Leetcode中的SQL题目练习(二) 175. Combine Two Tables 181. Employees Earning More Than Their Managers ?? 183. Customers Who Never Order 184. Department Highest Salary ??? 176. Second Highest Salary ?? 177. Nth Highest Salary 178. Rank Scores ?? 180. Consecutive Numbers 626. Exchange Seats ???
https://leetcode.com/problems/combine-two-tables/description/
Description
Person 表:
Column Name | Type |
---|---|
PersonId | int |
FirstName | varchar |
LastName | varchar |
Address 表:
Column Name | Type |
---|---|
AddressId | int |
PersonId | int |
City | varchar |
State | varchar |
查找 FirstName, LastName, City, State 数据,而不管一个用户有没有填地址信息。
solution
SELECT
FirstName,
LastName,
City,
State
FROM
Person P
LEFT JOIN Address A
ON P.PersonId = A.PersonId;
181. Employees Earning More Than Their Managers ??
https://leetcode.com/problems/employees-earning-more-than-their-managers/description/
Description
Employee 表:
Id | Name | Salary | ManagerId |
---|---|---|---|
1 | Joe | 70000 | 3 |
2 | Henry | 80000 | 4 |
3 | Sam | 60000 | NULL |
4 | Max | 90000 | NULL |
查找薪资大于其经理薪资的员工信息。
solution
SELECT
E1.NAME AS Employee
FROM
Employee E1
INNER JOIN Employee E2
ON E1.ManagerId = E2.Id
AND E1.Salary > E2.Salary;
183. Customers Who Never Order
https://leetcode.com/problems/customers-who-never-order/description/
Description
Curstomers 表:
Id | Name |
---|---|
1 | Joe |
2 | Henry |
3 | Sam |
4 | Max |
Orders 表:
Id | CustomerId |
---|---|
1 | 3 |
2 | 1 |
查找没有订单的顾客信息:
Customers |
---|
Henry |
Max |
soulution:
(1)子查询
SELECT
Name AS Customers
FROM
Customers
WHERE
Id NOT IN (SELECT CustomerId FROM Orders)
(2)
SELECT
C.Name AS Customers
FROM
Customers C
LEFT JOIN Orders O
ON C.Id = O.CustomerId
WHERE
O.CustomerId IS NULL;
184. Department Highest Salary ???
https://leetcode.com/problems/department-highest-salary/description/
Description
Employee 表:
Id | Name | Salary | DepartmentId |
---|---|---|---|
1 | Joe | 70000 | 1 |
2 | Henry | 80000 | 2 |
3 | Sam | 60000 | 2 |
4 | Max | 90000 | 1 |
Department 表:
Id | Name |
---|---|
1 | IT |
2 | Sales |
查找一个 Department 中收入最高者的信息:
Department | Employee | Salary |
---|---|---|
IT | Max | 90000 |
Sales | Henry | 80000 |
soulution:
SELECT
D.NAME Department,
E.NAME Employee,
E.Salary
FROM
Employee E,
Department D,
( SELECT DepartmentId, MAX( Salary ) Salary FROM Employee GROUP BY DepartmentId ) M
WHERE
E.DepartmentId = D.Id
AND E.DepartmentId = M.DepartmentId
AND E.Salary = M.Salary;
176. Second Highest Salary ??
https://leetcode.com/problems/second-highest-salary/description/
Description
Id | Salary |
---|---|
1 | 100 |
2 | 200 |
3 | 300 |
查找工资第二高的员工。
solution
SELECT
(SELECT DISTINCT
Salary
FROM
Employee
ORDER BY Salary DESC
LIMIT 1 OFFSET 1) AS SecondHighestSalary
177. Nth Highest Salary
https://leetcode.com/problems/nth-highest-salary/
Description
查找工资第 N 高的员工。
CREATE FUNCTION getNthHighestSalary ( N INT )
RETURNS INT
BEGIN
SET N = N - 1;
RETURN (
SELECT (
SELECT DISTINCT Salary
FROM Employee
ORDER BY Salary DESC
LIMIT N, 1 )
);
END
178. Rank Scores ??
https://leetcode.com/problems/rank-scores/description/
Description:
Id | Score |
---|---|
1 | 3.50 |
2 | 3.65 |
3 | 4.00 |
4 | 3.85 |
5 | 4.00 |
6 | 3.65 |
将得分排序,并统计排名。
Score | Rank |
---|---|
4.00 | 1 |
4.00 | 1 |
3.85 | 2 |
3.65 | 3 |
3.65 | 3 |
3.50 | 4 |
Solution:
(1)解法一:对于每一个分数,从表中找出有多少个大于或等于该分数的不重复分数,然后降序排列
select Score,
(select count(distinct Score) from Scores s2 where s2.Score >= s1.Score) Rank
from Scores s1
order by Score DESC;
(2)解法二:使用联结 join,条件是左表的分数小于等于右表的分数时,对右表的分数进行计数(即计算有几个不重复的分数大于自己,计算结果就是rank),然后根据id分组后,再根据分数降序排列
select s1.Score, count(distinct s2.Score) Rank
from Scores as s1 inner join Scores as s2 on s1.Score <= s2.Score
group by s1.Id
order by s1.Score DESC;
180. Consecutive Numbers
https://leetcode.com/problems/consecutive-numbers/description/
Description
数字表:
Id | Num |
---|---|
1 | 1 |
2 | 1 |
3 | 1 |
4 | 2 |
5 | 1 |
6 | 2 |
7 | 2 |
编写一个 SQL 查询,查找所有至少连续出现三次的数字。
ConsecutiveNums |
---|
1 |
solution:
(1)用l1分别和l2, l3内交,l1和l2的Id下一个位置比,l1和l3的下两个位置比,然后将Num都相同的数字返回即可
select distinct l1.Num ConsecutiveNums from Logs l1
join Logs l2 on l1.Id = l2.Id - 1
join Logs l3 on l1.Id = l3.Id - 2
where l1.Num = l2.Num and l2.Num = l3.Num;
(2)直接在三个表的实例中查找,然后把四个条件限定上,就可以返回正确结果了 ??
SELECT DISTINCT l1.Num FROM Logs l1, Logs l2, Logs l3
WHERE l1.Id = l2.Id - 1 AND l2.Id = l3.Id - 1
AND l1.Num = l2.Num AND l2.Num = l3.Num;
626. Exchange Seats ???
https://leetcode.com/problems/exchange-seats/description/
seat 表存储着座位对应的学生。
id | student |
---|---|
1 | Abbot |
2 | Doris |
3 | Emerson |
4 | Green |
5 | Jeames |
要求交换相邻座位的两个学生,如果最后一个座位是奇数,那么不交换这个座位上的学生。
id | student |
---|---|
1 | Doris |
2 | Abbot |
3 | Green |
4 | Emerson |
5 | Jeames |
solution
对照上表及其查询结果可以得知,当原id为奇数时,交换座位后的id变为id+1,当原id为偶数时,交换座位后的id变为id-1,另一个方面需要考虑的是,学生人数为奇数时,最后一个学生的id不变,故应当用子查询确定学生的人数,然后分情况讨论即可
select (case
when mod(id,2)!=0 and id!=counts then id+1
when mod(id,2)!=0 and id=counts then id
else id-1 end)as id,student
from seat,(select count(*)as counts from seat)as seat_counts
order by id;