leetcode sql 50题刷题攻略（下篇）_leetcode中sql题-爱代码爱编程

简介

刷题网址：高频 SQL 50 题（基础版） - 学习计划 - 力扣（LeetCode）全球极客挚爱的技术成长平台
上篇网址：Leetcode SQL 50题刷题攻略（上篇）-CSDN博客

正文开始

1789.员工的直属部门

select employee_id,department_id from (select employee_id,count(employee_id) as n,department_id from Employee group by employee_id) t where n=1
union
select employee_id,department_id from Employee where primary_flag = 'Y'

1. union的用法

   第一个select语句select employee_id,department_id from (select employee_id,count(employee_id) as n,department_id from Employee group by employee_id) t where n=1的结果：

   employee_id	department_id
   1	1
   3	3

   第二个select语句select employee_id,department_id from Employee where primary_flag = 'Y'的结果：

   employee_id	department_id
   2	1
   4	3

   通过union语句即SELECT column_a FROM table_a UNION SELECT column_b FROM table_b;合并的结果如下：

   employee_id	department_id
   1	1
   3	3
   2	1
   4	3

   需要注意的是，使用UNION操作符合并列时，两个表的列数和数据类型必须相同。

180.连续出现的数字

select distinct l1.num as ConsecutiveNums from Logs l1 left join Logs l2 on l1.id=l2.id+1 left join Logs l3 on l1.id=l3.id+2 where l1.num is not null and l1.num=l2.num and l1.num=l3.num

1. 很有意思的一道题，如何破解连续与计数的方式

1204.最后一个能进入巴士的人

select person_name from
(select person_id,person_name,sum(weight) over (order by turn) as sum_weight,turn from Queue order by turn) t where t.sum_weight<=1000 ORDER BY turn DESC  
LIMIT 1

1. 计算累计和

   原始数据：

   person_id	person_name	weight	turn
   5	Alice	250	1
   4	Bob	175	5
   3	Alex	350	2
   6	John Cena	400	3
   1	Winston	500	6
   2	Marie	200	4

   使用sum(…) over (order by …)来计算累加和select person_id,person_name,sum(weight) over (order by turn) as sum_weight,turn from Queue order by turn的结果：

   person_id	person_name	sum_weight	turn
   5	Alice	250	1
   3	Alex	600	2
   6	John Cena	1000	3
   2	Marie	1200	4
   4	Bob	1375	5
   1	Winston	1875	6

   即按turn的次序累加weight为sum_weight

2. 读取最后一行，使用ORDER BY turn DESC与LIMIT 1配合获取turn倒数第一行

1978.上级经理已离职的公司员工

select employee_id from Employees where salary < 30000 and manager_id not in
(select employee_id from Employees) order by employee_id

1. 如何查找一个经理id与所有的员工的id都不一样，即查找一个值与某一列都不一样，使用not in (select employee_id from Employees)来解决，也即SELECT * FROM table1 WHERE id NOT IN (SELECT id FROM table2)

626.换座位

SELECT  
case when id=(select max(id) from Seat) and id%2=1 then id
     when id%2=1 then id+1
     when id%2=0 then id-1
end as id,
student
FROM Seat order by id;

1. 查询子表使用XXX=(select...)的方式来查询

   SELECT column_name [, column_name ]
   FROM   table1 [, table2 ]
   WHERE  column_name OPERATOR
         (SELECT column_name [, column_name ]
         FROM table1 [, table2 ]
         [WHERE])
   

2. 使用case与when以及end来条件判断，免得使用union的连接操作：

   case when id=(select max(id) from Seat) and id%2=1 then id
        when id%2=1 then id+1
        when id%2=0 then id-1
   end as id
   

   也即：

   CASE WHEN condition1 THEN result1
        WHEN condition2 THEN result2
        ...
        ELSE resultN
   END
   

1341.电影评分

(select name as results from MovieRating m left join Users u on m.user_id=u.user_id group by m.user_id order by count(m.user_id) desc,name asc limit 1)
union all
(select title as results from MovieRating mr left join Movies m on mr.movie_id=m.movie_id where created_at>='2020-02-01' and created_at<='2020-02-29' group by mr.movie_id order by sum(rating)/count(rating) desc,title asc limit 1)

1. union会去掉重复项，union all不会

   比如union会返回：

   results
   Rebecca

   而union all会返回：

   results
   Rebecca
   Rebecca

2. order by可以使用sum，count等函数，结果可以不返回，排序的时候可以根据这些排序

1321.餐馆营业额变化增长

# 方式1，使用窗口函数
select visited_on,amount,round(amount/7,2) as average_amount from
(select max(visited_on) over (rows between 6 PRECEDING AND CURRENT ROW) as visited_on,sum(amount) over (rows between 6 PRECEDING AND CURRENT ROW) as amount from
(select visited_on,sum(amount) as amount from Customer group by visited_on) t) t1
having visited_on>=(select visited_on from Customer group by visited_on limit 6,1)
# 方式2，使用window子句
select visited_on,amount,round(amount/7,2) as average_amount from
(
select max(visited_on) over s as visited_on,sum(amount) over s as amount from
(select visited_on,sum(amount) as amount from Customer group by visited_on) t
window s as (rows between 6 PRECEDING AND CURRENT ROW)
) t1
having visited_on>=(select visited_on from Customer group by visited_on limit 6,1)

1. 

   通过上图可以得出一个伪SQL查询语句

   

   ！重点，where与having的作用是相同的，但是由于作用的时间段不一样，会导致having可以针对聚合计算结束的内容来进行判断，如下所示，having可以针对数学平均乘积筛选，而where是不能的，其只能针对原始数据筛选

   

2. sql操作的窗口函数

   窗口函数(Window Function)，又被叫做分析函数(Analytics Function)，通常在需要对数据进行分组汇总计算时使用，因此与聚集函数有一定的相似性。但与聚集函数不同的是，聚集函数通过对数据进行分组，仅能够输出分组汇总结果，而原始数据则无法展现在结果中。而窗口函数则可以同时将原始数据和聚集分析结果同时显示出来

   常见的形式如SUM(SCORE) OVER (PARTITION BY CLASSID ORDER BY SCORE ROWS BETWEEN 1 PRECEDING AND CURRENT ROW)，即上面一些题，用到了over啥的都是窗口函数

   PARTITION BY CLASSID意思是先按CLASSID分为很多大划分

   ORDER BY SCORE意思是在每个大划分下进行按照列SCORE来排序

   ROWS BETWEEN 1 PRECEDING AND CURRENT ROW意思是排序完成后，在每个大划分中进行，按照当前行与当前行前一行为一个窗口来进行小划分

   针对每个小划分来进行SUM(SCORE)操作

   具体如下图所示

   

   其中最后小划分的方式有如下格式

   RANGE|ROWS [BETWEEN] <rows_loc> [AND <rows_loc>]
   或
   RANGE|ROWS <rows_loc>
   

   中间的<rows_loc>可以为

   • UNBOUNDED PRECEDING表示该大划分的第一行
   • UNBOUNDED FOLLOWING表示该大划分的最后一行
   • CURRENT ROW表示当前行
   • PRECEDING表示从当前行往前数数量的行，其中不能包含变量，RANGE选项禁用
   • FOLLOWING表示从当前行往后数数量的行，其中不能包含变量，RANGE选项禁用

   举例：

   RANGE BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING 以该分组（也即大划分）所有元组为窗口

   RANGE BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW 以该分组起始行到当前行为窗口

   ROWS BETWEEN 10 PRECEDING AND 5 FOLLOWING 以该分组当前行前10行到后5行为窗口（不能超过起始行和结束行）

3. 窗口函数的window子句

   有时，一个查询会出现多个窗口函数，而内容还相同，那么可以使用window子句了，即使用一个window s as (rows between 6 PRECEDING AND CURRENT ROW)，然后其余要用到窗口函数的直接over s即可

4. limit的新用法

   limit n offset m表示从第m+1行开始读取n行，简单的写法可以写为limit m,n

585.2016年的投资

SELECT
    *
FROM
    insurance
WHERE
    insurance.TIV_2015 IN
    (
      SELECT
        TIV_2015
      FROM
        insurance
      GROUP BY TIV_2015
      HAVING COUNT(*) > 1
    )
    AND CONCAT(LAT, LON) IN
    (
      SELECT
        CONCAT(LAT, LON)
      FROM
        insurance
      GROUP BY LAT , LON
      HAVING COUNT(*) = 1
    )
;

1. having后可以接聚合函数，作用域上面一道题已经显示了having的运行位置
2. concat函数用于连接两个字符串形成一个字符串

185.部门工资前三高的所有员工

select d.name as Department, e.name as Employee, e.salary as Salary
from Employee e
         left join
     (select *
      from (select departmentId,
                   salary,
                   ROW_NUMBER() OVER (PARTITION BY departmentId ORDER BY salary DESC) AS rn
            FROM (select distinct departmentId,
                                  salary
                  from Employee) as t1) as t2
      where rn <= 3) t3
     on e.departmentId = t3.departmentId and e.salary = t3.salary
         left join
     Department d
     on e.departmentId = d.id
where t3.departmentId is not null

1. 如何将每个部门分开并且求每个部门的前三高的工资是多少，分为两步：

   1. 第一步将每个部门相同的薪水去掉

      select distinct departmentId,salary from Employee

      结果如下：

      departmentId	salary
      1	85000
      2	80000
      2	60000
      1	90000
      1	69000
      1	70000

   2. 第二部使用聚合函数中的ROW_NUMBER与窗口函数来分组标号

      select departmentId,
             salary,
             ROW_NUMBER() OVER (PARTITION BY departmentId ORDER BY salary DESC) AS rn
      FROM (select distinct departmentId,
                            salary
            from Employee) as t1
      

      结果如下：

      departmentId	salary	rn
      1	90000	1
      1	85000	2
      1	70000	3
      1	69000	4
      2	80000	1
      2	60000	2

      即按departmentId分组，每组按salary倒序，然后每组进行标号操作，标号操作后作为rn列

   3. 最后一个就是再套一个select取rn小于等于3的即可

1667.修复表中的名字

select user_id,concat(upper(left(name,1)),lower(substring(name, 2))) as name from Users order by user_id

1. mysql字符串截取常用函数有

   • left(str,length)

     从左边第一位开始截取指定长度字符串

   • right(str,length)

     从右边第一位开始截取指定长度字符串

   • substring(str,index,length)

     从指定开始位置截取指定长度字符串，如果不给length那么就从index截取后面全部长度，index可以为负数，负数就是相对于右边第一位的偏移

   • substr与mid函数的功能与substring类似

   • substring_index(str,dim,length)

     从指定字符位置开始截取指定长度字符串。

     length可正可负，若正那么就是相对于左边第length个dim，若负就是相对于右边第length个dim。正的取dim前所有字符串，负数取dim后所有字符串

     若找不到dim那么返回所有字符串

2. upper是字符串变大写，lower是字符串变小写

3. concat是连接两个字符串操作

1527.患某种疾病的患者

select * from Patients p where p.conditions REGEXP '\\bDIAB1.*'

1. mysql正则表达式使用p.conditions REGEXP '\\bDIAB1.*'即匹配了\bDIAB1.*的字符串都会被查询出，注意\b需要使用\\b才行

   \b是匹配一个单词边界

   \B是匹配一个非单词边界

   \w是匹配一个字母、数字或下划线。等价于 [A-Za-z0-9_]

   .表示匹配除换行符 \n 之外的任何单字符

   *表示零次或多次

   .*在一起就表示任意字符出现零次或多次

196.删除重复的电子邮箱

DELETE
FROM
	Person
WHERE
	id NOT IN 
(SELECT
		tmp.mid
	FROM
		(
			SELECT
				MIN(id) mid
			FROM
				Person
			GROUP BY
				email
		) tmp
)

1. 只保留重复数据id小的一个项，那么可以使用这个方法，即先分组求id最小，然后只删除不在这些id中的项即可

1484.按日期分组销售产品

select sell_date, count(sell_date) as num_sold,group_concat(product order by product) as products
from (select distinct sell_date, product from Activities) t
group by sell_date

1. group_concat与group by搭配使用，并且其内部能搭配使用order by

   sell_date	product
   2020-05-30	Headphone
   2020-06-01	Pencil
   2020-06-02	Mask
   2020-05-30	Basketball
   2020-06-01	Bible
   2020-06-02	Mask
   2020-05-30	T-Shirt

   第一步去掉重复项，使用distinct即select distinct sell_date, product from Activities

   sell_date	product
   2020-05-30	Headphone
   2020-06-01	Pencil
   2020-06-02	Mask
   2020-05-30	Basketball
   2020-06-01	Bible
   2020-05-30	T-Shirt

   第二步分组，使用count与group_concat，group_concat会自动使用逗号来连接，即可得到答案

   sell_date	num_sold	products
   2020-05-30	3	Basketball,Headphone,T-Shirt
   2020-06-01	2	Bible,Pencil
   2020-06-02	1	Mask

1327.列出指定时间段内所有的下单产品

select p.product_name, sum_unit as unit
from Products p
         left join
     (select product_id, order_date, sum(unit) as sum_unit
      from Orders
      where DATE_FORMAT(order_date, "%m") = 2
      group by DATE_FORMAT(order_date, "%Y年%m月"), product_id
      having sum_unit >= 100) t on p.product_id = t.product_id
where t.product_id is not null

1. 可以使用where DATE_FORMAT(order_date, "%m") = 2的方式将月份为2的找出来

1517.查找拥有有效邮箱的用户

select * from Users where mail regexp "^[a-zA-Z]+[\\w|\\-|\\.]*@leetcode\\.com"

1. 在mysql中使用正则表达式要使用\\来表示\即上面的^[a-zA-Z]+[\\w|\\-|\\.]*@leetcode\\.com实际正则表达式是^[a-zA-Z]+[\w|\-|\.]*@leetcode\.com
2. ^表示正则表达式的其实位置，如果不加这个那么.shapo@leetcode.com也会被匹配