175. Combine Two Tables
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;