如何快速掌握MYSQL？附LeetCode上出现频率最高的50道数据库题目详解

Posted 2021-10-16 报告，今天也有好好学习

如何才能快速掌握mysql？如何熟悉使用SQL以满足日常工作需求？

如果你目前啥也不会，只停留在知道SELECT用于查询的层面的话，又想要快速掌握MYSQL，那么刷题，并且过程中不会什么补什么，就是巩固和提升自己的SQL语言能力最快捷的方法。

LeetCode中有不少题是需要Plus会员才能查看并答题的，所以为避免以后会员过期无法再次查看以及加深自身对题目的理解，进行了MYSQL题目的一次大整理。

LeetCode上题目众多，而本篇筛选出了LetCode上出现频率最高的50道题目，并且按出现频率由高到低整理，每道题实现的具体方法和思路都贴在了代码注释，整理不易，希望大家能够做完这些题目或者看完这篇博客，并从中有所获。

PS：本篇博客主要是使用增删查改中的查询功能。

目录

• NO.1 简单 (175. 组合两个表)
• NO.2 简单 (176. 第二高的薪水)
• NO.3 中等 (184. 部门工资最高的员工)
• NO.4 中等 (177. 第N高的薪水)
• NO.5 中等 (180. 连续出现的数字)
• NO.6 困难 (185. 部门工资前三高的所有员工)
• NO.7 简单 (1777. 每家商店的产品价格)
• NO.8 中等 (178. 分数排名)
• NO.9 简单 (181. 超过经理收入的员工)
• NO.10 困难 (262. 行程和用户)
• NO.11 困难 (601. 体育馆的人流量)
• NO.12 简单 (196. 删除重复的电子邮箱)
• NO.13 简单 (1179. 重新格式化部门表)
• NO.14 简单 (182. 查找重复的电子邮箱)
• NO.15 简单 (197. 上升的温度)
• NO.16 中等 (626. 换座位)
• NO.17 简单 (620. 有趣的电影)
• NO.18 简单 (627. 变更性别)
• NO.19 简单 (183. 从不订购的客户)
• NO.20 简单 (511. 游戏玩法分析 I)
• NO.21 简单 (595. 大的国家)
• NO.22 简单 (1873. 计算特殊奖金)
• NO.23 困难 (1097. 游戏玩法分析 V)
• NO.24 困难 (569. 员工薪水中位数)
• NO.25 中等 (1841. League Statistics)
• NO.26 困难 (571. 给定数字的频率查询中位数)
• NO.27 困难 (618. 学生地理信息报告)
• NO.28 简单 (1083. 销售分析 II)
• NO.29 中等 (1205. 每月交易II)
• NO.30 中等 (1501. 可以放心投资的国家)
• NO.31 简单 (1821. 寻找今年具有正收入的客户)
• NO.32 中等 (1204. 最后一个能进入电梯的人)
• NO.33 简单 (1082. 销售分析 I )
• NO.34 中等 (1098. 小众书籍)
• NO.35 中等 (1270. 向公司CEO汇报工作的所有人)
• NO.36 简单 (597. 好友申请 I：总体通过率)
• NO.37 中等 (1193. 每月交易 I)
• NO.38 简单 (1280. 学生们参加各科测试的次数)
• NO.39 中等 (1715. 苹果和橘子的个数)
• NO.40 简单 (1809. 没有广告的剧集)
• NO.41 简单 (577. 员工奖金)
• NO.42 简单 (603. 连续空余座位)
• NO.43 中等 (1112. 每位学生的最高成绩)
• NO.44 中等 (1308. 不同性别每日分数总计)
• NO.45 中等 (570. 至少有5名直接下属的经理)
• NO.46 中等 (580. 统计各专业学生人数)
• NO.47 简单 (1407. 排名靠前的旅行者)
• NO.48 简单 (1795. 每个产品在不同商店的价格)
• NO.49 中等 (550. 游戏玩法分析 IV)
• NO.50 困难 (579. 查询员工的累计薪水)
• 结束语

NO.1 简单 (175. 组合两个表)

https://leetcode-cn.com/problems/combine-two-tables/

# 因为题目要求无论 person 是否有地址信息，所以不能用join，要用left join
select FirstName, LastName, City, State 
from Person 
left join Address on Person.PersonId=Address.PersonId

NO.2 简单 (176. 第二高的薪水)

https://leetcode-cn.com/problems/second-highest-salary/

# 将第一高的工资筛选掉，再次用MAX就可以取到第二高的工资了。
select max(Salary) as SecondHighestSalary from Employee
where Salary <>
(select max(Salary) from Employee)

NO.3 中等 (184. 部门工资最高的员工)

https://leetcode-cn.com/problems/department-highest-salary/

# 先用group by的方法求出部门最高薪资，再接着判断Employee表中哪些数据符合即可，
# 过程中会用到join来连接Employee表和Department表。
select d.Name Department, e.Name Employee, Salary
from  Employee e
join Department d on e.DepartmentId=d.Id
where (DepartmentId, Salary) in
(
    select DepartmentId, MAX(Salary) 
    FROM Employee
    group by DepartmentId
)

NO.4 中等 (177. 第N高的薪水)

https://leetcode-cn.com/problems/nth-highest-salary/

CREATE FUNCTION getNthHighestSalary(N INT) RETURNS INT
BEGIN
    declare m int;
    set m = N - 1;
    RETURN (
    # 这里主要是考LIMIT的用法：
    /*
    LIMIT 5 等价于 LIMIT 0, 5, 即查询(0+1)到(0+5)的数据，也就是1-5的数据。也就是说这里的0是查询的数据起点的前一位，5是查询的数量。
    LIMIT 2 OFFSET 3 等价于 LIMIT 3 2，即查询(3+1)到(3+2)的数据，也就是4-5的数据。也就是说这里的3是查询的数据起点的前一位，2是查询的数量。
    */
    # 那么要查排名第N的数据,那么就是LIMIT N-1, 1，所以是LIMIT m, 1
    # 最后记得去重
    SELECT IFNULL(
        (SELECT DISTINCT Salary
        FROM Employee 
        ORDER BY Salary DESC
        LIMIT m, 1)
    , NULL)
  );
END

NO.5 中等 (180. 连续出现的数字)

https://leetcode-cn.com/problems/consecutive-numbers/

# 题目要求连续三次及以上，我们只要求出连续三次的数字有哪些即可。
# 取三个Logs表，用来判断。
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 AND l2.Num=l3.Num

NO.6 困难 (185. 部门工资前三高的所有员工)

https://leetcode-cn.com/problems/department-top-three-salaries/

/*
# 第一种方法是利用窗口函数和dense_rank(),能够对各个部门的工资进行排序
SELECT Department,Employee,Salary
FROM
(
    SELECT
        d.Name AS Department,
        e.Name AS Employee,
        e.Salary AS Salary,
        DENSE_RANK() OVER (PARTITION BY e.DepartmentId ORDER BY e.Salary DESC) AS rk 
    FROM
        Employee e
        JOIN Department d
        ON e.DepartmentId = d.Id
) AS t
WHERE rk<=3
*/
# 第二种方法运用到的思想是：部门中如果不超过两个工资比我的工资高，那我肯定是前三名。
# 比方说如果我是第二名，那只有一个工资比我高（如我工资是8000，部门其他人是9000,7000,9000,6000）
SELECT 
    d.Name AS Department,
    e.Name AS Employee,
    e.Salary AS Salary
FROM
    Employee e
    JOIN Department d
    ON e.DepartmentId = d.Id
WHERE
    e.Id IN
    (
        SELECT
            e1.Id
        FROM
            Employee e1
            LEFT JOIN Employee e2
            on e1.DepartmentId = e2.DepartmentId AND e1.Salary < e2.Salary
        GROUP BY e1.Id
        HAVING COUNT(DISTINCT e2.Salary) <= 2
    )
    AND e.DepartmentId in (SELECT Id from Department)
    order by d.Id asc,e.Salary desc

NO.7 简单 (1777. 每家商店的产品价格)

https://leetcode-cn.com/problems/products-price-for-each-store/

# 同题1795. 每个产品在不同商店的价格，正好相反
# 这里主要是记得要用到聚合函数
# 除了用IF函数，同样的也可以用CASE
SELECT 
    product_id, 
    SUM(IF(store='store1', price, NULL)) store1,
    SUM(IF(store='store2', price, NULL)) store2,
    SUM(IF(store='store3', price, NULL)) store3
FROM
    Products
GROUP BY product_id

NO.8 中等 (178. 分数排名)

https://leetcode-cn.com/problems/rank-scores/

# 第一种方法：利用DENSE_RANK() 窗口函数
/*
SELECT 
    Score,
    DENSE_RANK() OVER (ORDER BY Score DESC) AS 'Rank'
FROM
    Scores
*/
# 第二种方法：主要思想是有几个分数（去重）大于等于我的分数，那这分数就是排第几。
SELECT 
    a.Score,
    (
        SELECT 
            COUNT(DISTINCT b.Score) 
        FROM 
            Scores b
        WHERE a.Score <= b.Score
    ) AS 'Rank'
FROM 
    Scores a
ORDER BY a.Score DESC

NO.9 简单 (181. 超过经理收入的员工)

https://leetcode-cn.com/problems/employees-earning-more-than-their-managers/

# 自连接，将经理和对应的员工处于同一行，再比较即可。
SELECT e1.Name AS 'Employee' 
FROM Employee e1 JOIN Employee e2 ON e1.ManagerId = e2.Id
WHERE e1.Salary > e2.Salary

NO.10 困难 (262. 行程和用户)

https://leetcode-cn.com/problems/trips-and-users/

# 解题思路：首先要将乘客或者司机是非禁止用户的行程筛选出来；
# 接着按日期分组，统计各个日期总订单数；
# 再判断按Status来统计其中取消的订单数量。
SELECT t.Request_at AS 'Day', 
    ROUND((SUM(IF(t.Status='completed', 0, 1))) /
    COUNT(t.Status), 2) AS `Cancellation Rate`
/*
    ROUND((SUM(IF(t.Status='cancelled_by_driver' or t.Status='cancelled_by_client', 1, 0))) /
    COUNT(t.Status), 2) AS `Cancellation Rate`
*/
FROM Trips t
JOIN Users u1 ON (t.Client_Id=u1.Users_Id AND u1.Banned='No')
JOIN Users u2 ON (t.Driver_Id=u2.Users_Id AND u2.Banned='No')
WHERE t.Request_at BETWEEN '2013-10-01' AND '2013-10-03'
GROUP BY t.Request_at

NO.11 困难 (601. 体育馆的人流量)

https://leetcode-cn.com/problems/human-traffic-of-stadium/

# 窗口函数yyds
# 首先是将人流量大于100的数据筛选出来，然后用窗口函数根据id进行排序（一定要是rank而不是dense_rank）
# 利用id-rank，这样就可以将连续的数据分成同一个组（都拥有同一个order1）
# 接着再利用窗口函数根据order1分组并统计各组数量，大于等于3的数据即为所需数据。
SELECT id, visit_date, people
FROM
(
    SELECT id, visit_date, people, COUNT(*) OVER(PARTITION BY order1) AS order2
    FROM
    (
        SELECT *, (id - rank() OVER(ORDER BY id)) AS order1 FROM
        (
            SELECT * FROM Stadium WHERE people >= 100
        ) AS t1
    ) AS t2
) AS t3
WHERE order2 >= 3

NO.12 简单 (196. 删除重复的电子邮箱)

https://leetcode-cn.com/problems/delete-duplicate-emails/

# 注意是要用到DELETE操作。
DELETE p1
FROM Person p1, Person p2 
WHERE p1.Email=p2.Email AND p1.Id>p2.Id

NO.13 简单 (1179. 重新格式化部门表)

https://leetcode-cn.com/problems/reformat-department-table/

# IF和CASE都可以，同时注意用上SUM函数。
/*
select id,
    SUM(if(month="Jan", revenue, NULL)) as "Jan_Revenue",
    SUM(if(month="Feb", revenue, NULL)) as "Feb_Revenue",
    SUM(if(month="Mar", revenue, NULL)) as "Mar_Revenue",
    SUM(if(month="Apr", revenue, NULL)) as "Apr_Revenue",
    SUM(if(month="May", revenue, NULL)) as "May_Revenue",
    SUM(if(month="Jun", revenue, NULL)) as "Jun_Revenue",
    SUM(if(month="Jul", revenue, NULL)) as "Jul_Revenue",
    SUM(if(month="Aug", revenue, NULL)) as "Aug_Revenue",
    SUM(if(month="Sep", revenue, NULL)) as "Sep_Revenue",
    SUM(if(month="Oct", revenue, NULL)) as "Oct_Revenue",
    SUM(if(month="Nov", revenue, NULL)) as "Nov_Revenue",
    SUM(if(month="Dec", revenue, NULL)) as "Dec_Revenue"
FROM Department
group by id
*/
select id,
    SUM(CASE WHEN month="Jan" THEN revenue ELSE NULL END) as "Jan_Revenue",
    SUM(CASE WHEN month="Feb" THEN revenue ELSE NULL END) as "Feb_Revenue",
    SUM(CASE WHEN month="Mar" THEN revenue ELSE NULL END) as "Mar_Revenue",
    SUM(CASE WHEN month="Apr" THEN revenue ELSE NULL END) as "Apr_Revenue",
    SUM(CASE WHEN month="May" THEN revenue ELSE NULL END) as "May_Revenue",
    SUM(CASE WHEN month="Jun" THEN revenue ELSE NULL END) as "Jun_Revenue",
    SUM(CASE WHEN month="Jul" THEN revenue ELSE NULL END) as "Jul_Revenue",
    SUM(CASE WHEN month="Aug" THEN revenue ELSE NULL END) as "Aug_Revenue",
    SUM(CASE WHEN month="Sep" THEN revenue ELSE NULL END) as "Sep_Revenue",
    SUM(CASE WHEN month="Oct" THEN revenue ELSE NULL END) as "Oct_Revenue",
    SUM(CASE WHEN month="Nov" THEN revenue ELSE NULL END) as "Nov_Revenue",
    SUM(CASE WHEN month="Dec" THEN revenue ELSE NULL END) as "Dec_Revenue"
FROM Department
group by id

NO.14 简单 (182. 查找重复的电子邮箱)

https://leetcode-cn.com/problems/duplicate-emails/

# 相同邮箱但不同Id即为重复数据（记得DISTINCT去重）
SELECT DISTINCT p1.Email
FROM Person p1, Person p2
WHERE p1.Email=p2.Email AND p1.Id<>p2.Id

NO.15 简单 (197. 上升的温度)

https://leetcode-cn.com/problems/rising-temperature/

# 两表之间相互比较即可
SELECT w1.id
FROM Weather w1, Weather w2
WHERE DATEDIFF(w1.recordDate,w2.recordDate)=1 AND w1.Temperature > w2.Temperature

NO.16 中等 (626. 换座位)

https://leetcode-cn.com/problems/exchange-seats/

# 关键在于如何将id转换，这里我用了^(异或)的方法
SELECT IF((SELECT COUNT(*) FROM seat)=id AND id%2<>0, id, (id+1)^1-1) AS id, student
FROM seat
ORDER BY id

NO.17 简单 (620. 有趣的电影)

https://leetcode-cn.com/problems/not-boring-movies/

# 没啥，就是WHERE的条件还有用到排序
SELECT *
FROM cinema
WHERE id%2=1 AND description<>'boring'
ORDER BY rating DESC

NO.18 简单 (627. 变更性别)

https://leetcode-cn.com/problems/swap-salary/

# 只能用UPDATE,然后结合CASE或者IF都可以。
UPDATE salary
SET
    sex = CASE sex
        WHEN 'm' THEN 'f'
        ELSE 'm'
    END;

NO.19 简单 (183. 从不订购的客户)

https://leetcode-cn.com/problems/customers-who-never-order/

# 判断哪些用户ID不在订单表里就行了
SELECT Name Customers FROM Customers
WHERE Id NOT IN (SELECT CustomerId FROM Orders)

NO.20 简单 (511. 游戏玩法分析 I)

https://leetcode-cn.com/problems/game-play-analysis-i/

# 按玩家ID分组，再取MIN(event_date)即可。
SELECT player_id, MIN(event_date) first_login
FROM Activity
GROUP BY player_id

NO.21 简单 (595. 大的国家)

https://leetcode-cn.com/problems/big-countries/

# 简单
SELECT name, population, area 
FROM World
WHERE area > 3000000 or population > 25000000

NO.22 简单 (1873. 计算特殊奖金)

https://leetcode-cn.com/problems/calculate-special-bonus/

# 简单
SELECT employee_id, IF(employee_id%2=1 AND LEFT(name, 1)<>'M', salary, 0) bonus
FROM Employees

NO.23 困难 (1097. 游戏玩法分析 V)

https://leetcode-cn.com/problems/game-play-analysis-v/

# 先取出所有安装日期，然后求出每个安装日期有几个玩家；
# 其中第二天接着玩的有谁(判断玩的日期和安装日期是否相差一天)，就可以了。
SELECT first_day AS install_dt, 
    COUNT(DISTINCT player_id) AS installs,
    ROUND((SUM(IF(DATEDIFF(event_date, first_day)=1, 1, 0)))/COUNT(DISTINCT player_id), 2) AS Day1_retention
FROM
(
    SELECT player_id, event_date, MIN(event_date) OVER(PARTITION BY player_id) AS first_day
    FROM Activity
) AS t1
GROUP BY first_day

NO.24 困难 (569. 员工薪水中位数)

https://leetcode-cn.com/problems/median-employee-salary/

# 题目什么东西？？不用任何内置函数，抱歉，窗口函数yyds
# 首先要按公司分组，然后对每个公司的员工工资排序（用ROW_NUMBER而不是RANK），然后通过中位数必定大于一半数同时小于一半数+1，得到了中位数。
SELECT Id, Company, Salary
FROM
(
    SELECT *, 
        ROW_NUMBER() OVER(PARTITION BY Company ORDER BY Salary) AS rk, 
        COUNT(*)  OVER(PARTITION BY Company) AS counts_cmy
    FROM Employee
) AS t
WHERE rK>=counts_cmy/2 AND rK<=counts_cmy/2+1

NO.25 中等 (1841. League Statistics)

https://leetcode-cn.com/problems/league-statistics/

# 关键点在于连接的时候可以用or，其余就是相加罢了，不难。
SELECT 
    t.team_name,
    COUNT(*) AS 'matches_played', 
    SUM(
        CASE 
            WHEN (t.team_id=m.home_team_id AND m.home_team_goals>m.away_team_goals) OR
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