update codes

2020-03-04 22:21:02 +08:00 · 2020-03-04 22:21:02 +08:00 · 1923e6e39b
parent 8e9e519adb
commit 1923e6e39b
24 changed files with 665 additions and 233 deletions
--- a/codes/mysql/Leetcode之SQL题/README.md
+++ b/codes/mysql/Leetcode之SQL题/README.md
@ -3,19 +3,21 @@
 > [题库地址](https://leetcode.com/problemset/database/)

 - **难度：简单**
-  - [组合两个表](easy/combine-two-tables.sql)
-  - [第二高的薪水](easy/second-highest-salary.sql)
-  - [超过经理收入的员工](easy/employees-earning-more-than-their-managers.sql)
-  - [查找重复的电子邮箱](easy/duplicate-emails.sql)
-  - [从不订购的客户](easy/customers-who-never-order.sql)
-  - [上升的温度](easy/rising-temperature.sql)
-  - [大的国家](easy/big-countries.sql)
-  - [超过5名学生的课](easy/classes-more-than-5-students.sql)
-  - [有趣的电影](easy/not-boring-movies.sql)
-  - [交换工资](easy/swap-salary.sql)
+  - [组合两个表](easy/组合两个表.sql)
+  - [第二高的薪水](easy/第二高的薪水.sql)
+  - [超过经理收入的员工](easy/超过经理收入的员工.sql)
+  - [查找重复的电子邮箱](easy/查找重复的电子邮箱.sql)
+  - [从不订购的客户](easy/从不订购的客户.sql)
+  - [删除重复的电子邮箱](easy/删除重复的电子邮箱.sql)
+  - [上升的温度](easy/上升的温度.sql)
+  - [大的国家](easy/大的国家.sql)
+  - [超过5名学生的课](easy/超过5名学生的课.sql)
+  - [有趣的电影](easy/有趣的电影.sql)
+  - [交换工资](easy/交换工资.sql)
+  - [重新格式化部门表](easy/重新格式化部门表.sql)
 - **难度：中等**
-  - [第 N 高的薪水](normal/nth-highest-salary.sql)
-  - [分数排名](normal/rank-scores.sql)
-  - [连续出现的数字](normal/consecutive-numbers.sql)
-  - [部门工资最高的员工](normal/department-highest-salary.sql)
-  - [换座位](normal/exchange-seats.sql)
+  - [第 N 高的薪水](normal/第N高的薪水.sql)
+  - [分数排名](normal/分数排名.sql)
+  - [连续出现的数字](normal/连续出现的数字.sql)
+  - [部门工资最高的员工](normal/部门工资最高的员工.sql)
+  - [换座位](normal/换座位.sql)
--- a/codes/mysql/Leetcode之SQL题/easy/duplicate-emails.sql
+++ b/codes/mysql/Leetcode之SQL题/easy/duplicate-emails.sql
@ -1,26 +0,0 @@
--    查找重复的电子邮箱
--
--    编写一个 SQL 查询，查找 Person 表中所有重复的电子邮箱。
--
--    示例：
--
--    +----+---------+
--    | Id | Email   |
--    +----+---------+
--    | 1  | a@b.com |
--    | 2  | c@d.com |
--    | 3  | a@b.com |
--    +----+---------+
--    根据以上输入，你的查询应返回以下结果：
--
--    +---------+
--    | Email   |
--    +---------+
--    | a@b.com |
--    +---------+
--    说明：所有电子邮箱都是小写字母。
-
-SELECT Email
-FROM Person
-GROUP BY Email
-HAVING COUNT(*) > 1;
--- a/codes/mysql/Leetcode之SQL题/easy/employees-earning-more-than-their-managers.sql
+++ b/codes/mysql/Leetcode之SQL题/easy/employees-earning-more-than-their-managers.sql
@ -1,40 +0,0 @@
--    超过经理收入的员工
--
--    Employee 表包含所有员工，他们的经理也属于员工。每个员工都有一个 Id，此外还有一列对应员工的经理的 Id。
--
--    +----+-------+--------+-----------+
--    | Id | Name  | Salary | ManagerId |
--    +----+-------+--------+-----------+
--    | 1  | Joe   | 70000  | 3         |
--    | 2  | Henry | 80000  | 4         |
--    | 3  | Sam   | 60000  | NULL      |
--    | 4  | Max   | 90000  | NULL      |
--    +----+-------+--------+-----------+
--    给定 Employee 表，编写一个 SQL 查询，该查询可以获取收入超过他们经理的员工的姓名。
--    在上面的表格中，Joe 是唯一一个收入超过他的经理的员工。
--
--    +----------+
--    | Employee |
--    +----------+
--    | Joe      |
--    +----------+
-
-- 以下 3 种解法，由上至下，处理速度越来越慢：
-- 第 1 种查询
-SELECT e1.Name AS Employee
-FROM Employee e1
-    INNER JOIN Employee e2
-    ON e1.ManagerId = e2.Id AND e1.Salary > e2.Salary;
-
-- 第 2 种解法
-SELECT e1.Name AS Employee
-FROM Employee e1, Employee  e2
-WHERE e1.ManagerId = e2.Id AND e1.Salary > e2.Salary;
-
-- 第 3 种查询
-SELECT e1.Name AS Employee
-FROM Employee e1
-WHERE
-e1.Salary > (SELECT e2.Salary
-FROM Employee e2
-WHERE e1.ManagerId = e2.Id);
--- a/codes/mysql/Leetcode之SQL题/easy/swap-salary.sql
+++ b/codes/mysql/Leetcode之SQL题/easy/swap-salary.sql
@ -1,29 +0,0 @@
--    【交换工资】
--
--    给定一个 salary表，如下所示，有m=男性 和 f=女性的值 。
--    交换所有的 f 和 m 值(例如，将所有 f 值更改为 m，反之亦然)。要求使用一个更新查询，并且没有中间临时表。
--
--    例如:
--
--    | id | name | sex | salary |
--    |----|------|-----|--------|
--    | 1  | A    | m   | 2500   |
--    | 2  | B    | f   | 1500   |
--    | 3  | C    | m   | 5500   |
--    | 4  | D    | f   | 500    |
--    运行你所编写的查询语句之后，将会得到以下表:
--
--    | id | name | sex | salary |
--    |----|------|-----|--------|
--    | 1  | A    | f   | 2500   |
--    | 2  | B    | m   | 1500   |
--    | 3  | C    | f   | 5500   |
--    | 4  | D    | m   | 500    |
-
-UPDATE SALARY
-SET SEX =
-    CASE SEX
-        WHEN 'm'
-        THEN 'f'
-        ELSE 'm'
-    END;
--- a/codes/mysql/Leetcode之SQL题/easy/rising-temperature.sql
+++ b/codes/mysql/Leetcode之SQL题/easy/rising-temperature.sql
@ -1,5 +1,7 @@
 --    上升的温度
 --
+--    @link https://leetcode-cn.com/problems/rising-temperature/
+--
 --    给定一个 Weather 表，编写一个 SQL 查询，来查找与之前（昨天的）日期相比温度更高的所有日期的 Id。
 --
 --    +---------+------------------+------------------+
@ -19,8 +21,23 @@
 --    |  4 |
 --    +----+

-SELECT w1.Id
-FROM Weather w1, Weather w2
-WHERE w1.RecordDate = DATE_ADD(w2.RecordDate,interval
-1 DAY )
-AND w1.Temperature > w2.Temperature;
+CREATE TABLE weather (
+    id          INT PRIMARY KEY AUTO_INCREMENT,
+    recorddate  TIMESTAMP,
+    temperature INT
+);
+
+INSERT INTO weather (recorddate, temperature)
+VALUES (TIMESTAMP('2015-01-01'), 10);
+INSERT INTO weather (recorddate, temperature)
+VALUES (TIMESTAMP('2015-01-02'), 25);
+INSERT INTO weather (recorddate, temperature)
+VALUES (TIMESTAMP('2015-01-03'), 20);
+INSERT INTO weather (recorddate, temperature)
+VALUES (TIMESTAMP('2015-01-04'), 30);
+
+-- 解题
+SELECT w1.id
+FROM weather w1, weather w2
+WHERE w1.recorddate = DATE_ADD(w2.recorddate, INTERVAL 1 DAY) AND w1.temperature > w2.temperature;
+
--- a/codes/mysql/Leetcode之SQL题/easy/交换工资.sql
+++ b/codes/mysql/Leetcode之SQL题/easy/交换工资.sql
@ -0,0 +1,53 @@
+-- 【交换工资】
+--
+-- @link https://leetcode-cn.com/problems/swap-salary/
+--
+-- 给定一个 salary 表，如下所示，有 m = 男性 和 f = 女性 的值。交换所有的 f 和 m 值（例如，将所有 f 值更改为 m，反之亦然）。要求只使用一个更新（Update）语句，并且没有中间的临时表。
+--
+-- 注意，您必只能写一个 Update 语句，请不要编写任何 Select 语句。
+--
+-- 例如：
+--
+--    | id | name | sex | salary |
+--    |----|------|-----|--------|
+--    | 1  | A    | m   | 2500   |
+--    | 2  | B    | f   | 1500   |
+--    | 3  | C    | m   | 5500   |
+--    | 4  | D    | f   | 500    |
+--
+--    运行你所编写的更新语句之后，将会得到以下表:
+--
+--    | id | name | sex | salary |
+--    |----|------|-----|--------|
+--    | 1  | A    | f   | 2500   |
+--    | 2  | B    | m   | 1500   |
+--    | 3  | C    | f   | 5500   |
+--    | 4  | D    | m   | 500    |
+
+CREATE TABLE IF NOT EXISTS salary (
+    id     INT PRIMARY KEY AUTO_INCREMENT,
+    name   CHAR(5),
+    sex    CHAR(1),
+    salary INT(10)
+);
+
+INSERT INTO salary(name, sex, salary)
+VALUES ('A', 'm', 2500);
+INSERT INTO salary(name, sex, salary)
+VALUES ('B', 'f', 1500);
+INSERT INTO salary(name, sex, salary)
+VALUES ('C', 'm', 5500);
+INSERT INTO salary(name, sex, salary)
+VALUES ('D', 'f', 500);
+
+-- 解题
+UPDATE salary
+SET sex =
+    CASE sex
+        WHEN 'm'
+            THEN 'f'
+        ELSE 'm'
+        END;
+
+SELECT *
+FROM salary;
--- a/codes/mysql/Leetcode之SQL题/easy/customers-who-never-order.sql
+++ b/codes/mysql/Leetcode之SQL题/easy/customers-who-never-order.sql
@ -1,5 +1,7 @@
 --    从不订购的客户
 --
+--    @link https://leetcode-cn.com/problems/customers-who-never-order/
+--
 --    某网站包含两个表，Customers 表和 Orders 表。编写一个 SQL 查询，找出所有从不订购任何东西的客户。
 --
 --    Customers 表：
@ -29,12 +31,30 @@
 --    | Max       |
 --    +-----------+

-SELECT Name AS Customers
-FROM Customers c
-WHERE c.Id NOT IN (SELECT DISTINCT CustomerId
-FROM Orders);
+CREATE TABLE IF NOT EXISTS customers (
+    id   INT PRIMARY KEY AUTO_INCREMENT,
+    name VARCHAR(20)
+);
+INSERT INTO customers(name)
+VALUES ('Joe');
+INSERT INTO customers(name)
+VALUES ('Henry');
+INSERT INTO customers(name)
+VALUES ('Sam');
+INSERT INTO customers(name)
+VALUES ('Max');

-SELECT Name AS Customers
-FROM Customers
-    INNER JOIN Orders
-    ON Customers.Id != Orders.CustomerId;
+CREATE TABLE IF NOT EXISTS orders (
+    id         INT PRIMARY KEY AUTO_INCREMENT,
+    customerid INT
+);
+INSERT INTO orders(customerid)
+VALUES (3);
+INSERT INTO orders(customerid)
+VALUES (1);
+
+-- 方法一
+SELECT name AS customers
+FROM customers c
+WHERE c.id NOT IN (SELECT DISTINCT customerid
+                   FROM orders);
--- a/codes/mysql/Leetcode之SQL题/easy/删除重复的电子邮箱.sql
+++ b/codes/mysql/Leetcode之SQL题/easy/删除重复的电子邮箱.sql
@ -0,0 +1,47 @@
+--     删除重复的电子邮箱
+--
+--     @link https://leetcode-cn.com/problems/delete-duplicate-emails/
+--
+--     编写一个 SQL 查询，来删除 Person 表中所有重复的电子邮箱，重复的邮箱里只保留 Id 最小 的那个。
+--
+--     +----+------------------+
+--     | Id | Email            |
+--     +----+------------------+
+--     | 1  | john@example.com |
+--     | 2  | bob@example.com  |
+--     | 3  | john@example.com |
+--     +----+------------------+
+--     Id 是这个表的主键。
+--     例如，在运行你的查询语句之后，上面的 Person 表应返回以下几行:
+--
+--     +----+------------------+
+--     | Id | Email            |
+--     +----+------------------+
+--     | 1  | john@example.com |
+--     | 2  | bob@example.com  |
+--     +----+------------------+
+--      
+--
+--     提示：
+--
+--     执行 SQL 之后，输出是整个 Person 表。
+--     使用 delete 语句。
+
+USE db_tutorial;
+
+CREATE TABLE IF NOT EXISTS person (
+    id    INT PRIMARY KEY AUTO_INCREMENT,
+    email VARCHAR(32)
+);
+
+INSERT INTO person (email)
+VALUES ('john@example.com');
+INSERT INTO person (email)
+VALUES ('bob@example.com');
+INSERT INTO person (email)
+VALUES ('john@example.com');
+
+-- 解题
+DELETE p1
+FROM person p1, person p2
+WHERE p1.id > p2.id AND p1.email = p2.email;
--- a/codes/mysql/Leetcode之SQL题/easy/big-countries.sql
+++ b/codes/mysql/Leetcode之SQL题/easy/big-countries.sql
@ -1,5 +1,7 @@
 --  大的国家
 --
+--  @link https://leetcode-cn.com/problems/big-countries/
+--
 --  这里有张 World 表
 --
 --  +-----------------+------------+------------+--------------+---------------+
@ -24,7 +26,35 @@
 --  | Algeria      | 37100000    | 2381741      |
 --  +--------------+-------------+--------------+

-SELECT NAME, POPULATION, AREA
-FROM World
-WHERE AREA > 3000000 OR POPULATION > 25000000;
+CREATE TABLE world (
+    name       VARCHAR(32) PRIMARY KEY,
+    continent  VARCHAR(32),
+    area       INT(10),
+    population INT(20),
+    gdp        INT(20)
+);

+INSERT INTO world
+VALUES ('Afghanistan', 'Asia', 652230, 25500100, 20343000);
+INSERT INTO world
+VALUES ('Albania', 'Europe', 28748, 2831741, 12960000);
+INSERT INTO world
+VALUES ('Algeria', 'Africa', 2381741, 37100000, 188681000);
+INSERT INTO world
+VALUES ('Andorra', 'Europe', 468, 78115, 3712000);
+INSERT INTO world
+VALUES ('Angola', 'Africa', 1246700, 20609294, 100990000);
+
+-- 方法一
+SELECT name, population, area
+FROM world
+WHERE area > 3000000 OR population > 25000000;
+
+-- 方法二
+SELECT name, population, area
+FROM world
+WHERE area > 3000000
+UNION
+SELECT name, population, area
+FROM world
+WHERE population > 25000000;
--- a/codes/mysql/Leetcode之SQL题/easy/not-boring-movies.sql
+++ b/codes/mysql/Leetcode之SQL题/easy/not-boring-movies.sql
@ -26,7 +26,27 @@
 --    |   1     | War       |   great 3D   |   8.9     |
 --    +---------+-----------+--------------+-----------+

+USE db_tutorial;
+CREATE TABLE IF NOT EXISTS cinema (
+    id          INT PRIMARY KEY AUTO_INCREMENT,
+    movie       VARCHAR(20),
+    description VARCHAR(20),
+    rating      DOUBLE
+);
+
+INSERT INTO cinema(movie, description, rating)
+VALUES ('War', 'great 3D', 8.9);
+INSERT INTO cinema(movie, description, rating)
+VALUES ('Science', 'fiction', 8.5);
+INSERT INTO cinema(movie, description, rating)
+VALUES ('irish', 'boring', 6.2);
+INSERT INTO cinema(movie, description, rating)
+VALUES ('Ice song', 'Fantacy', 8.6);
+INSERT INTO cinema(movie, description, rating)
+VALUES ('House card', 'Interesting', 9.1);
+
+-- 解题
 SELECT *
-FROM CINEMA
-WHERE DESCRIPTION != 'boring' AND ID % 2 = 1
+FROM cinema
+WHERE description != 'boring' AND id % 2 = 1
 ORDER BY rating DESC;
--- a/codes/mysql/Leetcode之SQL题/easy/查找重复的电子邮箱.sql
+++ b/codes/mysql/Leetcode之SQL题/easy/查找重复的电子邮箱.sql
@ -0,0 +1,35 @@
+-- --------------------------------------------------------------------------------------
+-- 查找重复的电子邮箱
+-- @link https://leetcode-cn.com/problems/duplicate-emails/
+-- @author Zhang Peng
+-- @date 2020/02/29
+-- ----------------------------------------------------------------------------------------
+
+USE db_tutorial;
+
+CREATE TABLE IF NOT EXISTS person (
+    id    INT PRIMARY KEY AUTO_INCREMENT,
+    email VARCHAR(32)
+);
+
+INSERT INTO person (email)
+VALUES ('a@b.com');
+INSERT INTO person (email)
+VALUES ('c@d.com');
+INSERT INTO person (email)
+VALUES ('a@b.com');
+
+-- 方法一
+SELECT email
+FROM (
+    SELECT email, COUNT(email) AS num
+    FROM person
+    GROUP BY email
+) AS statistic
+WHERE num > 1;
+
+-- 方法二
+SELECT email
+FROM person
+GROUP BY email
+HAVING count(email) > 1;
--- a/codes/mysql/Leetcode之SQL题/easy/second-highest-salary.sql
+++ b/codes/mysql/Leetcode之SQL题/easy/second-highest-salary.sql
@ -1,5 +1,7 @@
 --    第二高的薪水
 --
+--    @link https://leetcode-cn.com/problems/second-highest-salary/
+--
 --    编写一个 SQL 查询，获取 Employee 表中第二高的薪水（Salary） 。
 --
 --    +----+--------+
@ -17,7 +19,8 @@
 --    | 200                 |
 --    +---------------------+

-SELECT (SELECT DISTINCT SALARY
-    FROM Employee
-    ORDER BY SALARY DESC LIMIT 1,1)
-AS SecondHighestSalary;
+SELECT (SELECT DISTINCT salary
+        FROM employee
+        ORDER BY salary DESC
+        LIMIT 1,1)
+    AS secondhighestsalary;
--- a/codes/mysql/Leetcode之SQL题/easy/combine-two-tables.sql
+++ b/codes/mysql/Leetcode之SQL题/easy/combine-two-tables.sql
@ -1,5 +1,7 @@
 --    组合两个表
 --
+--    @link https://leetcode-cn.com/problems/combine-two-tables/
+--
 --    表1: Person
 --
 --    +-------------+---------+
@ -29,7 +31,8 @@
 --
 --    FirstName, LastName, City, State

-SELECT Person.FirstName, Person.LastName, Address.City, Address.State
-FROM Person
-    LEFT JOIN Address
-    ON Person.PersonId = Address.PersonId;
+
+SELECT person.firstname, person.lastname, address.city, address.state
+FROM person
+         LEFT JOIN address
+ON person.personid = address.personid;
--- a/codes/mysql/Leetcode之SQL题/easy/classes-more-than-5-students.sql
+++ b/codes/mysql/Leetcode之SQL题/easy/classes-more-than-5-students.sql
@ -29,7 +29,34 @@
 --    Note:
 --    学生在每个课中不应被重复计算。

-SELECT CLASS
-FROM COURSES
-GROUP BY CLASS
-HAVING COUNT(DISTINCT STUDENT)>4;
+USE db_tutorial;
+
+CREATE TABLE courses (
+    student VARCHAR(10) PRIMARY KEY,
+    class   VARCHAR(10)
+);
+
+INSERT INTO courses
+VALUES ('A', 'Math');
+INSERT INTO courses
+VALUES ('B', 'English');
+INSERT INTO courses
+VALUES ('C', 'Math');
+INSERT INTO courses
+VALUES ('D', 'Biology');
+INSERT INTO courses
+VALUES ('E', 'Math');
+INSERT INTO courses
+VALUES ('F', 'Computer');
+INSERT INTO courses
+VALUES ('G', 'Math');
+INSERT INTO courses
+VALUES ('H', 'Math');
+INSERT INTO courses
+VALUES ('I', 'Math');
+
+-- 解题
+SELECT class
+FROM courses
+GROUP BY class
+HAVING COUNT(DISTINCT student) >= 5;
--- a/codes/mysql/Leetcode之SQL题/easy/超过经理收入的员工.sql
+++ b/codes/mysql/Leetcode之SQL题/easy/超过经理收入的员工.sql
@ -0,0 +1,100 @@
+#     重新格式化部门表
+#
+#     @link https://leetcode-cn.com/problems/reformat-department-table/
+#
+#     部门表 Department：
+#
+#     +---------------+---------+
+#     | Column Name   | Type    |
+#     +---------------+---------+
+#     | id            | int     |
+#     | revenue       | int     |
+#     | month         | varchar |
+#     +---------------+---------+
+#     (id, month) 是表的联合主键。
+#     这个表格有关于每个部门每月收入的信息。
+#     月份（month）可以取下列值 ["Jan","Feb","Mar","Apr","May","Jun","Jul","Aug","Sep","Oct","Nov","Dec"]。
+#      
+#
+#     编写一个 SQL 查询来重新格式化表，使得新的表中有一个部门 id 列和一些对应 每个月 的收入（revenue）列。
+#
+#     查询结果格式如下面的示例所示：
+#
+#     Department 表：
+#     +------+---------+-------+
+#     | id   | revenue | month |
+#     +------+---------+-------+
+#     | 1    | 8000    | Jan   |
+#     | 2    | 9000    | Jan   |
+#     | 3    | 10000   | Feb   |
+#     | 1    | 7000    | Feb   |
+#     | 1    | 6000    | Mar   |
+#     +------+---------+-------+
+#
+#     查询得到的结果表：
+#     +------+-------------+-------------+-------------+-----+-------------+
+#     | id   | Jan_Revenue | Feb_Revenue | Mar_Revenue | ... | Dec_Revenue |
+#     +------+-------------+-------------+-------------+-----+-------------+
+#     | 1    | 8000        | 7000        | 6000        | ... | null        |
+#     | 2    | 9000        | null        | null        | ... | null        |
+#     | 3    | null        | 10000       | null        | ... | null        |
+#     +------+-------------+-------------+-------------+-----+-------------+
+#
+#     注意，结果表有 13 列 (1个部门 id 列 + 12个月份的收入列)。
+
+USE db_tutorial;
+
+CREATE TABLE IF NOT EXISTS department (
+    id      INT,
+    revenue INT,
+    month   VARCHAR(20)
+);
+
+INSERT INTO department
+VALUES (1, 8000, 'Jan');
+INSERT INTO department
+VALUES (2, 9000, 'Jan');
+INSERT INTO department
+VALUES (3, 10000, 'Feb');
+INSERT INTO department
+VALUES (1, 7000, 'Feb');
+INSERT INTO department
+VALUES (1, 6000, 'Mar');
+
+-- 解题
+
+SELECT id, revenue AS jan_revenue
+FROM department
+WHERE month = 'Jan';
+
+SELECT id,
+    SUM(CASE month WHEN 'Jan' THEN revenue END) jan_revenue,
+    SUM(CASE month WHEN 'Feb' THEN revenue END) feb_revenue,
+    SUM(CASE month WHEN 'Mar' THEN revenue END) mar_revenue,
+    SUM(CASE month WHEN 'Apr' THEN revenue END) apr_revenue,
+    SUM(CASE month WHEN 'May' THEN revenue END) may_revenue,
+    SUM(CASE month WHEN 'Jun' THEN revenue END) jun_revenue,
+    SUM(CASE month WHEN 'Jul' THEN revenue END) jul_revenue,
+    SUM(CASE month WHEN 'Aug' THEN revenue END) aug_revenue,
+    SUM(CASE month WHEN 'Sep' THEN revenue END) sep_revenue,
+    SUM(CASE month WHEN 'Oct' THEN revenue END) oct_revenue,
+    SUM(CASE month WHEN 'Nov' THEN revenue END) nov_revenue,
+    SUM(CASE month WHEN 'Dec' THEN revenue END) dec_revenue
+FROM department
+GROUP BY id;
+
+SELECT id,
+    SUM(IF(month = 'Jan', revenue, NULL)) jan_revenue,
+    SUM(IF(month = 'Feb', revenue, NULL)) feb_revenue,
+    SUM(IF(month = 'Mar', revenue, NULL)) mar_revenue,
+    SUM(IF(month = 'Apr', revenue, NULL)) apr_revenue,
+    SUM(IF(month = 'May', revenue, NULL)) may_revenue,
+    SUM(IF(month = 'Jun', revenue, NULL)) jun_revenue,
+    SUM(IF(month = 'Jul', revenue, NULL)) jul_revenue,
+    SUM(IF(month = 'Aug', revenue, NULL)) aug_revenue,
+    SUM(IF(month = 'Sep', revenue, NULL)) sep_revenue,
+    SUM(IF(month = 'Oct', revenue, NULL)) oct_revenue,
+    SUM(IF(month = 'Nov', revenue, NULL)) nov_revenue,
+    SUM(IF(month = 'Dec', revenue, NULL)) dec_revenue
+FROM department
+GROUP BY id;
--- a/codes/mysql/Leetcode之SQL题/easy/重新格式化部门表.sql
+++ b/codes/mysql/Leetcode之SQL题/easy/重新格式化部门表.sql
--- a/codes/mysql/Leetcode之SQL题/hard/部门工资前三高的所有员工.sql
+++ b/codes/mysql/Leetcode之SQL题/hard/部门工资前三高的所有员工.sql
@ -0,0 +1,75 @@
+--     部门工资前三高的所有员工
+--
+--     Employee 表包含所有员工信息，每个员工有其对应的工号 Id，姓名 Name，工资 Salary 和部门编号 DepartmentId 。
+--
+--     +----+-------+--------+--------------+
+--     | Id | Name  | Salary | DepartmentId |
+--     +----+-------+--------+--------------+
+--     | 1  | Joe   | 85000  | 1            |
+--     | 2  | Henry | 80000  | 2            |
+--     | 3  | Sam   | 60000  | 2            |
+--     | 4  | Max   | 90000  | 1            |
+--     | 5  | Janet | 69000  | 1            |
+--     | 6  | Randy | 85000  | 1            |
+--     | 7  | Will  | 70000  | 1            |
+--     +----+-------+--------+--------------+
+--     Department 表包含公司所有部门的信息。
+--
+--     +----+----------+
+--     | Id | Name     |
+--     +----+----------+
+--     | 1  | IT       |
+--     | 2  | Sales    |
+--     +----+----------+
+--     编写一个 SQL 查询，找出每个部门获得前三高工资的所有员工。例如，根据上述给定的表，查询结果应返回：
+--
+--     +------------+----------+--------+
+--     | Department | Employee | Salary |
+--     +------------+----------+--------+
+--     | IT         | Max      | 90000  |
+--     | IT         | Randy    | 85000  |
+--     | IT         | Joe      | 85000  |
+--     | IT         | Will     | 70000  |
+--     | Sales      | Henry    | 80000  |
+--     | Sales      | Sam      | 60000  |
+--     +------------+----------+--------+
+--     解释：
+--
+--     IT 部门中，Max 获得了最高的工资，Randy 和 Joe 都拿到了第二高的工资，Will 的工资排第三。销售部门（Sales）只有两名员工，Henry 的工资最高，Sam 的工资排第二。
+
+USE db_tutorial;
+CREATE TABLE IF NOT EXISTS employee (
+    id           INT PRIMARY KEY AUTO_INCREMENT,
+    name         VARCHAR(10),
+    salary       INT,
+    departmentid INT
+);
+
+INSERT INTO employee (name, salary, departmentid)
+VALUES ('Joe', 85000, 1);
+INSERT INTO employee (name, salary, departmentid)
+VALUES ('Henry', 80000, 2);
+INSERT INTO employee (name, salary, departmentid)
+VALUES ('Sam', 60000, 2);
+INSERT INTO employee (name, salary, departmentid)
+VALUES ('Max', 90000, 1);
+INSERT INTO employee (name, salary, departmentid)
+VALUES ('Janet', 69000, 1);
+INSERT INTO employee (name, salary, departmentid)
+VALUES ('Randy', 85000, 1);
+INSERT INTO employee (name, salary, departmentid)
+VALUES ('Will', 70000, 1);
+
+CREATE TABLE IF NOT EXISTS department (
+    id   INT PRIMARY KEY AUTO_INCREMENT,
+    name VARCHAR(10)
+);
+
+INSERT INTO department (name)
+VALUES ('IT');
+INSERT INTO department (name)
+VALUES ('Sale');
+
+SELECT *
+FROM employee
+WHERE departmentid = 1 LIMIT ;
--- a/codes/mysql/Leetcode之SQL题/normal/department-highest-salary.sql
+++ b/codes/mysql/Leetcode之SQL题/normal/department-highest-salary.sql
@ -1,52 +0,0 @@
--    部门工资最高的员工
--
--    Employee 表包含所有员工信息，每个员工有其对应的 Id, salary 和 department Id。
--
--    +----+-------+--------+--------------+
--    | Id | Name  | Salary | DepartmentId |
--    +----+-------+--------+--------------+
--    | 1  | Joe   | 70000  | 1            |
--    | 2  | Henry | 80000  | 2            |
--    | 3  | Sam   | 60000  | 2            |
--    | 4  | Max   | 90000  | 1            |
--    +----+-------+--------+--------------+
--    Department 表包含公司所有部门的信息。
--
--    +----+----------+
--    | Id | Name     |
--    +----+----------+
--    | 1  | IT       |
--    | 2  | Sales    |
--    +----+----------+
--    编写一个 SQL 查询，找出每个部门工资最高的员工。例如，根据上述给定的表格，Max 在 IT 部门有最高工资，
--    Henry 在 Sales 部门有最高工资。
--
--    +------------+----------+--------+
--    | Department | Employee | Salary |
--    +------------+----------+--------+
--    | IT         | Max      | 90000  |
--    | Sales      | Henry    | 80000  |
--    +------------+----------+--------+
-
-- 第 1 种解法
-SELECT D.Name AS Department , E.Name AS Employee , E.Salary
-FROM
-    Employee E,
-    (SELECT DepartmentId, MAX(Salary) AS max
-    FROM Employee
-    GROUP BY DepartmentId) T,
-    Department D
-WHERE E.DepartmentId = T.DepartmentId
-    AND E.Salary = T.MAX
-    AND E.DepartmentId = D.ID
-
-- 第 2 种解法
-SELECT D.Name AS Department , E.Name AS Employee , E.Salary
-FROM
-    Employee E,
-    Department D
-WHERE E.DepartmentId = D.ID
-    AND (DepartmentId, Salary) IN
-    (SELECT DepartmentId, MAX(Salary) AS max
-    FROM Employee
-    GROUP BY DepartmentId)
--- a/codes/mysql/Leetcode之SQL题/normal/nth-highest-salary.sql
+++ b/codes/mysql/Leetcode之SQL题/normal/nth-highest-salary.sql
@ -1,30 +0,0 @@
--    第N高的薪水
--
--    编写一个 SQL 查询，获取 Employee 表中第 n 高的薪水（Salary）。
--
--    +----+--------+
--    | Id | Salary |
--    +----+--------+
--    | 1  | 100    |
--    | 2  | 200    |
--    | 3  | 300    |
--    +----+--------+
--    例如上述 Employee 表，n = 2 时，应返回第二高的薪水 200。如果不存在第 n 高的薪水，那么查询应返回 null。
--
--    +------------------------+
--    | getNthHighestSalary(2) |
--    +------------------------+
--    | 200                    |
--    +------------------------+
-
-CREATE FUNCTION getNthHighestSalary(N int) RETURNS INT
-BEGIN
-    RETURN (
-      SELECT DISTINCT Salary
-    FROM Employee e
-    WHERE
-      N = (SELECT COUNT(DISTINCT Salary)
-    FROM Employee
-    WHERE Salary >= e.Salary)
-  );
-END
--- a/codes/mysql/Leetcode之SQL题/normal/分数排名.sql
+++ b/codes/mysql/Leetcode之SQL题/normal/分数排名.sql
@ -26,9 +26,30 @@
 --    | 3.50  | 4    |
 --    +-------+------+

-SELECT Score, (
-    SELECT COUNT(DISTINCT score)
-    FROM Scores
-    WHERE score >= s.score) AS Rank
-FROM Scores s
-ORDER BY Score DESC;
+USE db_tutorial;
+
+CREATE TABLE IF NOT EXISTS scores (
+    id    INT PRIMARY KEY AUTO_INCREMENT,
+    score DOUBLE
+);
+
+INSERT INTO scores (score)
+VALUES (3.50);
+INSERT INTO scores (score)
+VALUES (3.65);
+INSERT INTO scores (score)
+VALUES (4.00);
+INSERT INTO scores (score)
+VALUES (3.85);
+INSERT INTO scores (score)
+VALUES (4.00);
+INSERT INTO scores (score)
+VALUES (3.65);
+
+SELECT count(DISTINCT b.score)
+FROM scores b;
+
+SELECT a.score AS score,
+    (SELECT count(DISTINCT b.score) FROM scores b WHERE b.score >= a.score) AS ranking
+FROM scores a
+ORDER BY a.score DESC;
--- a/codes/mysql/Leetcode之SQL题/normal/exchange-seats.sql
+++ b/codes/mysql/Leetcode之SQL题/normal/exchange-seats.sql
--- a/codes/mysql/Leetcode之SQL题/normal/第N高的薪水.sql
+++ b/codes/mysql/Leetcode之SQL题/normal/第N高的薪水.sql
@ -0,0 +1,48 @@
+--    第N高的薪水
+--
+--    编写一个 SQL 查询，获取 Employee 表中第 n 高的薪水（Salary）。
+--
+--    +----+--------+
+--    | Id | Salary |
+--    +----+--------+
+--    | 1  | 100    |
+--    | 2  | 200    |
+--    | 3  | 300    |
+--    +----+--------+
+--    例如上述 Employee 表，n = 2 时，应返回第二高的薪水 200。如果不存在第 n 高的薪水，那么查询应返回 null。
+--
+--    +------------------------+
+--    | getNthHighestSalary(2) |
+--    +------------------------+
+--    | 200                    |
+--    +------------------------+
+
+USE db_tutorial;
+CREATE TABLE IF NOT EXISTS employee (
+    id     INT PRIMARY KEY AUTO_INCREMENT,
+    salary INT
+);
+
+INSERT INTO employee(salary)
+VALUES (100);
+INSERT INTO employee(salary)
+VALUES (200);
+INSERT INTO employee(salary)
+VALUES (300);
+
+SELECT DISTINCT salary
+FROM employee e
+WHERE 1 = (SELECT COUNT(DISTINCT salary)
+           FROM employee
+           WHERE salary >= e.salary);
+
+CREATE FUNCTION getNthHighestSalary(n INT) RETURNS INT
+BEGIN
+    RETURN (
+        SELECT DISTINCT salary
+        FROM employee e
+        WHERE n = (SELECT COUNT(DISTINCT salary)
+                   FROM employee
+                   WHERE salary >= e.salary)
+    );
+END
--- a/codes/mysql/Leetcode之SQL题/normal/连续出现的数字.sql
+++ b/codes/mysql/Leetcode之SQL题/normal/连续出现的数字.sql
@ -21,8 +21,30 @@
 --    | 1               |
 --    +-----------------+

-SELECT DISTINCT l1.num AS ConsecutiveNums
-FROM Logs l1
-    JOIN Logs l2 ON l1.id=l2.id-1
-    JOIN Logs l3 ON l1.id=l3.id-2
-WHERE l1.num=l2.num AND l2.num=l3.num;
+USE db_tutorial;
+
+CREATE TABLE IF NOT EXISTS logs (
+    id  INT PRIMARY KEY AUTO_INCREMENT,
+    num INT
+);
+
+INSERT INTO logs(num)
+VALUES (1);
+INSERT INTO logs(num)
+VALUES (1);
+INSERT INTO logs(num)
+VALUES (1);
+INSERT INTO logs(num)
+VALUES (2);
+INSERT INTO logs(num)
+VALUES (1);
+INSERT INTO logs(num)
+VALUES (2);
+INSERT INTO logs(num)
+VALUES (2);
+
+-- 解题
+SELECT DISTINCT (l1.num) AS 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;
+
--- a/codes/mysql/Leetcode之SQL题/normal/部门工资最高的员工.sql
+++ b/codes/mysql/Leetcode之SQL题/normal/部门工资最高的员工.sql
@ -0,0 +1,86 @@
+--    部门工资最高的员工
+--
+--    @link https://leetcode-cn.com/problems/department-highest-salary/
+--
+--    Employee 表包含所有员工信息，每个员工有其对应的 Id, salary 和 department Id。
+--
+--    +----+-------+--------+--------------+
+--    | Id | Name  | Salary | DepartmentId |
+--    +----+-------+--------+--------------+
+--    | 1  | Joe   | 70000  | 1            |
+--    | 2  | Henry | 80000  | 2            |
+--    | 3  | Sam   | 60000  | 2            |
+--    | 4  | Max   | 90000  | 1            |
+--    +----+-------+--------+--------------+
+--    Department 表包含公司所有部门的信息。
+--
+--    +----+----------+
+--    | Id | Name     |
+--    +----+----------+
+--    | 1  | IT       |
+--    | 2  | Sales    |
+--    +----+----------+
+--    编写一个 SQL 查询，找出每个部门工资最高的员工。例如，根据上述给定的表格，Max 在 IT 部门有最高工资，
+--    Henry 在 Sales 部门有最高工资。
+--
+--    +------------+----------+--------+
+--    | Department | Employee | Salary |
+--    +------------+----------+--------+
+--    | IT         | Max      | 90000  |
+--    | Sales      | Henry    | 80000  |
+--    +------------+----------+--------+
+
+USE db_tutorial;
+CREATE TABLE IF NOT EXISTS employee (
+    id           INT PRIMARY KEY AUTO_INCREMENT,
+    name         VARCHAR(10),
+    salary       INT,
+    departmentid INT
+);
+
+INSERT INTO employee (name, salary, departmentid)
+VALUES ('Joe', 70000, 1);
+INSERT INTO employee (name, salary, departmentid)
+VALUES ('Henry', 80000, 2);
+INSERT INTO employee (name, salary, departmentid)
+VALUES ('Sam', 60000, 2);
+INSERT INTO employee (name, salary, departmentid)
+VALUES ('Max', 90000, 1);
+
+CREATE TABLE IF NOT EXISTS department (
+    id   INT PRIMARY KEY AUTO_INCREMENT,
+    name VARCHAR(10)
+);
+
+INSERT INTO department (name)
+VALUES ('IT');
+INSERT INTO department (name)
+VALUES ('Sale');
+
+SELECT *
+FROM employee
+WHERE departmentid, salary IN
+(SELECT departmentid, MAX(salary)
+FROM employee
+GROUP BY departmentid);
+
+
+-- 第 1 种解法
+SELECT d.name AS department, e.name AS employee, e.salary
+FROM employee e,
+    (SELECT departmentid, MAX(salary) AS max
+     FROM employee
+     GROUP BY departmentid) t,
+    department d
+WHERE e.departmentid = t.departmentid AND e.salary = t.max AND e.departmentid = d.id;
+
+
+-- 第 2 种解法
+SELECT d.name AS department, e.name AS employee, e.salary
+FROM employee e,
+    department d
+WHERE e.departmentid = d.id AND
+        (departmentid, salary) IN
+        (SELECT departmentid, MAX(salary) AS max
+         FROM employee
+         GROUP BY departmentid);