［Leetcode］database 相关题目的总结

Combine Two Tables

https://leetcode.com/problems/combine-two-tables/

解题思路：

     合并table Person 和Address，其中提示说不管Address里有没有personId，Person里的记录都要显示全部。这是典型的 left join 功能。

select P.FirstName, P.LastName, A.City, A.State 
     from Person P 
          left outer join Address A 
         on P.PersonId = A.PersonId

知识点：left join (left outer join) 关键字从左表返回所有的行，即使在右表中没有匹配的行

Second Highest Salary

https://leetcode.com/problems/second-highest-salary/

解题思路：

     找到第二大的数字。第一反应就是先找到 max，然后从比max小的数里面找个最大的。

select max(Salary) from Employee
where Salary <
(select max(Salary) from Employee)

知识点：max 聚合函数，不能跟where一起，如果要把聚合函数作为条件，用having

Nth Highest Salary

https://leetcode.com/problems/nth-highest-salary/

解题思路：

     写个找第N大数字的函数，先排序再取数字。

CREATE FUNCTION getNthHighestSalary(N INT) RETURNS INT
BEGIN
  set N = N-1;
  RETURN (
      select distinct Salary from Employee order by Salary desc limit N,1

  );
END

知识点： limit N，相当于打印前N条； limit N,m， 打印从第N+1开始的m笔记录。计数从0开始；

limit N-1 是语法错误的，这里只能用 numeric

Rank Scores

https://leetcode.com/problems/rank-scores/

解题思路：

     对每一个Score，找出>=它的score的数量，这个数量就是rank

select Score ,
(select count(*) from (select distinct Score s from Scores) tmp where s >= Score)
from Scores
order by Score desc

知识点：在内层的select中 s >= Score  表示，对一个Score，每一个s都跟它比较一次。

相当于外面的select是外层for，内嵌的select是内层for的感觉

Employees Earning More Than Their Managers

https://leetcode.com/problems/employees-earning-more-than-their-managers/

解题思路：

     看成两个表e1和e2，找到满足e1.ManagerId = e2.Id 并且 e1.Salary > e2.Salary的记录

select e1.Name as Employee from Employee e1, Employee e2 
where e1.ManagerId = e2.Id 
and e1.Salary > e2.Salary

Duplicate Emails

https://leetcode.com/problems/duplicate-emails/

解题思路：

     找到group以后count > 1的记录

select Email from Person group by Email having count(Email) > 1

知识点：count作为合计函数，无法与where一起使用，所以出现了having

Customers Who Never Order

https://leetcode.com/problems/customers-who-never-order/  

解题思路：

     先 left join，再找到 CustomerId为空的记录

select C.Name as Customers 
from Customers C left join Orders O on C.Id = O.CustomerId
where O.CustomerId is NULL

Department Highest Salary

https://leetcode.com/problems/department-highest-salary/

解题思路：

     合并两个table，然后看salary是否是 max的那个

select D.Name as Department, E.Name as Employee, E.Salary 
from 
    Department D,
    Employee   E
where 
    D.Id = E.DepartmentId
and E.Salary in
(Select max(Salary) from Employee where DepartmentId = E.DepartmentId);

知识点： max

Department Top Three Salaries

https://leetcode.com/problems/department-top-three-salaries/

解题思路：

     先join，然后再来个select count，计算比当前salary大的数量，0表是当前salary最大，以此类推，找到 count ＝ 0，1，2的就是了

select D.Name as Department, E1.Name as Employee, E1.Salary
from 
    Employee E1
join Department D
    on E1.DepartmentId = D.Id
where 3 > (select count(*) from (select distinct Salary, DepartmentId from Employee) E2
    where E2.Salary > E1.Salary 
        and E1.DepartmentId = E2.DepartmentId)

知识点：count(*) , distinct , 3 > (table_numeric_column) 

Delete Duplicate Emails

https://leetcode.com/problems/delete-duplicate-emails/

解题思路：

     找到以email Group的最小id，然后删除所有不在这个id  集合里面的所有记录

delete from Person where Id not in (select A.Id from (select min(Id) as Id from Person Group by Email) A)

知识点：not in (table column )

Rising Temperature

https://leetcode.com/problems/rising-temperature/

解题思路：

     当成2个表来进行比较。条件满足 DATEDIFF(a.Data, b.Data) = 1 并且 a.温度 > b.温度

select a.Id from Weather as a, Weather as b where DATEDIFF(a.Date, b.Date) = 1 and a.Temperature > b.Temperature

知识点：DATEDIFF(a, b) = 1 当a-b=1时；

Trips and Users

https://leetcode.com/problems/trips-and-users/

解题思路：

     说不清楚，直接看指令吧

select t.Request_at Day, round(sum(t.Status like "cancelled_%") / count(*), 2) Rate from Trips t join Users u 
on t.Client_Id = u.Users_Id
where u.Banned = "No" and t.Request_at >="2013-10-01" and t.Request_at <= "2013-10-03"
group by t.Request_at

知识点：

round( , 2), sum(), like, count, group, where