leecode 数据库题

codes/leetcode/README.md

@@ -1,3 +0,0 @@
-# Leetcode 数据库篇题解
-
-> [题库地址](https://leetcode-cn.com/problemset/database/)

codes/mysql/leetcode-sql/README.md

@@ -0,0 +1,21 @@
+# Leetcode 数据库篇题解
+
+> [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)
+- **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)

codes/mysql/leetcode-sql/easy/big-countries.sql

@@ -1,3 +1,5 @@
+--  大的国家
+--
 --  这里有张 World 表
 --
 --  +-----------------+------------+------------+--------------+---------------+
@@ -22,6 +24,7 @@
 --  | Algeria      | 37100000    | 2381741      |
 --  +--------------+-------------+--------------+
 
-SELECT name, population, area FROM World
-WHERE area > 3000000 OR population > 25000000;
+SELECT NAME, POPULATION, AREA
+FROM World
+WHERE AREA > 3000000 OR POPULATION > 25000000;
 

codes/mysql/leetcode-sql/easy/classes-more-than-5-students.sql

@@ -1,3 +1,5 @@
+--    【超过5名学生的课】
+--
 --    有一个courses 表 ，有: student (学生) 和 class (课程)。
 --
 --    请列出所有超过或等于5名学生的课。
@@ -27,4 +29,7 @@
 --    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;
+SELECT CLASS
+FROM COURSES
+GROUP BY CLASS
+HAVING COUNT(DISTINCT STUDENT)>4;

codes/mysql/leetcode-sql/easy/combine-two-tables.sql

@@ -1,3 +1,5 @@
+--    组合两个表
+--
 --    表1: Person
 --
 --    +-------------+---------+

codes/mysql/leetcode-sql/easy/customers-who-never-order.sql

@@ -1,3 +1,5 @@
+--    从不订购的客户
+--
 --    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：
@@ -27,8 +29,10 @@
 --    | Max       |
 --    +-----------+
 
-SELECT Name AS Customers FROM Customers c
-WHERE c.Id NOT IN (SELECT DISTINCT CustomerId FROM Orders);
+SELECT Name AS Customers
+FROM Customers c
+WHERE c.Id NOT IN (SELECT DISTINCT CustomerId
+FROM Orders);
 
 SELECT Name AS Customers
 FROM Customers

codes/mysql/leetcode-sql/easy/duplicate-emails.sql

@@ -1,3 +1,5 @@
+--    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：
@@ -18,4 +20,7 @@
 --    +---------+
 --    说明：所有电子邮箱都是小写字母。
 
-SELECT Email FROM Person GROUP BY Email HAVING COUNT(*) > 1;
+SELECT Email
+FROM Person
+GROUP BY Email
+HAVING COUNT(*) > 1;

codes/mysql/leetcode-sql/easy/employees-earning-more-than-their-managers.sql

@@ -1,3 +1,5 @@
+--    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。
 --
 --    +----+-------+--------+-----------+
@@ -25,9 +27,14 @@ 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
+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);
+SELECT e1.Name AS Employee
+FROM Employee e1
+WHERE
+e1.Salary > (SELECT e2.Salary
+FROM Employee e2
+WHERE e1.ManagerId = e2.Id);

codes/mysql/leetcode-sql/easy/not-boring-movies.sql

@@ -1,3 +1,5 @@
+--    【有趣的电影】
+--
 --    某城市开了一家新的电影院，吸引了很多人过来看电影。该电影院特别注意用户体验，专门有个 LED显示板做电影推荐，上面公布着影评和相关电影描述。
 --
 --    作为该电影院的信息部主管，您需要编写一个 SQL查询，找出所有影片描述为非 boring (不无聊) 的并且 id 为奇数 的影片，结果请按等级 rating 排列。
@@ -24,6 +26,7 @@
 --    |   1     | War       |   great 3D   |   8.9     |
 --    +---------+-----------+--------------+-----------+
 
-SELECT * FROM cinema
-WHERE description != 'boring' AND id % 2 = 1
+SELECT *
+FROM CINEMA
+WHERE DESCRIPTION != 'boring' AND ID % 2 = 1
 ORDER BY rating DESC;

codes/mysql/leetcode-sql/easy/rising-temperature.sql

@@ -1,3 +1,5 @@
+--    上升的温度
+--
 --    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。
 --
 --    +---------+------------------+------------------+
@@ -17,6 +19,8 @@
 --    |  4 |
 --    +----+
 
-SELECT w1.Id FROM Weather w1, Weather w2
-WHERE w1.RecordDate = DATE_ADD(w2.RecordDate,interval 1 DAY )
+SELECT w1.Id
+FROM Weather w1, Weather w2
+WHERE w1.RecordDate = DATE_ADD(w2.RecordDate,interval
+1 DAY )
 AND w1.Temperature > w2.Temperature;

codes/mysql/leetcode-sql/easy/second-highest-salary.sql

@@ -1,3 +1,5 @@
+--    第二高的薪水
+--
 --    编写一个 SQL 查询，获取 Employee 表中第二高的薪水（Salary） 。
 --
 --    +----+--------+
@@ -15,7 +17,7 @@
 --    | 200                 |
 --    +---------------------+
 
-
-SELECT (SELECT DISTINCT salary FROM Employee
-ORDER BY salary DESC LIMIT 1,1)
+SELECT (SELECT DISTINCT SALARY
+    FROM Employee
+    ORDER BY SALARY DESC LIMIT 1,1)
 AS SecondHighestSalary;

codes/mysql/leetcode-sql/easy/swap-salary.sql

@@ -1,3 +1,5 @@
+--    【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。
 --
@@ -18,9 +20,9 @@
 --    | 3  | C    | f   | 5500   |
 --    | 4  | D    | m   | 500    |
 
-UPDATE salary
-SET sex =
-    CASE sex
+UPDATE SALARY
+SET SEX =
+    CASE SEX
         WHEN 'm'
         THEN 'f'
         ELSE 'm'

codes/mysql/leetcode-sql/normal/consecutive-numbers.sql

@@ -0,0 +1,28 @@
+--    连续出现的数字
+--
+--    编写一个 SQL 查询，查找所有至少连续出现三次的数字。
+--
+--    +----+-----+
+--    | Id | Num |
+--    +----+-----+
+--    | 1  |  1  |
+--    | 2  |  1  |
+--    | 3  |  1  |
+--    | 4  |  2  |
+--    | 5  |  1  |
+--    | 6  |  2  |
+--    | 7  |  2  |
+--    +----+-----+
+--    例如，给定上面的 Logs 表， 1 是唯一连续出现至少三次的数字。
+--
+--    +-----------------+
+--    | ConsecutiveNums |
+--    +-----------------+
+--    | 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;

codes/mysql/leetcode-sql/normal/department-highest-salary.sql

@@ -0,0 +1,52 @@
+--    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)

codes/mysql/leetcode-sql/normal/exchange-seats.sql

@@ -0,0 +1,43 @@
+--    换座位
+--
+--    小美是一所中学的信息科技老师，她有一张 seat 座位表，平时用来储存学生名字和与他们相对应的座位 id。
+--
+--    其中纵列的 id 是连续递增的
+--
+--    小美想改变相邻俩学生的座位。
+--
+--    你能不能帮她写一个 SQL query 来输出小美想要的结果呢？
+--
+--    示例：
+--
+--    +---------+---------+
+--    |    id   | student |
+--    +---------+---------+
+--    |    1    | Abbot   |
+--    |    2    | Doris   |
+--    |    3    | Emerson |
+--    |    4    | Green   |
+--    |    5    | Jeames  |
+--    +---------+---------+
+--    假如数据输入的是上表，则输出结果如下：
+--
+--    +---------+---------+
+--    |    id   | student |
+--    +---------+---------+
+--    |    1    | Doris   |
+--    |    2    | Abbot   |
+--    |    3    | Green   |
+--    |    4    | Emerson |
+--    |    5    | Jeames  |
+--    +---------+---------+
+--    注意：
+--
+--    如果学生人数是奇数，则不需要改变最后一个同学的座位。
+
+SELECT
+IF (id < (SELECT COUNT(*)
+FROM seat),
+IF (id mod 2=0, id-1, id+1),
+IF (id mod 2=0, id-1, id)) AS id, student
+FROM seat
+ORDER BY id ASC;

codes/mysql/leetcode-sql/normal/nth-highest-salary.sql

@@ -0,0 +1,30 @@
+--    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

codes/mysql/leetcode-sql/normal/rank-scores.sql

@@ -0,0 +1,34 @@
+--    分数排名
+--
+--    编写一个 SQL 查询来实现分数排名。如果两个分数相同，则两个分数排名（Rank）相同。
+--    请注意，平分后的下一个名次应该是下一个连续的整数值。换句话说，名次之间不应该有“间隔”。
+--
+--    +----+-------+
+--    | Id | Score |
+--    +----+-------+
+--    | 1  | 3.50  |
+--    | 2  | 3.65  |
+--    | 3  | 4.00  |
+--    | 4  | 3.85  |
+--    | 5  | 4.00  |
+--    | 6  | 3.65  |
+--    +----+-------+
+--    例如，根据上述给定的 Scores 表，你的查询应该返回（按分数从高到低排列）：
+--
+--    +-------+------+
+--    | Score | Rank |
+--    +-------+------+
+--    | 4.00  | 1    |
+--    | 4.00  | 1    |
+--    | 3.85  | 2    |
+--    | 3.65  | 3    |
+--    | 3.65  | 3    |
+--    | 3.50  | 4    |
+--    +-------+------+
+
+SELECT Score, (
+    SELECT COUNT(DISTINCT score)
+    FROM Scores
+    WHERE score >= s.score) AS Rank
+FROM Scores s
+ORDER BY Score DESC;