Max' );
INSERT INTO Orders ( Id, CustomerId ) INSERT INTO Orders ( Id, CustomerId )
VALUES VALUES
( '1', '3' ), ( 1, 3 ),
( '2', '1' ); ( 2, 1 );
``` ```
## Solution ## Solution
@ -192,8 +558,8 @@ VALUES
SELECT SELECT
C.NAME AS Customers C.NAME AS Customers
FROM FROM
Customers AS C Customers C
LEFT JOIN Orders AS O ON C.Id = O.CustomerId LEFT JOIN Orders O ON C.Id = O.CustomerId
WHERE WHERE
O.CustomerId IS NULL; O.CustomerId IS NULL;
``` ```
@ -204,7 +570,7 @@ WHERE
SELECT SELECT
C.NAME AS Customers C.NAME AS Customers
FROM FROM
Customers AS C Customers C
WHERE WHERE
C.Id NOT IN ( SELECT CustomerId FROM Orders ); C.Id NOT IN ( SELECT CustomerId FROM Orders );
``` ```
@ -277,16 +643,16 @@ VALUES
```sql ```sql
SELECT SELECT
D.NAME AS Department, D.NAME Department,
E.NAME AS Employee, E.NAME Employee,
E.Salary E.Salary
FROM FROM
Employee AS E, Employee E,
Department AS D, Department D,
( SELECT DepartmentId, MAX( Salary ) AS Salary FROM Employee GROUP BY DepartmentId ) AS M ( SELECT DepartmentId, MAX( Salary ) Salary FROM Employee GROUP BY DepartmentId ) M
WHERE WHERE
E.DepartmentId = D.Id E.DepartmentId = D.Id
AND E.DepartmentId = M.DepartmentId AND E.DepartmentId = M.DepartmentId
AND E.Salary = M.Salary; AND E.Salary = M.Salary;
``` ```
@ -327,9 +693,9 @@ IF
CREATE TABLE Employee ( Id INT, Salary INT ); CREATE TABLE Employee ( Id INT, Salary INT );
INSERT INTO Employee ( Id, Salary ) INSERT INTO Employee ( Id, Salary )
VALUES VALUES
( '1', '100' ), ( 1, 100 ),
( '2', '200' ), ( 2, 200 ),
( '3', '300' ); ( 3, 300 );
``` ```
## Solution ## Solution
@ -338,7 +704,7 @@ VALUES
```sql ```sql
SELECT SELECT
( SELECT DISTINCT Salary FROM Employee ORDER BY Salary DESC LIMIT 1, 1 ) AS SecondHighestSalary; ( SELECT DISTINCT Salary FROM Employee ORDER BY Salary DESC LIMIT 1, 1 ) SecondHighestSalary;
``` ```
# 177. Nth Highest Salary # 177. Nth Highest Salary
@ -407,12 +773,12 @@ IF
CREATE TABLE Scores ( Id INT, Score DECIMAL ( 3, 2 ) ); CREATE TABLE Scores ( Id INT, Score DECIMAL ( 3, 2 ) );
INSERT INTO Scores ( Id, Score ) INSERT INTO Scores ( Id, Score )
VALUES VALUES
( '1', '3.5' ), ( 1, 3.5 ),
( '2', '3.65' ), ( 2, 3.65 ),
( '3', '4.0' ), ( 3, 4.0 ),
( '4', '3.85' ), ( 4, 3.85 ),
( '5', '4.0' ), ( 5, 4.0 ),
( '6', '3.65' ); ( 6, 3.65 );
``` ```
## Solution ## Solution
@ -420,10 +786,10 @@ VALUES
```sql ```sql
SELECT SELECT
S1.score, S1.score,
COUNT( DISTINCT S2.score ) AS Rank COUNT( DISTINCT S2.score ) Rank
FROM FROM
Scores AS S1 Scores S1
INNER JOIN Scores AS S2 ON S1.score <= S2.score INNER JOIN Scores S2 ON S1.score <= S2.score
GROUP BY GROUP BY
S1.id S1.id
ORDER BY ORDER BY
@ -471,26 +837,26 @@ IF
CREATE TABLE LOGS ( Id INT, Num INT ); CREATE TABLE LOGS ( Id INT, Num INT );
INSERT INTO LOGS ( Id, Num ) INSERT INTO LOGS ( Id, Num )
VALUES VALUES
( '1', '1' ), ( 1, 1 ),
( '2', '1' ), ( 2, 1 ),
( '3', '1' ), ( 3, 1 ),
( '4', '2' ), ( 4, 2 ),
( '5', '1' ), ( 5, 1 ),
( '6', '2' ), ( 6, 2 ),
( '7', '2' ); ( 7, 2 );
``` ```
## Solution ## Solution
```sql ```sql
SELECT SELECT
DISTINCT L1.num AS ConsecutiveNums DISTINCT L1.num ConsecutiveNums
FROM FROM
Logs AS L1, Logs L1,
Logs AS L2, Logs L2,
Logs AS L3 Logs L3
WHERE L1.id = l2.id - 1 WHERE L1.id = l2.id - 1
AND L2.id = L3.id - 1 AND L2.id = L3.id - 1
AND L1.num = L2.num AND L1.num = L2.num
AND l2.num = l3.num; AND l2.num = l3.num;
``` ```