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Leetcode Database 18道题解题记录

2017-08-21 16:27 477 查看
Leetcode 数据库板块，不收费的18道SQL练习题解题记录，GitHub

175 Combine Two Tables

# Write your MySQL query statement below.
SELECT
FirstName,
LastName,
City,
State
FROM
Person
LEFT OUTER JOIN Address ON Person.PersonId = Address.PersonId;


176 Second Highest Salary

# Write your MySQL query statement below.
SELECT
IFNULL(
(
SELECT DISTINCT
Salary
FROM
Employee
ORDER BY
Salary DESC
LIMIT 1 OFFSET 1
),
NULL
) AS SecondHighestSalary;


使用了 
IFNULL(expr1, expr2)
 函数，将空值替换为NULL

177 Nth Highest Salary

CREATE FUNCTION getNthHighestSalary (N INT) RETURNS INT
BEGIN

DECLARE oset INT;
SET oset = N - 1;

RETURN (
# Write your MySQL query statement below.
SELECT
IFNULL(
(
SELECT DISTINCT
Salary
FROM
Employee
ORDER BY
Salary DESC
LIMIT oset, 1
),
NULL
)
);

END


limit后不能直接写N-1，因为limit后面不能接表达式；

178 Rank Scores

方法一：

# Write your MySQL query statement below
SELECT
Score,
(
SELECT
count(*)
FROM
(
SELECT DISTINCT
Score
FROM
Scores
) tmp
WHERE
tmp.Score >= Score
) As Rank
FROM
Scores
ORDER BY
Score DESC


对每一个Score，找出>=它的score的数量，这个数量就是Rank；

方法二：

# Write your MySQL query statement below
SELECT
Scores.Score AS Score,
COUNT(*) AS Rank
FROM
Scores
JOIN (
SELECT DISTINCT
Score
FROM
Scores
) Ranking ON Scores.Score <= Ranking.Score
GROUP BY
Scores.Id
ORDER BY
Scores.Score DESC;


求得每个Score与大于等于该Score的内连接表，再按照id分组，统计即可；

180. Consecutive Numbers

# Write your MySQL query statement below
SELECT DISTINCT
log1.Num AS ConsecutiveNums
FROM
LOGS log1,
LOGS log2,
LOGS log3
WHERE
log1.Id = log2.Id - 1
AND log2.Id = log3.Id - 1
AND log1.Num = log2.Num
AND log2.Num = log3.Num;


自联结查询

SELECT DISTINCT log1.Num
中的 
log1
可替换为 
log2
, 
log3
；

类似题目（更复杂）：601. Human Traffic of Stadium

181 Employees Earning More Than Their Managers

# Write your MySQL query statement below
SELECT
a.`Name` AS Employee
FROM
Employee AS a
INNER JOIN Employee AS b ON a.ManagerId = b.Id
AND a.Salary > b.Salary;


182. Duplicate Emails

方式一：

# Write your MySQL query statement below
SELECT
Email
FROM
Person
GROUP BY
Email
HAVING
count(Email) > 1


方式二：

# Write your MySQL query statement below
SELECT DISTINCT
a.Email
FROM
Person AS a
INNER JOIN Person AS b ON a.Id != b.Id
AND a.Email = b.Email;


183. Customers Who Never Order

# Write your MySQL query statement below
SELECT
Customers. NAME AS 'Customers'
FROM
Customers
WHERE
Customers.Id NOT IN (
SELECT DISTINCT
CustomerId
FROM
Orders
);


184 Department Highest Salary

SELECT
D.Name AS Department,
E.Name AS Employee,
E.Salary AS Salary
FROM
Employee E
INNER JOIN Department D ON E.DepartmentId = D.Id
AND (E.DepartmentId, E.Salary) IN (
SELECT
DepartmentId,
MAX(Salary) AS max
FROM
Employee
GROUP BY
DepartmentId
);


将Employee表按照DepartmentId分组，再将各组中Salary最高的对应Id和该Salary组成一个集合，即各Department中最高Salary的集合；

再逐一判断employee的id和Salary是否在该集合中即可；

185 Department Top Three Salaries

# Write your MySQL query statement below
SELECT
d. NAME AS 'Department',
e1. NAME AS 'Employee',
e1.Salary
FROM
Employee e1
INNER 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
ORDER BY
Department;


工资前三名的员工，可理解为：在该部门中比这三名员工中任一位工资高的人不超过3人，据此便可作为WHERE条件，加上 DepartmentId 连接，便可找出一个 Department 中工资前三的员工记录，最后再与 Department 表内连接，列出部门名称即可；

196 Delete Duplicate Emails

# Write your MySQL query statement below
DELETE p1
FROM
Person p1,
Person p2
WHERE
p1.Email = p2.Email
AND p1.Id > p2.Id;


197. Rising Temperature

# Write your MySQL query statement below
SELECT
w1.Id
FROM
Weather w1
INNER JOIN Weather w2 ON DATEDIFF(w1.Date, w2.Date) = 1
AND w1.Temperature > w2.Temperature;


使用 
DATEDIFF(expr1, expr2)
 函数

262. Trips and Users

# Write your MySQL query statement below
SELECT
t.Request_at AS `Day`,
ROUND(
SUM(IF (t.Status = 'completed', 0, 1)) / COUNT(*), 2
) AS `Cancellation Rate`
FROM
Trips t
INNER JOIN Users u ON t.Client_Id = u.Users_Id
WHERE
t.Request_at BETWEEN '2013-10-01'
AND '2013-10-03'
AND u.Banned = 'No'
AND u.Role = 'client'
GROUP BY
t.Request_at
ORDER BY
t.Request_at;


先筛选出符合条件的 trip 记录，然后按日期进行分组，再分别计算各组的 Cancellation Rate；

计算 Cancellation Rate 时，使用 
IF(expr1,expr2,expr3)
 函数、
ROUND()
函数、
sum()
函数；

595 Big Countries

# Write your MySQL query statement below
SELECT
NAME,
population,
area
FROM
World
WHERE
area > 3000000
OR population > 25000000;


596 Classes More Than 5 Students

# Write your MySQL query statement below
SELECT
class
FROM
courses
GROUP BY
class
HAVING
COUNT(DISTINCT student) >= 5;


WHERE
子句用于筛选行记录，筛选分组要使用 
HAVING
子句；

601 Human Traffic of Stadium

# Write your MySQL query statement below
SELECT DISTINCT
t1.*
FROM
stadium t1,
stadium t2,
stadium t3
WHERE
t1.people >= 100
AND t2.people >= 100
AND t3.people >= 100
AND (
(
t3.id - t2.id = 1
AND t2.id - t1.id = 1
AND t3.id - t1.id = 2
) -- id: t1 < t2 < t3
OR (
t2.id - t1.id = 1
AND t2.id - t3.id = 2
AND t1.id - t3.id = 1
) -- id: t3 < t1 < t2
OR (
t1.id - t2.id = 1
AND t1.id - t3.id = 2
AND t2.id - t3.id = 1
) -- id: t3 < t2 < t1
)
ORDER BY
t1.id;


自联结查询

由于需要列出匹配的所有记录，因此此处的t1需能匹配到第一天、第二天、第三天……直到连续最后一天的所有情况；

若 
SELECT DISTINCT t1.*
中的 
t1
替换成 
tx
，则 WHERE 子句中的条件需要相应地做出改变；

类似题目（简单）：180. Consecutive Numbers

620 Not Boring Movies

# Write your MySQL query statement below
SELECT
*
FROM
cinema
WHERE
MOD (id, 2) != 0
AND description != 'boring'
ORDER BY
rating DESC;


627 Swap Salary

# Write your MySQL query statement below
UPDATE salary
SET
sex = CASE sex
WHEN 'm' THEN 'f'
ELSE 'm'
END;


使用 
CASE WHEN
 流程控制
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