SQL LEETCODE 所有题（加锁题）多种解答总结

大家好，我最近总结了所有leetcode sql题的多种解法，跟大家分享一下。还有就是我差50分就能做有下载权限了，请大家让我体验一次起飞的感觉，谢谢！

Sqlzoo,modeanalytics,hackerank,leetcode
JOIN
        1. 标准句式之join+on/where
SELECT A.NAME
FROM EMPLOYEE A JOIN EMPLOYEE B
ON A.MANAERID=B.ID
WHERE A.Salary>b.Salary
On 用来限制join 的key
Where 用来限制最后呈现出的数据符合的条件
先开始就要把所有需要的fields在join another table之前都列出来
        2. 先开始列出多个表格时，用where
select distinct l1.num as ConsecutiveNums from logs l1,logs l2,logs l3
where l1.id=l2.id-1 and l2.id=l3.id-1
and l1.num=l2.num=l3.num

Select dep.name as department,emp.name as employee,emp.salary
From department dep,employee emp
Where emp.departmentid=dep.id
And emp.salary=(select max(salary) from employee e2 where e2.departmentid=dep.id)

 
JOIN的分类
如果我们想查找在第一个表里出现的并且在第二个表里也出现的（并且
Join by default是指inner join的
因此像上面那道题就可以直接用join 因为我们只关心inner table的内容
 
Left join会保存所以第一个table的内容 不管第二个table有没有匹配项目 通常可以用来查找第一个表里有但是第二个表里没有的东西
 
Example: 181 Employees Earning More Than their managers
select a.Name as Customers from Customers a
 join Orders b
on a.Id=b.CustomerId
where b.CustomerId is null
 
这样inner join return的结果是0 因为inner join的条件下没有customerid是null的情况
select a.Name as Customers from Customers a
LEFT join Orders b
on a.Id=b.CustomerId
where b.CustomerId is null
 
This one works. And if we change where b.CustomerId is null to where b.ID is null it also works but with slower performance
 
Select department highest salary



Option 1: 
SELECT D.Name AS Department ,E.Name AS Employee ,E.Salary
FROM
        Employee E,
        (SELECT DepartmentId,max(Salary) as max FROM Employee GROUP BY DepartmentId) T,
        Department D
WHERE E.DepartmentId = T.DepartmentId
  AND E.Salary = T.max
  AND E.DepartmentId = D.id
Option 2:
.. SELECT D.Name AS Department ,E.Name AS Employee ,E.Salary
from
        Employee E,
        Department D
WHERE E.DepartmentId = D.id
  AND (DepartmentId,Salary) in
  (SELECT DepartmentId,max(Salary) as max FROM Employee GROUP BY DepartmentId)



 
. 1point3acres.com RANK
SECOND HIGHEST SALARY


Option 1:
select max(salary) as SecondHighestSalary from
Employee
where Salary<(Select Max(Salary) from Employee)
 
Option 2:
select max(salary) as SecondHighestSalary
from Employee
where salary not in (
select max(salary)
from Employee
)
 
SQL BASICS
Find duplicated emails

# Write your MySQL query statement below
SELECT Email FROM (select
                  email,count(email) as num
                  from person
                  group by email) L
                  where num>1
Delete duplicate emails
Option 1:
DELETE FROM PERSON
WHERE ID NOT IN(SELECT MINID FROM (SELECT EMAIL,MIN(ID) AS MINID FROM PERSON
GROUP BY EMAIL) AS TMP)
Option 2:
Delete p1 from person p1,
Person p2
Where p1.email=p2.email and p1.id>p2.id


 无需要join只需要select符合条件的 (SUBQUERY)
The use of subquery


To display name, location,phone_number of the students from database table whose section is A

Answer:
SELECT NAME, LOCATION, PHONE_NUMBER FROM DATABASE
WHERE ROLL_NO IN
(SELECT ROLL_NO FROM STUDENT WHERE SECTION='A')







SELECT REQUEST_AT AS DAY,
ROUND(SUM(CASE WHEN STATUS LIKE 'cancelled%' THEN 1 ELSE 0 END)/COUNT(*),2) AS 'Cancellation Rate' from (
SELECT * FROM TRIPS T
WHERE
T.CLIENT_ID NOT IN (SELECT USER_ID FROM USERS WHERE BANED='Yes') and
T.DRIVER_ID NOT IN (SELECT USERS_ID FROM USERS WHERE BANNED='Yes') and
T.REQUEST_AT BETWEEN '2013-10-01' AND '2013-10-03') AS NEWT
GROUP BY RESUEST_AT



Answer 1:

select customers.name as 'Customers'
from customers
where customers.id not in
(
    select customerid from orders
);

Answer 2:
Select c.name as customers
From customers as c
Left join orders o
On c.id=o.customerid
Where o.customerid is null

注意这里不能用where 因为where 会only select 符合条件的（inner join相当于）会return nothing
select customers.name as customers from customers,orders
where customers.id=orders.customerid
and orders.customerid is null （错误答案）

下面一个例子同样不能用where语句要用join并选择合适的join的


Select d.dept_name, count(s.student_id) as student_number
From student s right join department d on s.dept_id=d.dept_id
Group by d.dept_name
Order by student_number desc, d.dept_name




How to join tables with itself (where /on两种写法都可以


Answer 1:
select e1.name as employee from employee e1
join employee e2
on e1.managerid=e2.id
where e1.salary>e2.salary

Answer 2:
SELECT
    *
FROM
    Employee AS a,
    Employee AS b
WHERE
    a.ManagerId = b.Id
        AND a.Salary > b.Salary

 
Given a Weather table, write a SQL query to find all dates' Ids with higher temperature compared to its previous (yesterday's) dates.
 
select p1.Id from Weather p1 join Weather p2
on datediff(p1.recorddate,p2.recorddate)=1 and p1.Temperature>p2.temperature
 
select p1.Id from Weather p1, Weather p2
where datediff(p1.recorddate,p2.recorddate)=1 and p1.Temperature>p2.temperature

Consecutive available seats 
当设计到连续的座位数（连续的天数）我们可以用table自己join自己 on 相差数为1 的方式，去得到我们想要的结果
. ----因为只有两个consecutive就满足条件，所以只能是join 两个table

Question:

Answer:
 select distinct a.seat_id
From cinema a
Join cinema b
On abs(a.seat_id-b.seat_id)=1
And a.free=true and b.free=true
Order by a.seat_id
 
Question:

Notice the difference between this one and the above one is that this one doesn't require you to join on abs value as this one is asking for the num instead of the id.


 Answer 1:
select distinct l1.num as ConsecutiveNums from logs l1,logs l2,logs l3
where l1.id=l2.id-1 and l2.id=l3.id-1
and l1.num=l2.num=l3.num

Answer 2:
Select t.num as consecutivenums
From
(select distinct a.num from
Logs a
Left join logs b on a.id=b.id-1
Left join logs c on a.id=c.id-2
Where a.num=b.num and a.num=c.num) t


Side knowledge:
The TO_DAYS() function returns the number of days between a date and year 0 (date "0000-00-00"). The TO_DAYS() function can be used only with dates within the Gregorian calendar
 
Answer 1:
SELECT WT1.ID
FROM WEATHER WT1 JOIN WEATHER WT2
ON TO_DAYS(WmT1.RECORDDATE)-TO_DAYS(WT2.RECORDDATE)=1
AND WT1.TEMPERATURE>WT2.TEMPERATURE

Answer 2:
SELECT WT1.ID
FROM WEATHER WT1 JOIN WEATHER WT2
ON TO_DAYS(WT1.RECORDDATE)-TO_DAYS(WT2.RECORDDATE)=1
WHERE WT1.TEMPERATURE>WT2.TEMPERATURE

Answer 3:
SELECT WT1.ID
FROM WEATHER WT1,WEATHER WT2
WHERE WT1.TEMPERATURE>WT2.TEMPERATURE AND
TO_DAYS(WT1.RECORDDATE)-TO_DAYS(WT2.RECORDDATE)=1


How to choose add number
Side Knowledge:
-baidu 1point3acresMySQL MOD() returns the remainder of a number divided by another number. This function also works on fractional values and returns the exact remainder. The function returns NULL when the value of divisor is 0

Question:
Please write a SQL query to output movies with an odd numbered ID and a description that is not 'boring'. Order the result by rating.
select * from cinema
where mod(id,2)=1 and description !='boring'
order by rating DESC
 
  
Select from temp table
Answer
Select class from (select count(distinct student) as num,class from courses
. 1point 3 acresGroup by class) as temp where num>=5  (notice, the parenthesis is before table name)


Difference between Where and On
Where is a part of the SELECT query as a whole, on is a part of each individual join
On can only refer to the fields of previously used tables

The on clause defines the relationship between the tables
The where clause describes which rows you are interested in
Many times you can swap them and still get the same result
However this is not always the case with a left outer join

As a rule of thumb, you should use columns that join your tables in On clauses and columns that are used for filtering in where clauses. This provides the best readability

Different types of the JOINs in SQL
Inner join: returns records that have matching values in both tables
Left(outer) join:return all records from the left table and the matched records from the right table
Right(outer) join: return all records from the right table and the matched records from the left table
Full(outer) join: return all records when there is a match in either left or right table

Managers with at least 5 direct reports




Option 1:
思路：可以通过inner join来确保另一个field满足条件
Select name from employee as t1 join
(select managerid from employee
Group by managerid
Having count(managerid)>=5) as t2
On t1.id=tj2.managerid

Option 2:
思路：可以不join而通过where限制另一个field的条件 (subquery)
Select name from employee
Where id in (select managerid from employee
Group by managerid
Having count(*)>=5)


The use of offset
Get nth salary
Answer:
CREATE FUNCTION getNthHighestSalary(N INT) RETURNS INT
BEGIN
DECLARE off_var INT;
SET off_var = N-1;
RETURN (
# Write your MySQL query statement below.
SELECT DISTINCT
salary
FROM
Employee
ORDER BY Salary DESC
LIMIT 1 OFFSET off_var
);
END

Offset will offset number of rows and start reading the table.

Ranking



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
            DepartmentId, MAX(Salary)
        FROM
            Employee
        GROUP BY DepartmentId
        )

It looks like we can't use e as a table name after from clause
And we need parenthesis when have two fields in where conditions
Where (employee.departmentid,salary)

Department Top 3 Salaries
在subquery里用自己和自己比较 （比较salary）


Option 1:
Select dep.Name as Department, emp.Name as Employee, emp.Salary from Department dep,
.--Employee emp where emp.DepartmentId=dep.Id and
(Select count(distinct Salary) From Employee where DepartmentId=dep.Id and Salary>emp.Salary)<3

Option 2:
select d.Name Department, e1.Name Employee, e1.Salary
from Employee e1
join Department d
on e1.DepartmentId = d.Id
where  (select count(distinct(e2.Salary))
                  from Employee e2
                  where e2.Salary > e1.Salary
                  and e1.DepartmentId = e2.DepartmentId
                  )<3;

Option 3: Oracle
SELECT Department, Employee, Salary FROM (
SELECT DENSE_RANK() OVER (PARTITION BY d.Name ORDER BY e.Salary DESC) AS Rank, d.Name AS Department, e.Name AS Employee, e.Salary, d.Id FROM Department d
JOIN (SELECT * FROM Employee) e ON d.Id = e.DepartmentId
)a
WHERE Rank <= 3
ORDER BY Id

Rank scores




Side knowledge:
DENSE_RANK computes the rank of a row in an ordered group of rows and returns the rank as a NUMBER . The ranks are consecutive integers beginning with 1. The largest rank value is the number of unique values returned by the query. ... Rows with equal values for the ranking criteria receive the same rank.

Answer:
SELECT SCORE, (SELECT COUNT(DISTTINCT SCORE) FROM SCORES WHERE SCORE>=S.SCORE) RANK
FROM SCORES S
ORDER BY SCORE DESC

Metric Calculation

注意在计算一个metric涉及到除法的时候，用ifnull用0代替null
注意在calculate题中，只是提取两个table里的内容不需要join两个table的话，可以直接select from 两个table而不需要让两个table join起来
Select ifnull(round(count(distinct requester_id,acceper_id)/count(distinct sender_i,send_to_id),2,),0) as accept_Rate
From rerquest_accepted,friend_request



思路：当遇到这种复杂问题时候，先考虑select 呈现最后想要的table，然后逐渐join 提取不同的fields
Answer:
Solve this problem by 3 steps as below
        1. Calculate the company's average salary in every month.
        Select avg(amount) as company_avg, date_format(pay_date,'%Y-%m') as pay_month from salary
        Group by date_format(pay_date,'%Y-%m')
        2. Calculate the each department's average salary in every month
        Select department_id,avg(amount) as department_avg, date_format(pay_date,'%Y-%m') as pay_month from salary
        Join employee on salary.employee_id=employee.employee_id
        Group by department_id,pay_month
        3. At last, combine the above two query and join them on department_salary.pay_month=company_salary.pay_month

        select department_salary.pay_month, department_id,
case
  when department_avg>company_avg then 'higher'
  when department_avg<company_avg then 'lower'
  else 'same'
end as comparison
from
(
  select department_id, avg(amount) as department_avg, date_format(pay_date, '%Y-%m') as pay_month
  from salary join employee on salary.employee_id = employee.employee_id
  group by department_id, pay_month
) as department_salary
join
(
  select avg(amount) as company_avg,  date_format(pay_date, '%Y-%m') as pay_month from salary group by date_format(pay_date, '%Y-%m')
) as company_salary
on department_salary.pay_month = company_salary.pay_month
       
        Winning candidate
       

思路：可以通过join的方式去找符合条件的id，也可以通过where语句去限制另一个table 找到符合条件的id

Option 1:
SELECT
    name AS 'Name'
FROM
    Candidate,,,
        JOIN
    (SELECT
        Candidateid
    FROM
        Vote
    GROUP BY Candidateid
    ORDER BY COUNT(*) DESC
    LIMIT 1) AS winner
WHERE
    Candidate.id = winner.Candidateid
;

Option 2:
Select name
From candidate
Where id=(select candidateid
From vote
Group by candidateid
Order by count(*) desc
Limit 1)

Option 3:
Select name From candidate join vote on
Candidate.id=vote.candidateid
Group by candidateid
Order by count(*) desc
Limit 1  (not efficient)

Find cumulative salary of an employee




select a.id, a.month, sum(b.salary) from employee a, employee b
where a.id = b.id and a.month >= b.month
and a.month < (select max(month) from employee c
where a.id = c.id)
group by a.id, a.month

One more way of doing it using ROW_NUMBER:
WITH s AS
(SELECT Id,Month,Salary,
Sum(Salary) OVER (PARTITION BY Id ORDER BY Month) as SumSal,
ROW_NUMBER() OVER (PARTITION BY id ORDER BY id ASC, month DESC) rn
FROM emp1)
SELECT Id,Month,SumSal as Salary
FROM s
WHERE rn > 1

Ifnull, if statement


Answer:
Select T.id,
If(isnull(T.p_id),'Root',if(T.id in (select p_id from tree),'Inner','Leaf')) Type
From tree T


Answer:
Select x,y,z
Case when x+y<=z OR
x+z<=y or
y+z<=x
Then 'no'
Else 'yes'
End as 'triangle'
From traingle

Get highest answer rate question
利用subquery自己制作一个表格来select



Answer:
Select question_id as survey_log
From
(select question_id, sum(case when action='show' then 1 else 0 end) as num_show,
Sum(case when action='answer' then 1 else 0 end) as num_answer
From survey_log
Group by question_id) as tbl
Order by (num_answer/num_show) DESC LIMIT1

How to deal with null value: 注意要考虑到null的情况也是<1000 ，但是系统不会记入


Answer:
SELECT NAME,BONUS
FROM EMPLOYEE LEFT JOIN BONUS
ON EMPLOYEE.EMPID=BONUS.EMPID
WHERE BONUS<1000 OR BONUS IS NULL

不可以用where as where is inner join
Select name,bonus from employee.bonus
Where …



Answer:
Select name from customer
Where referr_id<>2 or referee_id is null

Investments in 2016



Answer:
利用subquery
SELECT SUM(INSURANCE.TIV_2016) AS TIV_2016
FROM INSURANCE
WHERE INSURANCE.TIV_2015 IN   #注意语法 要先写这个field然后in 而不是where （select）
(SELECT
TIV_2015
FROM INSURANCE
GROUP BY TIV_2015
HAVING COUNT(*)>1
)
AND CONCAT(LAT,LON) IN
(SELECT
CONCAT(LAT,LON)
FROM INSURANCE
GROUP BY LAT,LON
HAVING COUNT(*)=1)

UDPATE Sentence in SQL

Answer:
UPDATE SALARY
SET
SEX=CASE SEX
WHEN 'M' THEN 'F'
ELSE 'M'
END;

hints：
当我们要选取second biggest number或者只出现了一次的biggest nb的时候，注意可以先select max 然后通过subquery的形式进一步限制条件



Answer:
SELECT
    MAX(num) AS num
FROM
    (SELECT
        num
    FROM
        number
    GROUP BY num
    HAVING COUNT(num) = 1) AS t




Answer:
SELECT
    America, Asia, Europe
FROM
    (SELECT @as:=0, @am:=0, @eu:=0) t,
    (SELECT
        @as:=@as + 1 AS asid, name AS Asia
    FROM
        student
    WHERE
        continent = 'Asia'
    ORDER BY Asia) AS t1
        RIGHT JOIN
    (SELECT
        @am:=@am + 1 AS amid, name AS America
    FROM
        student
    WHERE
        continent = 'America'
    ORDER BY America) AS t2 ON asid = amid
        LEFT JOIN
    (SELECT
        @eu:=@eu + 1 AS euid, name AS Europe
    FROM
        student
    WHERE
        continent = 'Europe'
    ORDER BY Europe) AS t3 ON amid = euid

Second degree follower

Basically is to count follower's follower
Option 1:
Select distinct follower, num from follow,
(select followee, count(distinct follower) as num from follow
Group by followee) t
Where follower=t.followee
Order by follower

Option 2:
Select f1.follower, count(distinct f2.follower) as num
From follow f1
Join follow f2 on f1.follower=f2.followee
Group by f1.follower
Order by f1.follower




Select min(abs(p1.x-p2.x)) as shortest from point as P1
JOIN POINT AS P2 ON P1.X <>P2.X



SELECT ROUND(SQRT(MIN(POW(A.X-B.X,2)+POW(A.Y-B.Y,2))),2) SHORTEST
FROM POINT_2D A,POINT_2D B
WHERE (A.X,A.Y)!=(B.X,B.Y)

MULTIPLE TABLES JOIN

两种做法：
三个表格的关系可能是，第一个table和第三个有common field 然后第二个和第三个有common field。这是可以先把第一个和第三个连接起来，然后用连接后的表和第二个join

找另一个表的info不再这个表上的可以考虑用subquery




Option 1:
Select s.name from salesperson s
Where s.sales_id not in (select o.sales_id from orders o left join company c on o.com_id=c.com_id
. ----Where c.name='RED')

Option 2:
SELET NAME FROM SALESPERSON WHERE NAME NOT IN(
SELECT S.NAME FROM SALESPERSON S,COMPANY C, ORDERS O
WHERE S.SALES_ID=O.SALES_ID AND C.COM_ID=O.COM_ID AND C.NAME='RED')

Friend requests: who has the most friends


Friend request example. No difference between requester and accepter then we can use union
Select id1 as id, count(id2) as num from
(select requester_id as id1, acccpter_id as id2 from request_accepted
Union
Select accepter_id as id1, requester_id as id2 from
Request_accepted) temp1
. Group by id1
Order by num desc limit 1



Answer:
Select customer_number from orders
Group by custoemr_number
Order by count(*) desc limit 1

How to join Consecutive numbers on id


注意：
join的关键不在于每个表格都要在不同的位置出现 在下题其实s2，s3并不重要，最重要的是s1的位置，s1一定要在第一位，第二位，第三位都出现，就完成了
Option
SELECT DISTINCT S1.*
FROM stadium S1
JOIN stadium S2     #注意join语句是可以直接join一个以上的表然后再一起写on条件
JOIN stadium S3
ON ((S1.id = S2.id - 1 AND S1.id = S3.id -2)  #s1<s2<s3    s1在末尾
OR (S3.id = S1.id - 1 AND S3.id = S2.id -2)    #s3<s1<s2    s1在中间
OR (S3.id = S2.id - 1 AND S3.id = S1.id -2))   #s3<s2<s1    s1在头顶
WHERE S1.people >= 100
AND S2.people >= 100
AND S3.people >= 100
ORDER BY S1.id;

Option 2
Select s1.* from stadium as s2, stadium as s2, stadium as s3
Where ((s1.id+1=s2.id
And s1.id+2=s3.id)
Or
(s1.id-1=s2.id
And s1.id+1=s3.id)
Or
(s1.id-2=s2.id
And s1.id-1=s3.id))
And s1.people>=100
And s2.people>=100
And s3.people>=100
Group by s1.id


补充内容 (2019-5-30 10:12):
求大家加大米~楼主还差一些分就可以下载了，感谢~

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感谢楼主分享~

楼主有看过发内容吗，怎么很多都乱七八糟的。比如：
Example: 181 Employees Earning More Than their managers
select a.Name as Customers from Customers a
join Orders b
on a.Id=b.CustomerId
where b.CustomerId is null
文不对题啊

感谢分享！

太感谢楼主的分享啦！

给楼主点个赞！

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