Solutions to ten questions of leetcode database

stay Before doing algorithm problems Found on the way ,LeetCode The database solution is introduced above , There are ten questions , So I did it in these two nights . The answer is secondary, and the main benefit is , Just review SQL How to write some query statements , Such as custom variables and common functions . The topics are relatively simple , Explain less , Mainly post questions and answers .

	#	Title	Acceptance	Difficulty
	175	Combine Two Tables	32.5%	Easy
	176	Second Highest Salary	23.8%	Easy
	177	Nth Highest Salary	14.1%	Medium
	178	Rank Scores	20.7%	Medium
	180	Consecutive Numbers	20.2%	Medium
	181	Employees Earning More Than Their Managers	44.2%	Easy
	182	Duplicate Emails	38.0%	Easy
	183	Customers Who Never Order	34.2%	Easy
	184	Department Highest Salary	19.2%	Medium
	185	Department Top Three Salaries	16.3%	Hard

Combine Two Tables

【 subject 】

Table: Person

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| PersonId    | int     |
| FirstName   | varchar |
| LastName    | varchar |
+-------------+---------+
PersonId is the primary key column for this table.

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Table: Address

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| AddressId   | int     |
| PersonId    | int     |
| City        | varchar |
| State       | varchar |
+-------------+---------+
AddressId is the primary key column for this table.

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Write a SQL query for a report that provides the following information for each person in the Person table, regardless if there is an address for each of those people:

FirstName, LastName, City, State

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【 answer 】

select p.FirstName, p.LastName, a.City, a.State from Person p left outer join Address a on p.PersonId=a.PersonId;

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Second Highest Salary

【 subject 】

Write a SQL query to get the second highest salary from the Employee table.

+----+--------+
| Id | Salary |
+----+--------+
| 1  | 100    |
| 2  | 200    |
| 3  | 300    |
+----+--------+

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For example, given the above Employee table, the second highest salary is 200. If there is no second highest salary, then the query should return null.

【 answer 】 The title is simple. , But if it's like me , I found that I don't remember several commonly used functions , You can review .

select IFNULL( (select e.Salary from Employee e group by e.Salary order by e.Salary desc limit 1, 1), NULL) SecondHighestSalary;

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Nth Highest Salary

【 subject 】

Write a SQL query to get the nth highest salary from the Employee table.

+----+--------+
| Id | Salary |
+----+--------+
| 1  | 100    |
| 2  | 200    |
| 3  | 300    |
+----+--------+

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For example, given the above Employee table, the nth highest salary where n = 2 is 200. If there is no nth highest salary, then the query should return null.

【 answer 】 The first n high , You have to use custom variables , I seldom use this thing , So I reviewed it first .

CREATE FUNCTION getNthHighestSalary(N INT) RETURNS INT
BEGIN
    RETURN (
        select IFNULL(Salary, NULL) Salary from (
            select @row_num := @row_num+1 Rank, Salary from (
                select Salary from Employee group by Salary desc
            ) t1 join (
                select @row_num := 0 from dual
            ) t2
        ) t where t.Rank=N
    );
END

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Rank Scores

【 subject 】

Write a SQL query to rank scores. If there is a tie between two scores, both should have the same ranking. Note that after a tie, the next ranking number should be the next consecutive integer value. In other words, there should be no “holes” between ranks.

+----+-------+
| Id | Score |
+----+-------+
| 1  | 3.50  |
| 2  | 3.65  |
| 3  | 4.00  |
| 4  | 3.85  |
| 5  | 4.00  |
| 6  | 3.65  |
+----+-------+

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For example, given the above Scores table, your query should generate the following report (order by highest score):

+-------+------+
| Score | Rank |
+-------+------+
| 4.00  | 1    |
| 4.00  | 1    |
| 3.85  | 2    |
| 3.65  | 3    |
| 3.65  | 3    |
| 3.50  | 4    |
+-------+------+

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【 answer 】 Another topic of custom variables , This pattern should be familiar , It's quite common . If you can't use “set @var_name=0;” Words （ Ask for one sentence SQL Get it done ）, That can be defined in the clause “select @var_name:=0”, Then use this variable outside it .

select s.Score, t.Rank from (
    select @row_num:[email protected]_num+1 Rank, Score from (
         select Score from Scores group by Score desc
    ) t1 join (
        select @row_num := 0 from dual
    ) t2
) t, Scores s where s.Score=t.Score group by Score desc, Rank asc, Id;

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Consecutive Numbers

【 subject 】

Write a SQL query to find all numbers that appear at least three times consecutively.

+----+-----+
| Id | Num |
+----+-----+
| 1  |  1  |
| 2  |  1  |
| 3  |  1  |
| 4  |  2  |
| 5  |  1  |
| 6  |  2  |
| 7  |  2  |
+----+-----+

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For example, given the above Logs table, 1 is the only number that appears consecutively for at least three times.

【 answer 】

select DISTINCT(l1.Num) from Logs l1, Logs l2, Logs l3 where l1.Id+1=l2.Id and l1.Id+2=l3.Id and l1.Num=l2.Num and l1.Num=l3.Num;

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Employees Earning More Than Their Managers

【 subject 】

The Employee table holds all employees including their managers. Every employee has an Id, and there is also a column for the manager Id.

+----+-------+--------+-----------+
| Id | Name  | Salary | ManagerId |
+----+-------+--------+-----------+
| 1  | Joe   | 70000  | 3         |
| 2  | Henry | 80000  | 4         |
| 3  | Sam   | 60000  | NULL      |
| 4  | Max   | 90000  | NULL      |
+----+-------+--------+-----------+

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Given the Employee table, write a SQL query that finds out employees who earn more than their managers. For the above table, Joe is the only employee who earns more than his manager.

+----------+
| Employee |
+----------+
| Joe      |
+----------+

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【 answer 】

select e.Name from Employee e, Employee m where e.ManagerId=m.Id and e.Salary>m.Salary;

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Duplicate Emails

【 subject 】

Write a SQL query to find all duplicate emails in a table named Person.

+----+---------+
| Id | Email   |
+----+---------+
| 1  | [email protected] |
| 2  | [email protected] |
| 3  | [email protected] |
+----+---------+

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For example, your query should return the following for the above table:

+---------+
| Email   |
+---------+
| [email protected] |
+---------+

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Note: All emails are in lowercase.

【 answer 】

select distinct(p.Email) from Person p, Person q where p.Id!=q.Id and p.Email=q.Email;

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Customers Who Never Order

【 subject 】

Suppose that a website contains two tables, the Customers table and the Orders table. Write a SQL query to find all customers who never order anything.

Table: Customers.

+----+-------+
| Id | Name  |
+----+-------+
| 1  | Joe   |
| 2  | Henry |
| 3  | Sam   |
| 4  | Max   |
+----+-------+

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Table: Orders.

+----+------------+
| Id | CustomerId |
+----+------------+
| 1  | 3          |
| 2  | 1          |
+----+------------+

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Using the above tables as example, return the following:

+-----------+
| Customers |
+-----------+
| Henry     |
| Max       |
+-----------+

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【 answer 】

select c.Name Customers from Customers c where c.Id not in (
    select CustomerId from Orders
)

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Department Highest Salary

【 subject 】

The Employee table holds all employees. Every employee has an Id, a salary, and there is also a column for the department Id.

+----+-------+--------+--------------+
| Id | Name  | Salary | DepartmentId |
+----+-------+--------+--------------+
| 1  | Joe   | 70000  | 1            |
| 2  | Henry | 80000  | 2            |
| 3  | Sam   | 60000  | 2            |
| 4  | Max   | 90000  | 1            |
+----+-------+--------+--------------+

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The Department table holds all departments of the company.

+----+----------+
| Id | Name     |
+----+----------+
| 1  | IT       |
| 2  | Sales    |
+----+----------+

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Write a SQL query to find employees who have the highest salary in each of the departments. For the above tables, Max has the highest salary in the IT department and Henry has the highest salary in the Sales department.

+------------+----------+--------+
| Department | Employee | Salary |
+------------+----------+--------+
| IT         | Max      | 90000  |
| Sales      | Henry    | 80000  |
+------------+----------+--------+

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【 answer 】

select d.Name Department, e.Name Employee, s.Salary from (
    select MAX(e.Salary) Salary, e.DepartmentId from Employee e, Department d where e.DepartmentId=d.Id group by e.DepartmentId
) s, Employee e, Department d where s.Salary=e.Salary and e.DepartmentId=d.Id and e.DepartmentId=s.DepartmentId;

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Department Top Three Salaries

【 subject 】

The Employee table holds all employees. Every employee has an Id, and there is also a column for the department Id.

+----+-------+--------+--------------+
| Id | Name  | Salary | DepartmentId |
+----+-------+--------+--------------+
| 1  | Joe   | 70000  | 1            |
| 2  | Henry | 80000  | 2            |
| 3  | Sam   | 60000  | 2            |
| 4  | Max   | 90000  | 1            |
| 5  | Janet | 69000  | 1            |
| 6  | Randy | 85000  | 1            |
+----+-------+--------+--------------+

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The Department table holds all departments of the company.

+----+----------+
| Id | Name     |
+----+----------+
| 1  | IT       |
| 2  | Sales    |
+----+----------+

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Write a SQL query to find employees who earn the top three salaries in each of the department. For the above tables, your SQL query should return the following rows.

+------------+----------+--------+
| Department | Employee | Salary |
+------------+----------+--------+
| IT         | Max      | 90000  |
| IT         | Randy    | 85000  |
| IT         | Joe      | 70000  |
| Sales      | Henry    | 80000  |
| Sales      | Sam      | 60000  |
+------------+----------+--------+

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【 answer 】 Take this question MySql It's actually a little difficult , If I use Oracle, I can rank() over(partition by xxx order by xxx desc) This is what happened , however MySql The frustration is that there is no such partition operation . But in the end, it was done with the help of three user-defined variables . The meaning is not explained , It's easy to understand ：

select Department, Employee, Salary from (
    select
        IF(@lastDep!=t1.Department, @count:=0, @count:[email protected]), IF(@lastDep=t1.Department and @lastSalary!=t1.Salary, @count:[email protected]+1, @count:[email protected]) Cnt,
        @lastDep:=t1.Department, @lastSalary:=t1.Salary,
        t1.Department, t1.Employee, t1.Salary
    from (
        select d.Name Department, e.Name Employee, e.Salary
        from Department d, Employee e where d.Id=e.DepartmentId order by Department asc, Salary desc
    ) t1, (
            select @lastDep:=null, @lastSalary:=0, @count:=0 from dual
    ) t2
) f where Cnt<3;

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After the event , I went to the discussion area , Find a beautiful answer , There are no custom variables , The key is distinct(Salary) Go to Heyuan Salary Compare , Filter out that the number of occurrences under this condition is greater than 3 The situation of ：

select D.Name as Department, E.Name as Employee, E.Salary as Salary 
  from Employee E, Department D
   where (select count(distinct(Salary)) from Employee 
           where DepartmentId = E.DepartmentId and Salary > E.Salary) in (0, 1, 2)
         and 
           E.DepartmentId = D.Id 
         order by E.DepartmentId, E.Salary DESC;

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