【持续更新】Leetcode SQL题目全解析（附建表sql）

Leetcode SQL题目全解析

越前须知（雾）
题目Q & A
- 175 组合两个表
- 181 超过经理收入的员工
- 182 查找重复电子邮箱
- 183 从不订购的用户
- 197 上升的温度
- 511 & 512 游戏玩法分析
- 577 员工奖金
- 584 查询用户推荐人
- 586 订单最多的客户
- 595 超过5名学生的课
- 597 好友申请通过率
- 603 连续空余座位
- 607 没有向某公司销售产品的销售员
- 610 判断三角形
- 613 两点间最近的距离
- 619 只出现一次的最大数
- 1050 合作过至少3次的演员和导演
- 1068 & 1069 产品销售
- 1075 & 1076 项目员工
- 1082 & 1083 & 1084 销售分析
- 1113 报告的记录
- 1141 查询近30天活跃用户数
- 1148 浏览文章
- 1173 即时食物配送
- 1211 查询结果质量和占比
- 1241 每个帖子评论人数
- 1251 查询售出单品平均售价
- 1270 向公司CEO汇报的人
- 1280 学生们参加各科测试的次数
- 1294 不同国家的天气类型
- 1303 查询团队人数

越前须知（雾）

本系列将详解Leetcode的SQL题库（附建表语句），采用DBMS为MySQL。
本系列仅提供博主个人解法，均经过MySQL检验，且解法有对应注意点，请放心食用。
本系列适合从入门到进阶选手；若希望系统学习SQL进阶知识，请移步博主主页。

题目Q & A

175 组合两个表

创表

create table Person
(personid int,firstname varchar(255), lastname varchar(255));create table Address
(addressid int,personid int, city varchar(255), state varchar(255));insert into Person values (1, 'San', 'Zhang');
insert into Person values (2, 'Si',  'Li') ;insert into Address values ('291', 1, 'New York City', 'New York');

Q：查询Person表中成员的名、姓、城市、州
A：以Person表为主，外连接

select firstname, lastname, city, statefrom Person left outer join Addresson Person.personid = Address.personid;

181 超过经理收入的员工

创表

Create table If Not Exists Employee
(Id int, Name varchar(255), Salary int, ManagerId int);Truncate table Employee;
insert into Employee (Id, Name, Salary, ManagerId) values ('1', 'Joe', '70000', '3');
insert into Employee (Id, Name, Salary, ManagerId) values ('2', 'Henry', '80000', '4');
insert into Employee (Id, Name, Salary, ManagerId) values ('3', 'Sam', '60000', null);
insert into Employee (Id, Name, Salary, ManagerId) values ('4', 'Max', '90000', null);

Q：查询工资超过其对应经理的员工名字
A：不是每个员工都有对应经理，内连接，匹配字段分别为经理id和员工id

select E1.namefrom Employee E1, Employee E2where E1.managerid = E2.id  -- 经理也是员工and E1.salary > E2.salary;

182 查找重复电子邮箱

创表

create table Email
(id char(4),email varchar(50));truncate table Email;
insert into Email values ('1', 'a@b.com');
insert into Email values ('2', 'c@d.com');
insert into Email values ('3', 'a@b.com');

Q：查询重复的电子邮箱
A：COUNT（*）>1 / 自连接

COUNT(*)

select emailfrom Emailgroup by emailhaving count(*) > 1;

自连接，相当于查询邮箱相同但ID不同的“错填”信息

select distinct E1.emailfrom Email E1, Email E2where E1.email = E2.emailand E1.id <> E2.id;

183 从不订购的用户

创表

create table Customers
(id char(4),name varchar(10));create table Orders
(id char(4),customerid char(4));insert into Customers values ('1', 'Joe');
insert into Customers values ('2', 'Henry');
insert into Customers values ('3', 'Sam');
insert into Customers values ('4', 'Max');insert into Orders values ('1', '3');
insert into Orders values ('2', '1');

Q：查询用户总表中没有订购记录的用户
A：从不订购 = 用户总表 - 订购过用户，NOT EXISTS

select id, namefrom Customers Cwhere not exists (select *    from Orders Owhere C.id = O.customerid);

197 上升的温度

创表

create table Weather(id int,recorddate date,temperature int);insert into Weather values (1, '2015-01-01', 10);
insert into Weather values (2, '2015-01-02', 25);
insert into Weather values (3, '2015-01-03', 20);
insert into Weather values (4, '2015-01-04', 30);

Q：查询与昨天/过去最近一天相比，温度升高的日期
A：对应时间轴连续和时间轴不连续解法

时间轴连续

EXISTS

select W1.idfrom Weather W1where exists (select *from Weather W2where W1.recorddate = W2.recorddate + 1and W1.temperature > W2.temperature);

自连接

select W1.idfrom Weather W1, Weather W2where W1.recorddate = W2.recorddate + 1and W1.temperature > W2.temperature;

DATEDIFF()

select W1.idfrom Weather W1, Weather W2where W1.temperature > W2.temperatureand datediff(W1.recorddate, W2.recorddate) = 1;

时间轴不连续

EXISTS，引入W3.recorddate与W1.recorddate的关系

select W1.idfrom Weather W1where exists (select * from Weather W2where W1.temperature > W2.temperatureand W2.recorddate = (select max(W3.recorddate)from Weather W3where W1.recorddate > W3.recorddate));

自连接

select W1.idfrom Weather W1, Weather W2where W1.temperature > W2.temperatureand W2.recorddate = (select max(W3.recorddate)from Weather W3where W1.recorddate > W3.recorddate);

511 & 512 游戏玩法分析

创表

create table Activity
(player_id int,device_id int,event_date date,games_played int,primary key (player_id, event_date));insert into Activity values (1, 2, '2016-03-01', 5);
insert into Activity values (1, 2, '2016-05-02', 6);
insert into Activity values (2, 3, '2017-06-25', 1);
insert into Activity values (3, 1, '2016-03-02', 0);
insert into Activity values (3, 4, '2018-07-03', 5);

Q：查询每个用户最早登陆时间
A：MIN（）

select player_id, min(event_date) as first_loginfrom Activitygroup by player_id;

Q：查询每个用户最早“打开游戏”的时间
A：需要筛选 games_played 大于0

select player_id, min(event_date) as first_loginfrom Activitywhere games_played > 0group by player_id;

Q：查询每个玩家首次登陆的设备名称
A：子查询 / 视图

子查询，筛选 event_date

select player_id, device_idfrom Activity A1where event_date = (select min(event_date)from Activity A2where A1.player_id = A2.player_idgroup by A2.player_id);

临时表

select A1.player_id, A1.device_idfrom Activity A1 inner join (select player_id, min(event_date) as first_datefrom Activitygroup by player_id) A2on A1.player_id = A2.player_idand A1.event_date = A2.first_date;

577 员工奖金

创表

create table Employee
(empId int,name varchar(20),supervisor int,salary int,primary key (empId));create table Bonus
(empId int,bonus int);insert into Employee values (1, 'John', 3, 1000);
insert into Employee values (2, 'Dan',  3, 2000);
insert into Employee values (3, 'Brad', null, 4000);
insert into Employee values (4, 'Thomas', 3, 4000);insert into Bonus values (2, 500);
insert into Bonus values (4, 2000);

Q：查询奖金不为 2000 的员工
A：需要包含 Bonus 表中未提及的员工1、3，以 Employee 为主表左连接

select name, bonusfrom Employee E left outer join Bonus Bon B.empId = E.empIdwhere B.bonus < 1000or B.bonus is null;   -- NULL无法用运算符比较

584 查询用户推荐人

创表

create table Customer
(id int,name varchar(10),referee_id int);insert into Customer values (1, 'Will', null);
insert into Customer values (2, 'Jane', null);
insert into Customer values (3, 'Alex', 2);
insert into Customer values (4, 'Bill', null);
insert into Customer values (5, 'Zack', 1);
insert into Customer values (6, 'Mark', 2);

Q：查询推荐人不是 2 的客户
A：注意 IS NULL 的记录

select name from Customer where referee_id <>2 or referee_id is null;  -- IS NULL

586 订单最多的客户

创表

create table Orders
(order_number int,customer_number int,order_date date,required_date date,shipped_date date,status char(15),comment varchar(200),primary key (order_number));insert into Orders values (1, 1, '2017-04-09', '2017-04-13', '2017-04-12', 'Closed', null);
insert into Orders values (2, 2, '2017-04-15', '2017-04-20', '2017-04-18', 'Closed', null);
insert into Orders values (3, 3, '2017-04-16', '2017-04-25', '2017-04-20', 'Closed', null);
insert into Orders values (4, 3, '2017-04-18', '2017-04-28', '2017-04-25', 'Closed', null);

Q：查询订单最多的客户
A：极值函数 / ORDER BY + LIMIT（仅用于无并列情况）

极值函数

select customer_numberfrom Ordersgroup by customer_numberhaving count(*) >= all (select count(*)from Ordersgroup by customer_number);

ORDER BY + LIMIT，存在缺陷，即并列情况不能全部显示

select customer_numberfrom Ordersgroup by customer_numberorder by count(*) desclimit 1;

595 超过5名学生的课

创表

create table if not exists Courses
(student char(4),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');

Q：查询选课学生5人及以上的课程名
A：COUNT（）但存在陷阱，感谢网友提醒