# SQL 练习
# 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 |
+--------------+-------------+--------------+
# Solution
SELECT name,
population,
area
FROM
World
WHERE
area > 3000000
OR population > 25000000;
# SQL Schema
SQL Schema 用于在本地环境下创建表结构并导入数据,从而方便在本地环境调试。
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' );
# 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 |
# Solution
两个相等的数异或的结果为 0,而 0 与任何一个数异或的结果为这个数。
sex 字段只有两个取值:'f' 和 'm',并且有以下规律:
'f' ^ ('m' ^ 'f') = 'm' ^ ('f' ^ 'f') = 'm'
'm' ^ ('m' ^ 'f') = 'f' ^ ('m' ^ 'm') = 'f'
因此将 sex 字段和 'm' ^ 'f' 进行异或操作,最后就能反转 sex 字段。
UPDATE salary
SET sex = CHAR ( ASCII(sex) ^ ASCII( 'm' ) ^ ASCII( 'f' ) );
# 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' );
# 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 |
+---------+-----------+--------------+-----------+
# Solution
SELECT
*
FROM
cinema
WHERE
id % 2 = 1
AND description != 'boring'
ORDER BY
rating DESC;
# 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 );
# 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 |
+---------+
# Solution
对 class 列进行分组之后,再使用 count 汇总函数统计每个分组的记录个数,之后使用 HAVING 进行筛选。HAVING 针对分组进行筛选,而 WHERE 针对每个记录(行)进行筛选。
SELECT
class
FROM
courses
GROUP BY
class
HAVING
count( DISTINCT student ) >= 5;
# 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' );
# 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 |
+---------+
# Solution
对 Email 进行分组,如果并使用 COUNT 进行计数统计,结果大于等于 2 的表示 Email 重复。
SELECT
Email
FROM
Person
GROUP BY
Email
HAVING
COUNT( * ) >= 2;
# 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' );
# 196. Delete Duplicate Emails
https://leetcode.com/problems/delete-duplicate-emails/description/
# Description
邮件地址表:
+----+---------+
| Id | Email |
+----+---------+
| 1 | john@example.com |
| 2 | bob@example.com |
| 3 | john@example.com |
+----+---------+
删除重复的邮件地址:
+----+------------------+
| Id | Email |
+----+------------------+
| 1 | john@example.com |
| 2 | bob@example.com |
+----+------------------+
# Solution
只保留相同 Email 中 Id 最小的那一个,然后删除其它的。
连接查询:
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 (opens new window)
# SQL Schema
与 182 相同。
# 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。
# Solution
涉及到 Person 和 Address 两个表,在对这两个表执行连接操作时,因为要保留 Person 表中的信息,即使在 Address 表中没有关联的信息也要保留。此时可以用左外连接,将 Person 表放在 LEFT JOIN 的左边。
SELECT
FirstName,
LastName,
City,
State
FROM
Person P
LEFT JOIN Address A
ON P.PersonId = A.PersonId;
# 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' );
# 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。
# Solution
SELECT
E1.NAME AS Employee
FROM
Employee E1
INNER JOIN Employee E2
ON E1.ManagerId = E2.Id
AND E1.Salary > E2.Salary;
# 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 );
# 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 |
+-----------+
# 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
);
# 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 );
# 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 |
+------------+----------+--------+
# 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;
# 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' );
# 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。
# 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;
# 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 );
# 177. Nth Highest Salary
# Description
查找工资第 N 高的员工。
# 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
# SQL Schema
同 176。
# 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 |
+-------+------+
# Solution
要统计某个 score 的排名,只要统计大于等于该 score 的 score 数量。
Id | score | 大于等于该 score 的 score 数量 | 排名 |
---|---|---|---|
1 | 4.1 | 3 | 3 |
2 | 4.2 | 2 | 2 |
3 | 4.3 | 1 | 1 |
使用连接操作找到某个 score 对应的大于等于其值的记录:
SELECT
*
FROM
Scores S1
INNER JOIN Scores S2
ON S1.score <= S2.score
ORDER BY
S1.score DESC, S1.Id;
S1.Id | S1.score | S2.Id | S2.score |
---|---|---|---|
3 | 4.3 | 3 | 4.3 |
2 | 4.2 | 2 | 4.2 |
2 | 4.2 | 3 | 4.3 |
1 | 4.1 | 1 | 4.1 |
1 | 4.1 | 2 | 4.2 |
1 | 4.1 | 3 | 4.3 |
可以看到每个 S1.score 都有对应好几条记录,我们再进行分组,并统计每个分组的数量作为 'Rank'
SELECT
S1.score 'Score',
COUNT(*) 'Rank'
FROM
Scores S1
INNER JOIN Scores S2
ON S1.score <= S2.score
GROUP BY
S1.id, S1.score
ORDER BY
S1.score DESC, S1.Id;
score | Rank |
---|---|
4.3 | 1 |
4.2 | 2 |
4.1 | 3 |
上面的解法看似没问题,但是对于以下数据,它却得到了错误的结果:
Id | score |
---|---|
1 | 4.1 |
2 | 4.2 |
3 | 4.2 |
score | Rank |
---|---|
4.2 | 2 |
4.2 | 2 |
4.1 | 3 |
而我们希望的结果为:
score | Rank |
---|---|
4.2 | 1 |
4.2 | 1 |
4.1 | 2 |
连接情况如下:
S1.Id | S1.score | S2.Id | S2.score |
---|---|---|---|
2 | 4.2 | 3 | 4.2 |
2 | 4.2 | 2 | 4.2 |
3 | 4.2 | 3 | 4.2 |
3 | 4.2 | 2 | 4.1 |
1 | 4.1 | 3 | 4.2 |
1 | 4.1 | 2 | 4.2 |
1 | 4.1 | 1 | 4.1 |
我们想要的结果是,把分数相同的放在同一个排名,并且相同分数只占一个位置,例如上面的分数,Id=2 和 Id=3 的记录都有相同的分数,并且最高,他们并列第一。而 Id=1 的记录应该排第二名,而不是第三名。所以在进行 COUNT 计数统计时,我们需要使用 COUNT( DISTINCT S2.score ) 从而只统计一次相同的分数。
SELECT
S1.score '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;
# SQL Schema
DROP TABLE
IF
EXISTS Scores;
CREATE TABLE Scores ( Id INT, Score DECIMAL ( 3, 2 ) );
INSERT INTO Scores ( Id, Score )
VALUES
( 1, 4.1 ),
( 2, 4.1 ),
( 3, 4.2 ),
( 4, 4.2 ),
( 5, 4.3 ),
( 6, 4.3 );
# 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 |
+-----------------+
# 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;
# 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 );
# 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 |
+---------+---------+
# Solution
使用多个 union。
## 处理偶数 id,让 id 减 1
## 例如 2,4,6,... 变成 1,3,5,...
SELECT
s1.id - 1 AS id,
s1.student
FROM
seat s1
WHERE
s1.id MOD 2 = 0 UNION
## 处理奇数 id,让 id 加 1。但是如果最大的 id 为奇数,则不做处理
## 例如 1,3,5,... 变成 2,4,6,...
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
## 如果最大的 id 为奇数,单独取出这个数
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;
# 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' );