【sql】leetcode习题 (共 42 题)

【175】Combine Two Tables （2018年11月23日，开始集中review基础）

Table: Person
+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| PersonId    | int     |
| FirstName   | varchar |
| LastName    | varchar |
+-------------+---------+
PersonId is the primary key column for this table.
Table: Address
+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| AddressId   | int     |
| PersonId    | int     |
| City        | varchar |
| State       | varchar |
+-------------+---------+
AddressId is the primary key column for this table. 
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 

题解：因为题目要求说person表里面有的项目即使address表里没有也需要展示，所以用 left join

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

【176】Second Highest Salary （第二高的工资）（2018年11月23日）

Write a SQL query to get the second highest salary from the Employee table.
+----+--------+
| Id | Salary |
+----+--------+
| 1  | 100    |
| 2  | 200    |
| 3  | 300    |
+----+--------+

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

注意，题目有个要求，如果没有第二高的工资要返回 null，而不是空条目。还有一个问题就是如果表里只有两条，但是两条的工资都是100， 这个需要返回 null，不是 100，所以要用 distinct

我一开始写成了如下，但是没有第二高的工资要返回 null 这个条件不满足。所以 WA。

select Salary as SecondHighestSalary from Employee order by Salary desc limit 1, 1;

后来看了答案，答案说要重新搞一张表。（limit 1，1 和 limit 1 offset 1 是等价的）

select (select distinct Salary from Employee order by Salary desc limit 1 offset 1) as SecondHighestSalary;

【177】Nth Highest Salary （2018年11月23日）

Write a SQL query to get the nth highest salary from the Employee table.
+----+--------+
| Id | Salary |
+----+--------+
| 1  | 100    |
| 2  | 200    |
| 3  | 300    |
+----+--------+

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.
+------------------------+
| getNthHighestSalary(2) |
+------------------------+
| 200                    |
+------------------------+

题意就是返回第 N 高的工资。和上面一题很像。注意点就是不能直接写 limit N-1, 1 语法会出错。

CREATE FUNCTION getNthHighestSalary(N INT) RETURNS INT
BEGIN
#limit m, n 表示的是从第 m 个条目(m is 0 based)开始的 n 条, 如果下面直接用 N-1 的话不行的，语法错误。
  DECLARE M INT;
  SET M = N - 1;
  RETURN (
      select (select distinct Salary from Employee order by Salary desc limit M, 1)
  );
END

【178】Rank Scores （2018年11月23日）

【180】Consecutive Numbers （2018年11月23日）

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  |
+----+-----+

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

题解：本来不会写，后来找了题解：https://my.oschina.net/Tsybius2014/blog/494823

可以用 select， 也可以 join， 还有一种通解的写法（如果把 3 延长到 N怎么办）

select distinct L1.Num as ConsecutiveNums
from Logs L1, Logs L2, Logs L3
where (L1.Id + 1 = L2.Id AND L1.Num = L2.Num) AND (L1.Id + 2 = L3.Id AND L2.Num = L3.Num) 

【181】Employees Earning More Than Their Managers

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      |
+----+-------+--------+-----------+

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      |
+----------+

题解：别名的使用。

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

【182】Duplicate Emails （group by ... having ..子句， 2018年11月23日）

Write a SQL query to find all duplicate emails in a table named Person.
+----+---------+
| Id | Email   |
+----+---------+
| 1  | [e-mail] |
| 2  | [e-mail] |
| 3  | [e-mail] |
+----+---------+

For example, your query should return the following for the above table:
+---------+
| Email   |
+---------+
| [e-mail] |
+---------+
Note: All emails are in lowercase.

题解：

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

【183】Customers Who Never Order

【184】Department Highest Salary

【185】Department Top Three Salaries

【196】Delete Duplicate Emails

【197】Rising Temperature

【262】Trips and Users

【569】Median Employee Salary

【570】Managers with at Least 5 Direct Reports

【571】Find Median Given Frequency of Numbers

【574】Winning Candidate (2018年11月24日)

Table: Candidate
+-----+---------+
| id  | Name    |
+-----+---------+
| 1   | A       |
| 2   | B       |
| 3   | C       |
| 4   | D       |
| 5   | E       |
+-----+---------+
Table: Vote
+-----+--------------+
| id  | CandidateId  |
+-----+--------------+
| 1   |     2        |
| 2   |     4        |
| 3   |     3        |
| 4   |     2        |
| 5   |     5        |
+-----+--------------+
id is the auto-increment primary key,
CandidateId is the id appeared in Candidate table.
Write a sql to find the name of the winning candidate, the above example will return the winner B.
+------+
| Name |
+------+
| B    |
+------+
Notes:
You may assume there is no tie, in other words there will be at most one winning candidate.

题解：注意 group by 和 order by 一起使用的时候，order by 中的列必须要出现在 group by 中。（solution里面还有别的方法）

select Name
from Candidate
where id = (select CandidateId from Vote group by CandidateId order by count(CandidateId) desc limit 1);                   

【577】Employee Bonus （2018年11月23日）

Select all employee's name and bonus whose bonus is < 1000.
Table:Employee
+-------+--------+-----------+--------+
| empId |  name  | supervisor| salary |
+-------+--------+-----------+--------+
|   1   | John   |  3        | 1000   |
|   2   | Dan    |  3        | 2000   |
|   3   | Brad   |  null     | 4000   |
|   4   | Thomas |  3        | 4000   |
+-------+--------+-----------+--------+
empId is the primary key column for this table.
Table: Bonus
+-------+-------+
| empId | bonus |
+-------+-------+
| 2     | 500   |
| 4     | 2000  |
+-------+-------+
empId is the primary key column for this table.
Example ouput:
+-------+-------+
| name  | bonus |
+-------+-------+
| John  | null  |
| Dan   | 500   |
| Brad  | null  |
+-------+-------+

题解：left join语句，还用到了 sql 的三值逻辑（true, false, unkown）

select emp.name as name, bon.bonus as bonus
from Employee as emp
left join Bonus as bon on emp.empId = bon.empId
where bon.bonus < 1000 or bon.bonus is null

【578】Get Highest Answer Rate Question

【579】Find Cumulative Salary of an Employee

【580】Count Student Number in Departments

【584】Find Customer Referee （2018年11月23日）

Given a table customer holding customers information and the referee.
+------+------+-----------+
| id   | name | referee_id|
+------+------+-----------+
|    1 | Will |      NULL |
|    2 | Jane |      NULL |
|    3 | Alex |         2 |
|    4 | Bill |      NULL |
|    5 | Zack |         1 |
|    6 | Mark |         2 |
+------+------+-----------+
Write a query to return the list of customers NOT referred by the person with id '2'.
For the sample data above, the result is:
+------+
| name |
+------+
| Will |
| Jane |
| Bill |
| Zack |
+------+

题解：本题需要了解的知识点是， sql 的三值逻辑: true, false, unkown。一切和 null 比较的值都是 unkown, 包括 null 本身。所以 sql 提供了 'is null' 和 'is not null' 这两个关键词。

select name from customer where referee_id <> 2 or referee_id is null;

如果条件中只有 referee_id <> 2 这一个条件的话，那么只会返回 Zack 这一个结果。

【585】Investments in 2016

【586】Customer Placing the Largest Number of Orders

【595】Big Countries

【596】Classes More Than 5 Students

【597】Friend Requests I: Overall Acceptance Rate

【601】Human Traffic of Stadium

【602】Friend Requests II: Who Has the Most Friends

【603】Consecutive Available Seats

【607】Sales Person

【608】Tree Node

【610】Triangle Judgement

【612】Shortest Distance in a Plane

【613】Shortest Distance in a Line

【614】Second Degree Follower

【615】Average Salary: Departments VS Company

【618】Students Report By Geography

【619】Biggest Single Number

【620】Not Boring Movies

【626】Exchange Seats