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423
docs/notes/Leetcode-Database 题解.md
View File

@ -46,9 +46,22 @@ https://leetcode.com/problems/big-countries/description/
+--------------+-------------+--------------+ +--------------+-------------+--------------+
``` ```
## Solution
```sql
SELECT name,
population,
area
FROM
World
WHERE
area > 3000000
OR population > 25000000;
```
## SQL Schema ## SQL Schema
SQL Schema 用于在本地环境下创建表结构并导入数据，从而方便在本地环境解答。 SQL Schema 用于在本地环境下创建表结构并导入数据，从而方便在本地环境调试。
```sql ```sql
DROP TABLE DROP TABLE
@ -64,19 +77,6 @@ VALUES
( 'Angola', 'Africa', '1246700', '20609294', '1009900000' ); ( 'Angola', 'Africa', '1246700', '20609294', '1009900000' );
``` ```
## Solution
```sql
SELECT name,
population,
area
FROM
World
WHERE
area > 3000000
OR population > 25000000;
```
# 627. Swap Salary # 627. Swap Salary
https://leetcode.com/problems/swap-salary/description/ https://leetcode.com/problems/swap-salary/description/
@ -103,6 +103,24 @@ https://leetcode.com/problems/swap-salary/description/
| 4 | D | m | 500 | | 4 | D | m | 500 |
``` ```
## Solution
两个相等的数异或的结果为 0，而 0 与任何一个数异或的结果为这个数。
sex 字段只有两个取值：'f' 和 'm'，并且有以下规律：
```
'f' ^ ('m' ^ 'f') = 'm' ^ ('f' ^ 'f') = 'm'
'm' ^ ('m' ^ 'f') = 'f' ^ ('m' ^ 'm') = 'f'
```
因此将 sex 字段和 'm' ^ 'f' 进行异或操作，最后就能反转 sex 字段。
```sql
UPDATE salary
SET sex = CHAR ( ASCII(sex) ^ ASCII( 'm' ) ^ ASCII( 'f' ) );
```
## SQL Schema ## SQL Schema
```sql ```sql
@ -118,22 +136,6 @@ VALUES
( '4', 'D', 'f', '500' ); ( '4', 'D', 'f', '500' );
``` ```
## Solution
使用异或操作，两个相等的数异或的结果为 0，而 0 与任何一个数异或的结果为这个数。
```
'f' ^ 'm' ^ 'f' = 'm'
'm' ^ 'm' ^ 'f' = 'f'
```
```sql
UPDATE salary
SET sex = CHAR ( ASCII(sex) ^ ASCII( 'm' ) ^ ASCII( 'f' ) );
```
# 620. Not Boring Movies # 620. Not Boring Movies
https://leetcode.com/problems/not-boring-movies/description/ https://leetcode.com/problems/not-boring-movies/description/
@ -164,6 +166,20 @@ https://leetcode.com/problems/not-boring-movies/description/
+---------+-----------+--------------+-----------+ +---------+-----------+--------------+-----------+
``` ```
## Solution
```sql
SELECT
*
FROM
cinema
WHERE
id % 2 = 1
AND description != 'boring'
ORDER BY
rating DESC;
```
## SQL Schema ## SQL Schema
```sql ```sql
@ -180,20 +196,6 @@ VALUES
( 5, 'House card', 'Interesting', 9.1 ); ( 5, 'House card', 'Interesting', 9.1 );
``` ```
## Solution
```sql
SELECT
*
FROM
cinema
WHERE
id % 2 = 1
AND description != 'boring'
ORDER BY
rating DESC;
```
# 596. Classes More Than 5 Students # 596. Classes More Than 5 Students
https://leetcode.com/problems/classes-more-than-5-students/description/ https://leetcode.com/problems/classes-more-than-5-students/description/
@ -226,6 +228,21 @@ https://leetcode.com/problems/classes-more-than-5-students/description/
+---------+ +---------+
``` ```
## Solution
对 class 列进行分组之后，再使用 count 汇总函数统计数量，统计之后使用 having 进行过滤。
```sql
SELECT
class
FROM
courses
GROUP BY
class
HAVING
count( DISTINCT student ) >= 5;
```
## SQL Schema ## SQL Schema
```sql ```sql
@ -246,21 +263,6 @@ VALUES
( 'I', 'Math' ); ( 'I', 'Math' );
``` ```
## Solution
对 class 列进行分组之后，再使用 count 汇总函数统计数量，统计之后使用 having 进行过滤。
```sql
SELECT
class
FROM
courses
GROUP BY
class
HAVING
count( DISTINCT student ) >= 5;
```
# 182. Duplicate Emails # 182. Duplicate Emails
https://leetcode.com/problems/duplicate-emails/description/ https://leetcode.com/problems/duplicate-emails/description/
@ -289,20 +291,6 @@ https://leetcode.com/problems/duplicate-emails/description/
+---------+ +---------+
``` ```
## SQL Schema
```sql
DROP TABLE
IF
EXISTS Person;
CREATE TABLE Person ( Id INT, Email VARCHAR ( 255 ) );
INSERT INTO Person ( Id, Email )
VALUES
( 1, 'a@b.com' ),
( 2, 'c@d.com' ),
( 3, 'a@b.com' );
```
## Solution ## Solution
对 Email 进行分组，如果相同 Email 的数量大于等于 2，则表示该 Email 重复。 对 Email 进行分组，如果相同 Email 的数量大于等于 2，则表示该 Email 重复。
@ -318,6 +306,21 @@ HAVING
COUNT( * ) >= 2; COUNT( * ) >= 2;
``` ```
## SQL Schema
```sql
DROP TABLE
IF
EXISTS Person;
CREATE TABLE Person ( Id INT, Email VARCHAR ( 255 ) );
INSERT INTO Person ( Id, Email )
VALUES
( 1, 'a@b.com' ),
( 2, 'c@d.com' ),
( 3, 'a@b.com' );
```
# 196. Delete Duplicate Emails # 196. Delete Duplicate Emails
https://leetcode.com/problems/delete-duplicate-emails/description/ https://leetcode.com/problems/delete-duplicate-emails/description/
@ -347,10 +350,6 @@ https://leetcode.com/problems/delete-duplicate-emails/description/
+----+------------------+ +----+------------------+
``` ```
## SQL Schema
与 182 相同。
## Solution ## Solution
只保留相同 Email 中 Id 最小的那一个，然后删除其它的。 只保留相同 Email 中 Id 最小的那一个，然后删除其它的。
@ -389,6 +388,12 @@ WHERE
参考：[pMySQL Error 1093 - Can't specify target table for update in FROM clause](https://stackoverflow.com/questions/45494/mysql-error-1093-cant-specify-target-table-for-update-in-from-clause) 参考：[pMySQL Error 1093 - Can't specify target table for update in FROM clause](https://stackoverflow.com/questions/45494/mysql-error-1093-cant-specify-target-table-for-update-in-from-clause)
## SQL Schema
与 182 相同。
# 175. Combine Two Tables # 175. Combine Two Tables
https://leetcode.com/problems/combine-two-tables/description/ https://leetcode.com/problems/combine-two-tables/description/
@ -424,6 +429,22 @@ AddressId is the primary key column for this table.
查找 FirstName, LastName, City, State 数据，而不管一个用户有没有填地址信息。 查找 FirstName, LastName, City, State 数据，而不管一个用户有没有填地址信息。
## Solution
涉及到 Person 和 Address 两个表，在对这两个表执行连接操作时，因为要保留 Person 表中的信息，即使在 Address 表中没有关联的信息也要保留。此时可以用左外连接，将 Person 表放在 LEFT JOIN 的左边。
```sql
SELECT
FirstName,
LastName,
City,
State
FROM
Person P
LEFT JOIN Address A
ON P.PersonId = A.PersonId;
```
## SQL Schema ## SQL Schema
```sql ```sql
@ -443,22 +464,6 @@ VALUES
( 1, 2, 'New York City', 'New York' ); ( 1, 2, 'New York City', 'New York' );
``` ```
## Solution
涉及到 Person 和 Address 两个表，在对这两个表执行连接操作时，因为要保留 Person 表中的信息，即使在 Address 表中没有关联的信息也要保留。此时可以用左外连接，将 Person 表放在 LEFT JOIN 的左边。
```sql
SELECT
FirstName,
LastName,
City,
State
FROM
Person P
LEFT JOIN Address A
ON P.PersonId = A.PersonId;
```
# 181. Employees Earning More Than Their Managers # 181. Employees Earning More Than Their Managers
https://leetcode.com/problems/employees-earning-more-than-their-managers/description/ https://leetcode.com/problems/employees-earning-more-than-their-managers/description/
@ -480,6 +485,18 @@ Employee 表：
查找薪资大于其经理薪资的员工信息。 查找薪资大于其经理薪资的员工信息。
## Solution
```sql
SELECT
E1.NAME AS Employee
FROM
Employee E1
INNER JOIN Employee E2
ON E1.ManagerId = E2.Id
AND E1.Salary > E2.Salary;
```
## SQL Schema ## SQL Schema
```sql ```sql
@ -495,18 +512,6 @@ VALUES
( 4, 'Max', 90000, NULL ); ( 4, 'Max', 90000, NULL );
``` ```
## Solution
```sql
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 # 183. Customers Who Never Order
https://leetcode.com/problems/customers-who-never-order/description/ https://leetcode.com/problems/customers-who-never-order/description/
@ -548,29 +553,6 @@ Orders 表：
+-----------+ +-----------+
``` ```
## SQL Schema
```sql
DROP TABLE
IF
EXISTS Customers;
CREATE TABLE Customers ( Id INT, NAME VARCHAR ( 255 ) );
DROP TABLE
IF
EXISTS Orders;
CREATE TABLE Orders ( Id INT, CustomerId INT );
INSERT INTO Customers ( Id, NAME )
VALUES
( 1, 'Joe' ),
( 2, 'Henry' ),
( 3, 'Sam' ),
( 4, 'Max' );
INSERT INTO Orders ( Id, CustomerId )
VALUES
( 1, 3 ),
( 2, 1 );
```
## Solution ## Solution
左外链接 左外链接
@ -597,6 +579,29 @@ WHERE
Id NOT IN ( SELECT CustomerId FROM Orders ); Id NOT IN ( SELECT CustomerId FROM Orders );
``` ```
## SQL Schema
```sql
DROP TABLE
IF
EXISTS Customers;
CREATE TABLE Customers ( Id INT, NAME VARCHAR ( 255 ) );
DROP TABLE
IF
EXISTS Orders;
CREATE TABLE Orders ( Id INT, CustomerId INT );
INSERT INTO Customers ( Id, NAME )
VALUES
( 1, 'Joe' ),
( 2, 'Henry' ),
( 3, 'Sam' ),
( 4, 'Max' );
INSERT INTO Orders ( Id, CustomerId )
VALUES
( 1, 3 ),
( 2, 1 );
```
# 184. Department Highest Salary # 184. Department Highest Salary
https://leetcode.com/problems/department-highest-salary/description/ https://leetcode.com/problems/department-highest-salary/description/
@ -638,25 +643,6 @@ Department 表：
+------------+----------+--------+ +------------+----------+--------+
``` ```
## SQL Schema
```sql
DROP TABLE IF EXISTS Employee;
CREATE TABLE Employee ( Id INT, NAME VARCHAR ( 255 ), Salary INT, DepartmentId INT );
DROP TABLE IF EXISTS Department;
CREATE TABLE Department ( Id INT, NAME VARCHAR ( 255 ) );
INSERT INTO Employee ( Id, NAME, Salary, DepartmentId )
VALUES
( 1, 'Joe', 70000, 1 ),
( 2, 'Henry', 80000, 2 ),
( 3, 'Sam', 60000, 2 ),
( 4, 'Max', 90000, 1 );
INSERT INTO Department ( Id, NAME )
VALUES
( 1, 'IT' ),
( 2, 'Sales' );
```
## Solution ## Solution
创建一个临时表，包含了部门员工的最大薪资。可以对部门进行分组，然后使用 MAX() 汇总函数取得最大薪资。 创建一个临时表，包含了部门员工的最大薪资。可以对部门进行分组，然后使用 MAX() 汇总函数取得最大薪资。
@ -678,6 +664,26 @@ WHERE
AND E.Salary = M.Salary; AND E.Salary = M.Salary;
``` ```
## SQL Schema
```sql
DROP TABLE IF EXISTS Employee;
CREATE TABLE Employee ( Id INT, NAME VARCHAR ( 255 ), Salary INT, DepartmentId INT );
DROP TABLE IF EXISTS Department;
CREATE TABLE Department ( Id INT, NAME VARCHAR ( 255 ) );
INSERT INTO Employee ( Id, NAME, Salary, DepartmentId )
VALUES
( 1, 'Joe', 70000, 1 ),
( 2, 'Henry', 80000, 2 ),
( 3, 'Sam', 60000, 2 ),
( 4, 'Max', 90000, 1 );
INSERT INTO Department ( Id, NAME )
VALUES
( 1, 'IT' ),
( 2, 'Sales' );
```
# 176. Second Highest Salary # 176. Second Highest Salary
https://leetcode.com/problems/second-highest-salary/description/ https://leetcode.com/problems/second-highest-salary/description/
@ -706,6 +712,15 @@ https://leetcode.com/problems/second-highest-salary/description/
没有找到返回 null 而不是不返回数据。 没有找到返回 null 而不是不返回数据。
## Solution
为了在没有查找到数据时返回 null，需要在查询结果外面再套一层 SELECT。
```sql
SELECT
( SELECT DISTINCT Salary FROM Employee ORDER BY Salary DESC LIMIT 1, 1 ) SecondHighestSalary;
```
## SQL Schema ## SQL Schema
```sql ```sql
@ -720,25 +735,12 @@ VALUES
( 3, 300 ); ( 3, 300 );
``` ```
## Solution
为了在没有查找到数据时返回 null，需要在查询结果外面再套一层 SELECT。
```sql
SELECT
( SELECT DISTINCT Salary FROM Employee ORDER BY Salary DESC LIMIT 1, 1 ) SecondHighestSalary;
```
# 177. Nth Highest Salary # 177. Nth Highest Salary
## Description ## Description
查找工资第 N 高的员工。 查找工资第 N 高的员工。
## SQL Schema
同 176。
## Solution ## Solution
```sql ```sql
@ -750,6 +752,11 @@ RETURN ( SELECT ( SELECT DISTINCT Salary FROM Employee ORDER BY Salary DESC LIMI
END END
``` ```
## SQL Schema
同 176。
# 178. Rank Scores # 178. Rank Scores
https://leetcode.com/problems/rank-scores/description/ https://leetcode.com/problems/rank-scores/description/
@ -786,23 +793,6 @@ https://leetcode.com/problems/rank-scores/description/
+-------+------+ +-------+------+
``` ```
## SQL Schema
```sql
DROP TABLE
IF
EXISTS Scores;
CREATE TABLE Scores ( Id INT, Score DECIMAL ( 3, 2 ) );
INSERT INTO Scores ( Id, Score )
VALUES
( 1, 3.5 ),
( 2, 3.65 ),
( 3, 4.0 ),
( 4, 3.85 ),
( 5, 4.0 ),
( 6, 3.65 );
```
## Solution ## Solution
要统计某个 score 的排名，只要统计大于该 score 的 score 数量，然后加 1。 要统计某个 score 的排名，只要统计大于该 score 的 score 数量，然后加 1。
@ -852,6 +842,23 @@ ORDER BY
S1.score DESC; S1.score DESC;
``` ```
## SQL Schema
```sql
DROP TABLE
IF
EXISTS Scores;
CREATE TABLE Scores ( Id INT, Score DECIMAL ( 3, 2 ) );
INSERT INTO Scores ( Id, Score )
VALUES
( 1, 3.5 ),
( 2, 3.65 ),
( 3, 4.0 ),
( 4, 3.85 ),
( 5, 4.0 ),
( 6, 3.65 );
```
# 180. Consecutive Numbers # 180. Consecutive Numbers
https://leetcode.com/problems/consecutive-numbers/description/ https://leetcode.com/problems/consecutive-numbers/description/
@ -884,6 +891,21 @@ https://leetcode.com/problems/consecutive-numbers/description/
+-----------------+ +-----------------+
``` ```
## Solution
```sql
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;
```
## SQL Schema ## SQL Schema
```sql ```sql
@ -902,21 +924,6 @@ VALUES
( 7, 2 ); ( 7, 2 );
``` ```
## Solution
```sql
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;
```
# 626. Exchange Seats # 626. Exchange Seats
https://leetcode.com/problems/exchange-seats/description/ https://leetcode.com/problems/exchange-seats/description/
@ -951,22 +958,6 @@ seat 表存储着座位对应的学生。
+---------+---------+ +---------+---------+
``` ```
## SQL Schema
```sql
DROP TABLE
IF
EXISTS seat;
CREATE TABLE seat ( id INT, student VARCHAR ( 255 ) );
INSERT INTO seat ( id, student )
VALUES
( '1', 'Abbot' ),
( '2', 'Doris' ),
( '3', 'Emerson' ),
( '4', 'Green' ),
( '5', 'Jeames' );
```
## Solution ## Solution
使用多个 union。 使用多个 union。
@ -1004,6 +995,22 @@ ORDER BY
id; id;
``` ```
## SQL Schema
```sql
DROP TABLE
IF
EXISTS seat;
CREATE TABLE seat ( id INT, student VARCHAR ( 255 ) );
INSERT INTO seat ( id, student )
VALUES
( '1', 'Abbot' ),
( '2', 'Doris' ),
( '3', 'Emerson' ),
( '4', 'Green' ),
( '5', 'Jeames' );
```

423
notes/Leetcode-Database 题解.md
View File

@ -46,9 +46,22 @@ https://leetcode.com/problems/big-countries/description/
+--------------+-------------+--------------+ +--------------+-------------+--------------+
``` ```
## Solution
```sql
SELECT name,
population,
area
FROM
World
WHERE
area > 3000000
OR population > 25000000;
```
## SQL Schema ## SQL Schema
SQL Schema 用于在本地环境下创建表结构并导入数据，从而方便在本地环境解答。 SQL Schema 用于在本地环境下创建表结构并导入数据，从而方便在本地环境调试。
```sql ```sql
DROP TABLE DROP TABLE
@ -64,19 +77,6 @@ VALUES
( 'Angola', 'Africa', '1246700', '20609294', '1009900000' ); ( 'Angola', 'Africa', '1246700', '20609294', '1009900000' );
``` ```
## Solution
```sql
SELECT name,
population,
area
FROM
World
WHERE
area > 3000000
OR population > 25000000;
```
# 627. Swap Salary # 627. Swap Salary
https://leetcode.com/problems/swap-salary/description/ https://leetcode.com/problems/swap-salary/description/
@ -103,6 +103,24 @@ https://leetcode.com/problems/swap-salary/description/
| 4 | D | m | 500 | | 4 | D | m | 500 |
``` ```
## Solution
两个相等的数异或的结果为 0，而 0 与任何一个数异或的结果为这个数。
sex 字段只有两个取值：'f' 和 'm'，并且有以下规律：
```
'f' ^ ('m' ^ 'f') = 'm' ^ ('f' ^ 'f') = 'm'
'm' ^ ('m' ^ 'f') = 'f' ^ ('m' ^ 'm') = 'f'
```
因此将 sex 字段和 'm' ^ 'f' 进行异或操作，最后就能反转 sex 字段。
```sql
UPDATE salary
SET sex = CHAR ( ASCII(sex) ^ ASCII( 'm' ) ^ ASCII( 'f' ) );
```
## SQL Schema ## SQL Schema
```sql ```sql
@ -118,22 +136,6 @@ VALUES
( '4', 'D', 'f', '500' ); ( '4', 'D', 'f', '500' );
``` ```
## Solution
使用异或操作，两个相等的数异或的结果为 0，而 0 与任何一个数异或的结果为这个数。
```
'f' ^ 'm' ^ 'f' = 'm'
'm' ^ 'm' ^ 'f' = 'f'
```
```sql
UPDATE salary
SET sex = CHAR ( ASCII(sex) ^ ASCII( 'm' ) ^ ASCII( 'f' ) );
```
# 620. Not Boring Movies # 620. Not Boring Movies
https://leetcode.com/problems/not-boring-movies/description/ https://leetcode.com/problems/not-boring-movies/description/
@ -164,6 +166,20 @@ https://leetcode.com/problems/not-boring-movies/description/
+---------+-----------+--------------+-----------+ +---------+-----------+--------------+-----------+
``` ```
## Solution
```sql
SELECT
*
FROM
cinema
WHERE
id % 2 = 1
AND description != 'boring'
ORDER BY
rating DESC;
```
## SQL Schema ## SQL Schema
```sql ```sql
@ -180,20 +196,6 @@ VALUES
( 5, 'House card', 'Interesting', 9.1 ); ( 5, 'House card', 'Interesting', 9.1 );
``` ```
## Solution
```sql
SELECT
*
FROM
cinema
WHERE
id % 2 = 1
AND description != 'boring'
ORDER BY
rating DESC;
```
# 596. Classes More Than 5 Students # 596. Classes More Than 5 Students
https://leetcode.com/problems/classes-more-than-5-students/description/ https://leetcode.com/problems/classes-more-than-5-students/description/
@ -226,6 +228,21 @@ https://leetcode.com/problems/classes-more-than-5-students/description/
+---------+ +---------+
``` ```
## Solution
对 class 列进行分组之后，再使用 count 汇总函数统计数量，统计之后使用 having 进行过滤。
```sql
SELECT
class
FROM
courses
GROUP BY
class
HAVING
count( DISTINCT student ) >= 5;
```
## SQL Schema ## SQL Schema
```sql ```sql
@ -246,21 +263,6 @@ VALUES
( 'I', 'Math' ); ( 'I', 'Math' );
``` ```
## Solution
对 class 列进行分组之后，再使用 count 汇总函数统计数量，统计之后使用 having 进行过滤。
```sql
SELECT
class
FROM
courses
GROUP BY
class
HAVING
count( DISTINCT student ) >= 5;
```
# 182. Duplicate Emails # 182. Duplicate Emails
https://leetcode.com/problems/duplicate-emails/description/ https://leetcode.com/problems/duplicate-emails/description/
@ -289,20 +291,6 @@ https://leetcode.com/problems/duplicate-emails/description/
+---------+ +---------+
``` ```
## SQL Schema
```sql
DROP TABLE
IF
EXISTS Person;
CREATE TABLE Person ( Id INT, Email VARCHAR ( 255 ) );
INSERT INTO Person ( Id, Email )
VALUES
( 1, 'a@b.com' ),
( 2, 'c@d.com' ),
( 3, 'a@b.com' );
```
## Solution ## Solution
对 Email 进行分组，如果相同 Email 的数量大于等于 2，则表示该 Email 重复。 对 Email 进行分组，如果相同 Email 的数量大于等于 2，则表示该 Email 重复。
@ -318,6 +306,21 @@ HAVING
COUNT( * ) >= 2; COUNT( * ) >= 2;
``` ```
## SQL Schema
```sql
DROP TABLE
IF
EXISTS Person;
CREATE TABLE Person ( Id INT, Email VARCHAR ( 255 ) );
INSERT INTO Person ( Id, Email )
VALUES
( 1, 'a@b.com' ),
( 2, 'c@d.com' ),
( 3, 'a@b.com' );
```
# 196. Delete Duplicate Emails # 196. Delete Duplicate Emails
https://leetcode.com/problems/delete-duplicate-emails/description/ https://leetcode.com/problems/delete-duplicate-emails/description/
@ -347,10 +350,6 @@ https://leetcode.com/problems/delete-duplicate-emails/description/
+----+------------------+ +----+------------------+
``` ```
## SQL Schema
与 182 相同。
## Solution ## Solution
只保留相同 Email 中 Id 最小的那一个，然后删除其它的。 只保留相同 Email 中 Id 最小的那一个，然后删除其它的。
@ -389,6 +388,12 @@ WHERE
参考：[pMySQL Error 1093 - Can't specify target table for update in FROM clause](https://stackoverflow.com/questions/45494/mysql-error-1093-cant-specify-target-table-for-update-in-from-clause) 参考：[pMySQL Error 1093 - Can't specify target table for update in FROM clause](https://stackoverflow.com/questions/45494/mysql-error-1093-cant-specify-target-table-for-update-in-from-clause)
## SQL Schema
与 182 相同。
# 175. Combine Two Tables # 175. Combine Two Tables
https://leetcode.com/problems/combine-two-tables/description/ https://leetcode.com/problems/combine-two-tables/description/
@ -424,6 +429,22 @@ AddressId is the primary key column for this table.
查找 FirstName, LastName, City, State 数据，而不管一个用户有没有填地址信息。 查找 FirstName, LastName, City, State 数据，而不管一个用户有没有填地址信息。
## Solution
涉及到 Person 和 Address 两个表，在对这两个表执行连接操作时，因为要保留 Person 表中的信息，即使在 Address 表中没有关联的信息也要保留。此时可以用左外连接，将 Person 表放在 LEFT JOIN 的左边。
```sql
SELECT
FirstName,
LastName,
City,
State
FROM
Person P
LEFT JOIN Address A
ON P.PersonId = A.PersonId;
```
## SQL Schema ## SQL Schema
```sql ```sql
@ -443,22 +464,6 @@ VALUES
( 1, 2, 'New York City', 'New York' ); ( 1, 2, 'New York City', 'New York' );
``` ```
## Solution
涉及到 Person 和 Address 两个表，在对这两个表执行连接操作时，因为要保留 Person 表中的信息，即使在 Address 表中没有关联的信息也要保留。此时可以用左外连接，将 Person 表放在 LEFT JOIN 的左边。
```sql
SELECT
FirstName,
LastName,
City,
State
FROM
Person P
LEFT JOIN Address A
ON P.PersonId = A.PersonId;
```
# 181. Employees Earning More Than Their Managers # 181. Employees Earning More Than Their Managers
https://leetcode.com/problems/employees-earning-more-than-their-managers/description/ https://leetcode.com/problems/employees-earning-more-than-their-managers/description/
@ -480,6 +485,18 @@ Employee 表：
查找薪资大于其经理薪资的员工信息。 查找薪资大于其经理薪资的员工信息。
## Solution
```sql
SELECT
E1.NAME AS Employee
FROM
Employee E1
INNER JOIN Employee E2
ON E1.ManagerId = E2.Id
AND E1.Salary > E2.Salary;
```
## SQL Schema ## SQL Schema
```sql ```sql
@ -495,18 +512,6 @@ VALUES
( 4, 'Max', 90000, NULL ); ( 4, 'Max', 90000, NULL );
``` ```
## Solution
```sql
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 # 183. Customers Who Never Order
https://leetcode.com/problems/customers-who-never-order/description/ https://leetcode.com/problems/customers-who-never-order/description/
@ -548,29 +553,6 @@ Orders 表：
+-----------+ +-----------+
``` ```
## SQL Schema
```sql
DROP TABLE
IF
EXISTS Customers;
CREATE TABLE Customers ( Id INT, NAME VARCHAR ( 255 ) );
DROP TABLE
IF
EXISTS Orders;
CREATE TABLE Orders ( Id INT, CustomerId INT );
INSERT INTO Customers ( Id, NAME )
VALUES
( 1, 'Joe' ),
( 2, 'Henry' ),
( 3, 'Sam' ),
( 4, 'Max' );
INSERT INTO Orders ( Id, CustomerId )
VALUES
( 1, 3 ),
( 2, 1 );
```
## Solution ## Solution
左外链接 左外链接
@ -597,6 +579,29 @@ WHERE
Id NOT IN ( SELECT CustomerId FROM Orders ); Id NOT IN ( SELECT CustomerId FROM Orders );
``` ```
## SQL Schema
```sql
DROP TABLE
IF
EXISTS Customers;
CREATE TABLE Customers ( Id INT, NAME VARCHAR ( 255 ) );
DROP TABLE
IF
EXISTS Orders;
CREATE TABLE Orders ( Id INT, CustomerId INT );
INSERT INTO Customers ( Id, NAME )
VALUES
( 1, 'Joe' ),
( 2, 'Henry' ),
( 3, 'Sam' ),
( 4, 'Max' );
INSERT INTO Orders ( Id, CustomerId )
VALUES
( 1, 3 ),
( 2, 1 );
```
# 184. Department Highest Salary # 184. Department Highest Salary
https://leetcode.com/problems/department-highest-salary/description/ https://leetcode.com/problems/department-highest-salary/description/
@ -638,25 +643,6 @@ Department 表：
+------------+----------+--------+ +------------+----------+--------+
``` ```
## SQL Schema
```sql
DROP TABLE IF EXISTS Employee;
CREATE TABLE Employee ( Id INT, NAME VARCHAR ( 255 ), Salary INT, DepartmentId INT );
DROP TABLE IF EXISTS Department;
CREATE TABLE Department ( Id INT, NAME VARCHAR ( 255 ) );
INSERT INTO Employee ( Id, NAME, Salary, DepartmentId )
VALUES
( 1, 'Joe', 70000, 1 ),
( 2, 'Henry', 80000, 2 ),
( 3, 'Sam', 60000, 2 ),
( 4, 'Max', 90000, 1 );
INSERT INTO Department ( Id, NAME )
VALUES
( 1, 'IT' ),
( 2, 'Sales' );
```
## Solution ## Solution
创建一个临时表，包含了部门员工的最大薪资。可以对部门进行分组，然后使用 MAX() 汇总函数取得最大薪资。 创建一个临时表，包含了部门员工的最大薪资。可以对部门进行分组，然后使用 MAX() 汇总函数取得最大薪资。
@ -678,6 +664,26 @@ WHERE
AND E.Salary = M.Salary; AND E.Salary = M.Salary;
``` ```
## SQL Schema
```sql
DROP TABLE IF EXISTS Employee;
CREATE TABLE Employee ( Id INT, NAME VARCHAR ( 255 ), Salary INT, DepartmentId INT );
DROP TABLE IF EXISTS Department;
CREATE TABLE Department ( Id INT, NAME VARCHAR ( 255 ) );
INSERT INTO Employee ( Id, NAME, Salary, DepartmentId )
VALUES
( 1, 'Joe', 70000, 1 ),
( 2, 'Henry', 80000, 2 ),
( 3, 'Sam', 60000, 2 ),
( 4, 'Max', 90000, 1 );
INSERT INTO Department ( Id, NAME )
VALUES
( 1, 'IT' ),
( 2, 'Sales' );
```
# 176. Second Highest Salary # 176. Second Highest Salary
https://leetcode.com/problems/second-highest-salary/description/ https://leetcode.com/problems/second-highest-salary/description/
@ -706,6 +712,15 @@ https://leetcode.com/problems/second-highest-salary/description/
没有找到返回 null 而不是不返回数据。 没有找到返回 null 而不是不返回数据。
## Solution
为了在没有查找到数据时返回 null，需要在查询结果外面再套一层 SELECT。
```sql
SELECT
( SELECT DISTINCT Salary FROM Employee ORDER BY Salary DESC LIMIT 1, 1 ) SecondHighestSalary;
```
## SQL Schema ## SQL Schema
```sql ```sql
@ -720,25 +735,12 @@ VALUES
( 3, 300 ); ( 3, 300 );
``` ```
## Solution
为了在没有查找到数据时返回 null，需要在查询结果外面再套一层 SELECT。
```sql
SELECT
( SELECT DISTINCT Salary FROM Employee ORDER BY Salary DESC LIMIT 1, 1 ) SecondHighestSalary;
```
# 177. Nth Highest Salary # 177. Nth Highest Salary
## Description ## Description
查找工资第 N 高的员工。 查找工资第 N 高的员工。
## SQL Schema
同 176。
## Solution ## Solution
```sql ```sql
@ -750,6 +752,11 @@ RETURN ( SELECT ( SELECT DISTINCT Salary FROM Employee ORDER BY Salary DESC LIMI
END END
``` ```
## SQL Schema
同 176。
# 178. Rank Scores # 178. Rank Scores
https://leetcode.com/problems/rank-scores/description/ https://leetcode.com/problems/rank-scores/description/
@ -786,23 +793,6 @@ https://leetcode.com/problems/rank-scores/description/
+-------+------+ +-------+------+
``` ```
## SQL Schema
```sql
DROP TABLE
IF
EXISTS Scores;
CREATE TABLE Scores ( Id INT, Score DECIMAL ( 3, 2 ) );
INSERT INTO Scores ( Id, Score )
VALUES
( 1, 3.5 ),
( 2, 3.65 ),
( 3, 4.0 ),
( 4, 3.85 ),
( 5, 4.0 ),
( 6, 3.65 );
```
## Solution ## Solution
要统计某个 score 的排名，只要统计大于该 score 的 score 数量，然后加 1。 要统计某个 score 的排名，只要统计大于该 score 的 score 数量，然后加 1。
@ -852,6 +842,23 @@ ORDER BY
S1.score DESC; S1.score DESC;
``` ```
## SQL Schema
```sql
DROP TABLE
IF
EXISTS Scores;
CREATE TABLE Scores ( Id INT, Score DECIMAL ( 3, 2 ) );
INSERT INTO Scores ( Id, Score )
VALUES
( 1, 3.5 ),
( 2, 3.65 ),
( 3, 4.0 ),
( 4, 3.85 ),
( 5, 4.0 ),
( 6, 3.65 );
```
# 180. Consecutive Numbers # 180. Consecutive Numbers
https://leetcode.com/problems/consecutive-numbers/description/ https://leetcode.com/problems/consecutive-numbers/description/
@ -884,6 +891,21 @@ https://leetcode.com/problems/consecutive-numbers/description/
+-----------------+ +-----------------+
``` ```
## Solution
```sql
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;
```
## SQL Schema ## SQL Schema
```sql ```sql
@ -902,21 +924,6 @@ VALUES
( 7, 2 ); ( 7, 2 );
``` ```
## Solution
```sql
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;
```
# 626. Exchange Seats # 626. Exchange Seats
https://leetcode.com/problems/exchange-seats/description/ https://leetcode.com/problems/exchange-seats/description/
@ -951,22 +958,6 @@ seat 表存储着座位对应的学生。
+---------+---------+ +---------+---------+
``` ```
## SQL Schema
```sql
DROP TABLE
IF
EXISTS seat;
CREATE TABLE seat ( id INT, student VARCHAR ( 255 ) );
INSERT INTO seat ( id, student )
VALUES
( '1', 'Abbot' ),
( '2', 'Doris' ),
( '3', 'Emerson' ),
( '4', 'Green' ),
( '5', 'Jeames' );
```
## Solution ## Solution
使用多个 union。 使用多个 union。
@ -1004,6 +995,22 @@ ORDER BY
id; id;
``` ```
## SQL Schema
```sql
DROP TABLE
IF
EXISTS seat;
CREATE TABLE seat ( id INT, student VARCHAR ( 255 ) );
INSERT INTO seat ( id, student )
VALUES
( '1', 'Abbot' ),
( '2', 'Doris' ),
( '3', 'Emerson' ),
( '4', 'Green' ),
( '5', 'Jeames' );
```