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CyC2018
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@ -3,7 +3,10 @@
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* [181. Employees Earning More Than Their Managers](#181-employees-earning-more-than-their-managers)
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* [183. Customers Who Never Order](#183-customers-who-never-order)
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* [184. Department Highest Salary](#184-department-highest-salary)
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* [未完待续...](#未完待续)
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* [176. Second Highest Salary](#176-second-highest-salary)
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* [177. Nth Highest Salary](#177-nth-highest-salary)
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* [178. Rank Scores](#178-rank-scores)
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* [180. Consecutive Numbers](#180-consecutive-numbers)
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<!-- GFM-TOC -->
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@ -45,9 +48,13 @@ AddressId is the primary key column for this table.
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## SQL Schema
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```sql
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DROP TABLE Person;
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DROP TABLE
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IF
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EXISTS Person;
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CREATE TABLE Person ( PersonId INT, FirstName VARCHAR ( 255 ), LastName VARCHAR ( 255 ) );
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DROP TABLE Address;
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DROP TABLE
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IF
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EXISTS Address;
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CREATE TABLE Address ( AddressId INT, PersonId INT, City VARCHAR ( 255 ), State VARCHAR ( 255 ) );
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INSERT INTO Person ( PersonId, LastName, FirstName )
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VALUES
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@ -62,9 +69,14 @@ VALUES
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使用左外连接。
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```sql
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SELECT FirstName, LastName, City, State
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FROM Person AS P LEFT JOIN Address AS A
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ON P.PersonId = A.PersonId;
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SELECT
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FirstName,
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LastName,
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City,
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State
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FROM
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Person AS P
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LEFT JOIN Address AS A ON P.PersonId = A.PersonId;
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```
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# 181. Employees Earning More Than Their Managers
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@ -86,12 +98,12 @@ Employee 表：
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+----+-------+--------+-----------+
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```
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查找所有员工，它们的薪资大于其经理薪资。
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查找所有员工，他们的薪资大于其经理薪资。
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## SQL Schema
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```sql
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DROP TABLE Employee;
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DROP TABLE IF EXISTS Employee;
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CREATE TABLE Employee ( Id INT, NAME VARCHAR ( 255 ), Salary INT, ManagerId INT );
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INSERT INTO Employee ( Id, NAME, Salary, ManagerId )
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VALUES
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@ -104,9 +116,12 @@ VALUES
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## Solution
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```sql
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SELECT E1.Name AS Employee
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FROM Employee AS E1, Employee AS E2
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WHERE E1.ManagerId = E2.Id AND E1.Salary > E2.Salary;
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SELECT
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E1.NAME AS Employee
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FROM
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Employee AS E1
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INNER JOIN Employee AS E2 ON E1.ManagerId = E2.Id
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AND E1.Salary > E2.Salary;
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```
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# 183. Customers Who Never Order
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@ -153,9 +168,9 @@ Orders 表：
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## SQL Schema
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```sql
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DROP TABLE Customers;
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DROP TABLE IF EXISTS Customers;
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CREATE TABLE Customers ( Id INT, NAME VARCHAR ( 255 ) );
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DROP TABLE Orders;
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DROP TABLE IF EXISTS Orders;
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CREATE TABLE Orders ( Id INT, CustomerId INT );
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INSERT INTO Customers ( Id, NAME )
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VALUES
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@ -174,21 +189,24 @@ VALUES
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左外链接
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```sql
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SELECT C.Name AS Customers
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FROM Customers AS C LEFT JOIN Orders AS O
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ON C.Id = O.CustomerId
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WHERE O.CustomerId IS NULL;
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SELECT
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C.NAME AS Customers
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FROM
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Customers AS C
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LEFT JOIN Orders AS O ON C.Id = O.CustomerId
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WHERE
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O.CustomerId IS NULL;
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```
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子查询
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```sql
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SELECT C.Name AS Customers
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FROM Customers AS C
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WHERE C.Id NOT IN (
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SELECT CustomerId
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FROM Orders
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);
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SELECT
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C.NAME AS Customers
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FROM
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Customers AS C
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WHERE
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C.Id NOT IN ( SELECT CustomerId FROM Orders );
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```
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# 184. Department Highest Salary
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@ -235,9 +253,9 @@ Department 表：
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## SQL Schema
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```sql
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DROP TABLE Employee;
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DROP TABLE IF EXISTS Employee;
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CREATE TABLE Employee ( Id INT, NAME VARCHAR ( 255 ), Salary INT, DepartmentId INT );
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DROP TABLE Department;
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DROP TABLE IF EXISTS Department;
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CREATE TABLE Department ( Id INT, NAME VARCHAR ( 255 ) );
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INSERT INTO Employee ( Id, NAME, Salary, DepartmentId )
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VALUES
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@ -253,15 +271,226 @@ VALUES
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## Solution
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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。
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之后使用连接将找到一个部门中薪资等于临时表中最大薪资的员工。
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```sql
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SELECT D.Name AS Department, E.Name AS Employee, E.Salary
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FROM Employee AS E, Department AS D,
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(SELECT DepartmentId, MAX(Salary) AS Salary
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FROM Employee
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GROUP BY DepartmentId) AS M
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WHERE E.DepartmentId = D.Id
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AND E.DepartmentId = M.DepartmentId
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SELECT
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D.NAME AS Department,
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E.NAME AS Employee,
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E.Salary
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FROM
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Employee AS E,
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Department AS D,
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( SELECT DepartmentId, MAX( Salary ) AS Salary FROM Employee GROUP BY DepartmentId ) AS M
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WHERE
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E.DepartmentId = D.Id
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AND E.DepartmentId = M.DepartmentId
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AND E.Salary = M.Salary;
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```
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# 未完待续...
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# 176. Second Highest Salary
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https://leetcode.com/problems/second-highest-salary/description/
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## Description
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```html
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+----+--------+
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| Id | Salary |
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+----+--------+
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| 1 | 100 |
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| 2 | 200 |
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| 3 | 300 |
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+----+--------+
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```
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查找工资第二高的员工。
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```html
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+---------------------+
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| SecondHighestSalary |
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+---------------------+
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| 200 |
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+---------------------+
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```
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如果没有找到，那么就返回 null 而不是不返回数据。
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## SQL Schema
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```sql
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DROP TABLE
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IF
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EXISTS Employee;
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CREATE TABLE Employee ( Id INT, Salary INT );
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INSERT INTO Employee ( Id, Salary )
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VALUES
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( '1', '100' ),
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( '2', '200' ),
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( '3', '300' );
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```
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## Solution
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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。
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```sql
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SELECT
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( SELECT DISTINCT Salary FROM Employee ORDER BY Salary DESC LIMIT 1, 1 ) AS SecondHighestSalary;
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```
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# 177. Nth Highest Salary
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## Description
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查找工资第 N 高的员工。
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## SQL Schema
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同 176。
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## Solution
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```sql
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CREATE FUNCTION getNthHighestSalary ( N INT ) RETURNS INT BEGIN
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SET N = N - 1;
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RETURN ( SELECT ( SELECT DISTINCT Salary FROM Employee ORDER BY Salary DESC LIMIT N, 1 ) );
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END
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```
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# 178. Rank Scores
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https://leetcode.com/problems/rank-scores/description/
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## Description
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得分表：
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```html
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+----+-------+
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| Id | Score |
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+----+-------+
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| 1 | 3.50 |
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| 2 | 3.65 |
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| 3 | 4.00 |
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| 4 | 3.85 |
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| 5 | 4.00 |
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| 6 | 3.65 |
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+----+-------+
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```
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将得分排序，并统计排名。
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```html
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+-------+------+
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| Score | Rank |
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+-------+------+
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| 4.00 | 1 |
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| 4.00 | 1 |
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| 3.85 | 2 |
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| 3.65 | 3 |
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| 3.65 | 3 |
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| 3.50 | 4 |
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+-------+------+
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```
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## SQL Schema
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```sql
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DROP TABLE
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IF
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EXISTS Scores;
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CREATE TABLE Scores ( Id INT, Score DECIMAL ( 3, 2 ) );
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INSERT INTO Scores ( Id, Score )
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VALUES
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( '1', '3.5' ),
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( '2', '3.65' ),
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( '3', '4.0' ),
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( '4', '3.85' ),
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( '5', '4.0' ),
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( '6', '3.65' );
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```
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## Solution
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```sql
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SELECT
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S1.score,
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COUNT( DISTINCT S2.score ) AS Rank
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FROM
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Scores AS S1
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INNER JOIN Scores AS S2 ON S1.score <= S2.score
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GROUP BY
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S1.id
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ORDER BY
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S1.score DESC;
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```
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# 180. Consecutive Numbers
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https://leetcode.com/problems/consecutive-numbers/description/
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## Description
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数字表：
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```html
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+----+-----+
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| Id | Num |
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+----+-----+
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| 1 | 1 |
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| 2 | 1 |
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| 3 | 1 |
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| 4 | 2 |
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| 5 | 1 |
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| 6 | 2 |
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| 7 | 2 |
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+----+-----+
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```
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查找连续出现三次的数字。
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```html
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+-----------------+
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| ConsecutiveNums |
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+-----------------+
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| 1 |
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+-----------------+
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```
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## SQL Schema
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```sql
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DROP TABLE
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IF
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EXISTS LOGS;
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CREATE TABLE LOGS ( Id INT, Num INT );
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INSERT INTO LOGS ( Id, Num )
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VALUES
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( '1', '1' ),
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( '2', '1' ),
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( '3', '1' ),
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( '4', '2' ),
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( '5', '1' ),
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( '6', '2' ),
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( '7', '2' );
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```
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## Solution
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```sql
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SELECT
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DISTINCT L1.num AS ConsecutiveNums
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FROM
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Logs AS L1,
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Logs AS L2,
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Logs AS L3
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WHERE L1.id = l2.id - 1
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AND L2.id = L3.id - 1
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AND L1.num = L2.num
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AND l2.num = l3.num;
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```
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