练习一

思路：先将两表连结，再在分组中取最大值后连结

select e.employee,e.salary,d.department_name department
from Employee as e
join Department as d
on e.departmentid=d.department_id;

之后就不会了，连结给我整懵了。。。

正确答案：

SELECT
    Department.name AS 'Department',
    Employee.name AS 'Employee',
    Salary
FROM
    Employee
        JOIN
    Department ON Employee.DepartmentId = Department.Id
WHERE
    (Employee.DepartmentId , Salary) IN 
    (   SELECT --有可能有多个员工同时拥有最高工资，所以最好在这个查询中不包含雇员名字的信息
            DepartmentId, MAX(Salary)
        FROM
            Employee
        GROUP BY DepartmentId
	)
;

有句话特别重要：有可能有多个员工同时拥有最高工资，所以最好在这个查询中不包含雇员名字的信息。

练习二

思路：偶数上移一位，奇数下移一位，如果最后一位是奇数，则不变。

我想用if实现，感觉if比case...when...then写起来更简便。

我瞎写的  区分奇偶使用了%2，即取余；如果n%2=0，则说明n是偶数。

select if (id%2=0,id=replace(id,id,id-1),id=replace(id,id,id+1))as id,student
from seat;

leecode评论区答案，膜！希望有朝一日我也这么牛X

select if (id%2=0, id-1, if(id=(select count(id) from seat),id,id+1)) as id , student
from seat
order by id asc;

我写的太麻烦了，用replace简直大材小用了，答主的if嵌套用的好熟练。。。 

MYSQL中IF：

IF(X,A,B)

如果X为True，则返回A，否则，返回B

 练习三

思路：dense_rank排序+窗口函数

select score as "Score",DENSE_RANK() over(order by score DESC) as "Rank"
from Scores;

让我不太李姐的是leecode的重要提示：对于 MySQL 解决方案，如果要转义用作列名的保留字，可以在关键字之前和之后使用撇号。例如 `Rank`，或许李姐会理解？

练习四

select count(case when a.Num >= b.Num then count()+1 else count()=0 end) as ConsecutiveNums
from Logs as a
left join Logs as b
on a.Num >= b.Num
;