LeetCode 数据库刷题

发现LeetCode还能做数据库的题，最近比较闲，就一起做了。题目还挺难的，中等的基本就不会了，题解大法好。

175. 组合两个表

思路：left join即可

SELECT
	p.FirstName,
	p.LastName,
	a.City,
	a.State
FROM
	Person p
LEFT JOIN Address a ON p.PersonId = a.PersonId

175.第二高的薪水

思路：学习到了ifnull这个函数，当查询结果为空时，返回定义的结果。distinct去重

SELECT
	ifnull(
		(
			SELECT DISTINCT
				Salary
			FROM
				Employee
			ORDER BY
				Salary DESC
			LIMIT 1,
			1
		),
		NULL
	) AS SecondHighestSalary;

177. 第N高的薪水

思路：和前一题差不多

CREATE FUNCTION getNthHighestSalary (N INT) RETURNS INT
BEGIN

SET N = N - 1;

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


END

178. 分数排名

思路：首先查出所有分数，并按照分数排序，排名通过count函数来得到。考虑统计所有大于等于该分数的人数，就是该分数的排名。

SELECT
	a.score score,
	(
		SELECT
			count(DISTINCT b.score)
		FROM
			scores b
		WHERE
			b.score >= a.score
	) rank
FROM
	scores a
ORDER BY
	a.score DESC

180. 连续出现的数字

思路：题目怎么说，sql怎么写。id用于判断连续，num用于判断相等，distinct去重

SELECT DISTINCT
	l1.num 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
AND l2.num = l3.num

181. 超过经理收入的员工

思路：题目怎么说，sql怎么写

SELECT
	a. NAME Employee
FROM
	Employee a,
	Employee b
WHERE
	a.managerid = b.id
AND a.salary > b.salary

182. 查找重复的电子邮箱

思路：使用having筛选group后的数据

SELECT
	Email
FROM
	Person
GROUP BY
	Email
HAVING
	count(Email) > 1

183. 从不订购的客户

思路：left join之后，查询customerid为空的人

SELECT
	c. NAME Customers
FROM
	Customers c
LEFT JOIN Orders o ON c.id = o.customerid
WHERE
	customerid IS NULL

184. 部门工资最高的员工

思路：先将两张表inner join，得到交集部分。然后对于employee表，group by之后，选出最高的salary和departmentid。使用in语句查询出名字，不直接查出名字是因为可能有多个相同的最高工资。

SELECT
	b. NAME Department,
	a. NAME Employee,
	a.salary Salary
FROM
	(
		employee a
		INNER JOIN department b ON a.departmentid = b.id
	)
WHERE
	(a.departmentid, a.salary) IN (
		SELECT
			c.departmentid,
			max(salary)
		FROM
			employee c
		GROUP BY
			c.departmentid
	)

196. 删除重复的电子邮箱

思路：将Email和自身根据email和id两个条件inner join，查出id较大的重复数据，然后删除即可

DELETE p1.*
FROM
	person p1
INNER JOIN person p2 ON p1.email = p2.email
AND p1.id > p2.id

197. 上升的温度

思路：主要学习了DATEDIFF的使用，也可以使用date_sub，date_sub好像比较慢
DATEDIFF：

SELECT
	w2.id
FROM
	Weather w1,
	Weather w2
WHERE
	DATEDIFF(
		w2.RecordDate,
		w1.RecordDate
	) = 1
AND w1.Temperature < w2.Temperature

date_sub：

SELECT
	w2.id
FROM
	Weather w1,
	Weather w2
WHERE
	w1.RecordDate = date_sub(w2.RecordDate, INTERVAL 1 DAY)
AND w1.Temperature < w2.Temperature