1099 lines
22 KiB
Java
1099 lines
22 KiB
Java
<!-- GFM-TOC -->
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* [595. Big Countries](#595-big-countries)
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* [627. Swap Salary](#627-swap-salary)
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* [620. Not Boring Movies](#620-not-boring-movies)
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* [596. Classes More Than 5 Students](#596-classes-more-than-5-students)
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* [182. Duplicate Emails](#182-duplicate-emails)
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* [196. Delete Duplicate Emails](#196-delete-duplicate-emails)
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* [175. Combine Two Tables](#175-combine-two-tables)
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* [181. Employees Earning More Than Their Managers](#181-employees-earning-more-than-their-managers)
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* [183. Customers Who Never Order](#183-customers-who-never-order)
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* [184. Department Highest Salary](#184-department-highest-salary)
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* [176. Second Highest Salary](#176-second-highest-salary)
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* [177. Nth Highest Salary](#177-nth-highest-salary)
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* [178. Rank Scores](#178-rank-scores)
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* [180. Consecutive Numbers](#180-consecutive-numbers)
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* [626. Exchange Seats](#626-exchange-seats)
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<!-- GFM-TOC -->
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# 595. Big Countries
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https://leetcode.com/problems/big-countries/description/
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## Description
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```html
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+-----------------+------------+------------+--------------+---------------+
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| name | continent | area | population | gdp |
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+-----------------+------------+------------+--------------+---------------+
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| Afghanistan | Asia | 652230 | 25500100 | 20343000 |
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| Albania | Europe | 28748 | 2831741 | 12960000 |
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| Algeria | Africa | 2381741 | 37100000 | 188681000 |
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| Andorra | Europe | 468 | 78115 | 3712000 |
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| Angola | Africa | 1246700 | 20609294 | 100990000 |
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+-----------------+------------+------------+--------------+---------------+
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```
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查找面积超过 3,000,000 或者人口数超过 25,000,000 的国家。
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```html
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+--------------+-------------+--------------+
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| name | population | area |
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+--------------+-------------+--------------+
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| Afghanistan | 25500100 | 652230 |
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| Algeria | 37100000 | 2381741 |
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+--------------+-------------+--------------+
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```
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## Solution
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```sql
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SELECT name,
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population,
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area
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FROM
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World
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WHERE
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area > 3000000
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OR population > 25000000;
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```
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## SQL Schema
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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。
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```sql
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DROP TABLE
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IF
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EXISTS World;
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CREATE TABLE World ( NAME VARCHAR ( 255 ), continent VARCHAR ( 255 ), area INT, population INT, gdp INT );
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INSERT INTO World ( NAME, continent, area, population, gdp )
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VALUES
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( 'Afghanistan', 'Asia', '652230', '25500100', '203430000' ),
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( 'Albania', 'Europe', '28748', '2831741', '129600000' ),
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( 'Algeria', 'Africa', '2381741', '37100000', '1886810000' ),
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( 'Andorra', 'Europe', '468', '78115', '37120000' ),
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( 'Angola', 'Africa', '1246700', '20609294', '1009900000' );
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```
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# 627. Swap Salary
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https://leetcode.com/problems/swap-salary/description/
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## Description
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```html
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| id | name | sex | salary |
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|----|------|-----|--------|
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| 1 | A | m | 2500 |
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| 2 | B | f | 1500 |
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| 3 | C | m | 5500 |
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| 4 | D | f | 500 |
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```
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只用一个 SQL 查询，将 sex 字段反转。
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```html
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| id | name | sex | salary |
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|----|------|-----|--------|
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| 1 | A | f | 2500 |
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| 2 | B | m | 1500 |
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| 3 | C | f | 5500 |
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| 4 | D | m | 500 |
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```
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## Solution
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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。
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sex 字段只有两个取值：'f' 和 'm'，并且有以下规律：
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```
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'f' ^ ('m' ^ 'f') = 'm' ^ ('f' ^ 'f') = 'm'
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'm' ^ ('m' ^ 'f') = 'f' ^ ('m' ^ 'm') = 'f'
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```
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因此将 sex 字段和 'm' ^ 'f' 进行异或操作，最后就能反转 sex 字段。
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```sql
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UPDATE salary
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SET sex = CHAR ( ASCII(sex) ^ ASCII( 'm' ) ^ ASCII( 'f' ) );
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```
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## SQL Schema
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```sql
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DROP TABLE
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IF
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EXISTS salary;
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CREATE TABLE salary ( id INT, NAME VARCHAR ( 100 ), sex CHAR ( 1 ), salary INT );
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INSERT INTO salary ( id, NAME, sex, salary )
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VALUES
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( '1', 'A', 'm', '2500' ),
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( '2', 'B', 'f', '1500' ),
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( '3', 'C', 'm', '5500' ),
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( '4', 'D', 'f', '500' );
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```
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# 620. Not Boring Movies
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https://leetcode.com/problems/not-boring-movies/description/
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## Description
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```html
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+---------+-----------+--------------+-----------+
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| id | movie | description | rating |
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+---------+-----------+--------------+-----------+
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| 1 | War | great 3D | 8.9 |
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| 2 | Science | fiction | 8.5 |
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| 3 | irish | boring | 6.2 |
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| 4 | Ice song | Fantacy | 8.6 |
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| 5 | House card| Interesting| 9.1 |
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+---------+-----------+--------------+-----------+
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```
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查找 id 为奇数，并且 description 不是 boring 的电影，按 rating 降序。
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```html
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+---------+-----------+--------------+-----------+
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| id | movie | description | rating |
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+---------+-----------+--------------+-----------+
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| 5 | House card| Interesting| 9.1 |
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| 1 | War | great 3D | 8.9 |
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+---------+-----------+--------------+-----------+
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```
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## Solution
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```sql
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SELECT
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*
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FROM
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cinema
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WHERE
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id % 2 = 1
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AND description != 'boring'
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ORDER BY
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rating DESC;
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```
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## SQL Schema
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```sql
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DROP TABLE
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IF
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EXISTS cinema;
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CREATE TABLE cinema ( id INT, movie VARCHAR ( 255 ), description VARCHAR ( 255 ), rating FLOAT ( 2, 1 ) );
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INSERT INTO cinema ( id, movie, description, rating )
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VALUES
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( 1, 'War', 'great 3D', 8.9 ),
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( 2, 'Science', 'fiction', 8.5 ),
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( 3, 'irish', 'boring', 6.2 ),
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( 4, 'Ice song', 'Fantacy', 8.6 ),
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( 5, 'House card', 'Interesting', 9.1 );
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```
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# 596. Classes More Than 5 Students
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https://leetcode.com/problems/classes-more-than-5-students/description/
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## Description
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```html
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+---------+------------+
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| student | class |
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+---------+------------+
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| A | Math |
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| B | English |
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| C | Math |
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| D | Biology |
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| E | Math |
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| F | Computer |
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| G | Math |
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| H | Math |
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| I | Math |
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+---------+------------+
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```
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查找有五名及以上 student 的 class。
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```html
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+---------+
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| class |
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+---------+
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| Math |
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+---------+
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```
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## Solution
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对 class 列进行分组之后，再使用 count 汇总函数统计每个分组的记录个数，之后使用 HAVING 进行筛选。HAVING 针对分组进行筛选，而 WHERE 针对每个记录（行）进行筛选。
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```sql
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SELECT
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class
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FROM
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courses
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GROUP BY
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class
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HAVING
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count( DISTINCT student ) >= 5;
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```
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## SQL Schema
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```sql
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DROP TABLE
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IF
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EXISTS courses;
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CREATE TABLE courses ( student VARCHAR ( 255 ), class VARCHAR ( 255 ) );
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INSERT INTO courses ( student, class )
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VALUES
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( 'A', 'Math' ),
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( 'B', 'English' ),
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( 'C', 'Math' ),
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( 'D', 'Biology' ),
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( 'E', 'Math' ),
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( 'F', 'Computer' ),
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( 'G', 'Math' ),
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( 'H', 'Math' ),
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( 'I', 'Math' );
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```
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# 182. Duplicate Emails
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https://leetcode.com/problems/duplicate-emails/description/
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## Description
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邮件地址表：
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```html
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+----+---------+
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| Id | Email |
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+----+---------+
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| 1 | a@b.com |
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| 2 | c@d.com |
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| 3 | a@b.com |
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+----+---------+
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```
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查找重复的邮件地址：
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```html
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+---------+
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| Email |
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+---------+
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| a@b.com |
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+---------+
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```
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## Solution
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对 Email 进行分组，如果并使用 COUNT 进行计数统计，结果大于等于 2 的表示 Email 重复。
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```sql
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SELECT
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Email
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FROM
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Person
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GROUP BY
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Email
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HAVING
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COUNT( * ) >= 2;
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```
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## SQL Schema
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```sql
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DROP TABLE
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IF
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EXISTS Person;
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CREATE TABLE Person ( Id INT, Email VARCHAR ( 255 ) );
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INSERT INTO Person ( Id, Email )
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VALUES
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( 1, 'a@b.com' ),
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( 2, 'c@d.com' ),
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( 3, 'a@b.com' );
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```
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# 196. Delete Duplicate Emails
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|
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https://leetcode.com/problems/delete-duplicate-emails/description/
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## Description
|
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邮件地址表：
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```html
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+----+---------+
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| Id | Email |
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+----+---------+
|
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| 1 | john@example.com |
|
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| 2 | bob@example.com |
|
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| 3 | john@example.com |
|
||
+----+---------+
|
||
```
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|
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删除重复的邮件地址：
|
||
|
||
```html
|
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+----+------------------+
|
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| Id | Email |
|
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+----+------------------+
|
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| 1 | john@example.com |
|
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| 2 | bob@example.com |
|
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+----+------------------+
|
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```
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|
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## Solution
|
||
|
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只保留相同 Email 中 Id 最小的那一个，然后删除其它的。
|
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|
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连接查询：
|
||
|
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```sql
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DELETE p1
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FROM
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Person p1,
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Person p2
|
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WHERE
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p1.Email = p2.Email
|
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AND p1.Id > p2.Id
|
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```
|
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|
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子查询：
|
||
|
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```sql
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DELETE
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FROM
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Person
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WHERE
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id NOT IN (
|
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SELECT id
|
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FROM (
|
||
SELECT min( id ) AS id
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FROM Person
|
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GROUP BY email
|
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) AS m
|
||
);
|
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```
|
||
|
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应该注意的是上述解法额外嵌套了一个 SELECT 语句，如果不这么做，会出现错误：You can't specify target table 'Person' for update in FROM clause。以下演示了这种错误解法。
|
||
|
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```sql
|
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DELETE
|
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FROM
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Person
|
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WHERE
|
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id NOT IN (
|
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SELECT min( id ) AS id
|
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FROM Person
|
||
GROUP BY email
|
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);
|
||
```
|
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|
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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
|
||
|
||
https://leetcode.com/problems/combine-two-tables/description/
|
||
|
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## Description
|
||
|
||
Person 表：
|
||
|
||
```html
|
||
+-------------+---------+
|
||
| Column Name | Type |
|
||
+-------------+---------+
|
||
| PersonId | int |
|
||
| FirstName | varchar |
|
||
| LastName | varchar |
|
||
+-------------+---------+
|
||
PersonId is the primary key column for this table.
|
||
```
|
||
|
||
Address 表：
|
||
|
||
```html
|
||
+-------------+---------+
|
||
| Column Name | Type |
|
||
+-------------+---------+
|
||
| AddressId | int |
|
||
| PersonId | int |
|
||
| City | varchar |
|
||
| State | varchar |
|
||
+-------------+---------+
|
||
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。
|
||
|
||
## 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,
|
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LastName,
|
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City,
|
||
State
|
||
FROM
|
||
Person P
|
||
LEFT JOIN Address A
|
||
ON P.PersonId = A.PersonId;
|
||
```
|
||
|
||
## SQL Schema
|
||
|
||
```sql
|
||
DROP TABLE
|
||
IF
|
||
EXISTS Person;
|
||
CREATE TABLE Person ( PersonId INT, FirstName VARCHAR ( 255 ), LastName VARCHAR ( 255 ) );
|
||
DROP TABLE
|
||
IF
|
||
EXISTS Address;
|
||
CREATE TABLE Address ( AddressId INT, PersonId INT, City VARCHAR ( 255 ), State VARCHAR ( 255 ) );
|
||
INSERT INTO Person ( PersonId, LastName, FirstName )
|
||
VALUES
|
||
( 1, 'Wang', 'Allen' );
|
||
INSERT INTO Address ( AddressId, PersonId, City, State )
|
||
VALUES
|
||
( 1, 2, 'New York City', 'New York' );
|
||
```
|
||
|
||
# 181. Employees Earning More Than Their Managers
|
||
|
||
https://leetcode.com/problems/employees-earning-more-than-their-managers/description/
|
||
|
||
## Description
|
||
|
||
Employee 表：
|
||
|
||
```html
|
||
+----+-------+--------+-----------+
|
||
| 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
|
||
|
||
```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
|
||
DROP TABLE
|
||
IF
|
||
EXISTS Employee;
|
||
CREATE TABLE Employee ( Id INT, NAME VARCHAR ( 255 ), Salary INT, ManagerId INT );
|
||
INSERT INTO Employee ( Id, NAME, Salary, ManagerId )
|
||
VALUES
|
||
( 1, 'Joe', 70000, 3 ),
|
||
( 2, 'Henry', 80000, 4 ),
|
||
( 3, 'Sam', 60000, NULL ),
|
||
( 4, 'Max', 90000, NULL );
|
||
```
|
||
|
||
# 183. Customers Who Never Order
|
||
|
||
https://leetcode.com/problems/customers-who-never-order/description/
|
||
|
||
## Description
|
||
|
||
Customers 表：
|
||
|
||
```html
|
||
+----+-------+
|
||
| Id | Name |
|
||
+----+-------+
|
||
| 1 | Joe |
|
||
| 2 | Henry |
|
||
| 3 | Sam |
|
||
| 4 | Max |
|
||
+----+-------+
|
||
```
|
||
|
||
Orders 表：
|
||
|
||
```html
|
||
+----+------------+
|
||
| Id | CustomerId |
|
||
+----+------------+
|
||
| 1 | 3 |
|
||
| 2 | 1 |
|
||
+----+------------+
|
||
```
|
||
|
||
查找没有订单的顾客信息：
|
||
|
||
```html
|
||
+-----------+
|
||
| Customers |
|
||
+-----------+
|
||
| Henry |
|
||
| Max |
|
||
+-----------+
|
||
```
|
||
|
||
## Solution
|
||
|
||
左外链接
|
||
|
||
```sql
|
||
SELECT
|
||
C.Name AS Customers
|
||
FROM
|
||
Customers C
|
||
LEFT JOIN Orders O
|
||
ON C.Id = O.CustomerId
|
||
WHERE
|
||
O.CustomerId IS NULL;
|
||
```
|
||
|
||
子查询
|
||
|
||
```sql
|
||
SELECT
|
||
Name AS Customers
|
||
FROM
|
||
Customers
|
||
WHERE
|
||
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
|
||
|
||
https://leetcode.com/problems/department-highest-salary/description/
|
||
|
||
## Description
|
||
|
||
Employee 表：
|
||
|
||
```html
|
||
+----+-------+--------+--------------+
|
||
| Id | Name | Salary | DepartmentId |
|
||
+----+-------+--------+--------------+
|
||
| 1 | Joe | 70000 | 1 |
|
||
| 2 | Henry | 80000 | 2 |
|
||
| 3 | Sam | 60000 | 2 |
|
||
| 4 | Max | 90000 | 1 |
|
||
+----+-------+--------+--------------+
|
||
```
|
||
|
||
Department 表：
|
||
|
||
```html
|
||
+----+----------+
|
||
| Id | Name |
|
||
+----+----------+
|
||
| 1 | IT |
|
||
| 2 | Sales |
|
||
+----+----------+
|
||
```
|
||
|
||
查找一个 Department 中收入最高者的信息：
|
||
|
||
```html
|
||
+------------+----------+--------+
|
||
| Department | Employee | Salary |
|
||
+------------+----------+--------+
|
||
| IT | Max | 90000 |
|
||
| Sales | Henry | 80000 |
|
||
+------------+----------+--------+
|
||
```
|
||
|
||
## 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。
|
||
|
||
之后使用连接找到一个部门中薪资等于临时表中最大薪资的员工。
|
||
|
||
```sql
|
||
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;
|
||
```
|
||
|
||
## 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
|
||
|
||
https://leetcode.com/problems/second-highest-salary/description/
|
||
|
||
## Description
|
||
|
||
```html
|
||
+----+--------+
|
||
| Id | Salary |
|
||
+----+--------+
|
||
| 1 | 100 |
|
||
| 2 | 200 |
|
||
| 3 | 300 |
|
||
+----+--------+
|
||
```
|
||
|
||
查找工资第二高的员工。
|
||
|
||
```html
|
||
+---------------------+
|
||
| SecondHighestSalary |
|
||
+---------------------+
|
||
| 200 |
|
||
+---------------------+
|
||
```
|
||
|
||
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
|
||
DROP TABLE
|
||
IF
|
||
EXISTS Employee;
|
||
CREATE TABLE Employee ( Id INT, Salary INT );
|
||
INSERT INTO Employee ( Id, Salary )
|
||
VALUES
|
||
( 1, 100 ),
|
||
( 2, 200 ),
|
||
( 3, 300 );
|
||
```
|
||
|
||
# 177. Nth Highest Salary
|
||
|
||
## Description
|
||
|
||
查找工资第 N 高的员工。
|
||
|
||
## Solution
|
||
|
||
```sql
|
||
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
|
||
```
|
||
|
||
## SQL Schema
|
||
|
||
同 176。
|
||
|
||
|
||
# 178. Rank Scores
|
||
|
||
https://leetcode.com/problems/rank-scores/description/
|
||
|
||
## Description
|
||
|
||
得分表：
|
||
|
||
```html
|
||
+----+-------+
|
||
| 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。
|
||
|
||
```html
|
||
+-------+------+
|
||
| Score | Rank |
|
||
+-------+------+
|
||
| 4.00 | 1 |
|
||
| 4.00 | 1 |
|
||
| 3.85 | 2 |
|
||
| 3.65 | 3 |
|
||
| 3.65 | 3 |
|
||
| 3.50 | 4 |
|
||
+-------+------+
|
||
```
|
||
|
||
## Solution
|
||
|
||
要统计某个 score 的排名，只要统计大于等于该 score 的 score 数量。
|
||
|
||
| Id | score | 大于等于该 score 的 score 数量 | 排名 |
|
||
| :---: | :---: | :---: | :---: |
|
||
| 1 | 4.1 | 3 | 3 |
|
||
| 2 | 4.2 | 2 | 2 |
|
||
| 3 | 4.3 | 1 | 1 |
|
||
|
||
使用连接操作找到某个 score 对应的大于等于其值的记录：
|
||
|
||
```sql
|
||
SELECT
|
||
*
|
||
FROM
|
||
Scores S1
|
||
INNER JOIN Scores S2
|
||
ON S1.score <= S2.score
|
||
ORDER BY
|
||
S1.score DESC, S1.Id;
|
||
```
|
||
|
||
| S1.Id | S1.score | S2.Id | S2.score |
|
||
| :---: | :---: | :---: | :---: |
|
||
|3| 4.3| 3 |4.3|
|
||
|2| 4.2| 2| 4.2|
|
||
|2| 4.2 |3 |4.3|
|
||
|1| 4.1 |1| 4.1|
|
||
|1| 4.1 |2| 4.2|
|
||
|1| 4.1 |3| 4.3|
|
||
|
||
可以看到每个 S1.score 都有对应好几条记录，我们再进行分组，并统计每个分组的数量作为 'Rank'
|
||
|
||
```sql
|
||
SELECT
|
||
S1.score 'Score',
|
||
COUNT(*) 'Rank'
|
||
FROM
|
||
Scores S1
|
||
INNER JOIN Scores S2
|
||
ON S1.score <= S2.score
|
||
GROUP BY
|
||
S1.id, S1.score
|
||
ORDER BY
|
||
S1.score DESC, S1.Id;
|
||
```
|
||
|
||
| score | Rank |
|
||
| :---: | :---: |
|
||
| 4.3 | 1 |
|
||
| 4.2 | 2 |
|
||
| 4.1 | 3 |
|
||
|
||
上面的解法看似没问题，但是对于以下数据，它却得到了错误的结果：
|
||
|
||
| Id | score |
|
||
| :---: | :---: |
|
||
| 1 | 4.1 |
|
||
| 2 | 4.2 |
|
||
| 3 | 4.2 |
|
||
|
||
| score | Rank |
|
||
| :---: | :--: |
|
||
| 4.2 | 2 |
|
||
| 4.2 | 2 |
|
||
| 4.1 | 3 |
|
||
|
||
而我们希望的结果为：
|
||
|
||
| score | Rank |
|
||
| :---: | :--: |
|
||
| 4.2 | 1 |
|
||
| 4.2 | 1 |
|
||
| 4.1 | 2 |
|
||
|
||
连接情况如下：
|
||
|
||
| S1.Id | S1.score | S2.Id | S2.score |
|
||
| :---: | :------: | :---: | :------: |
|
||
| 2 | 4.2 | 3 | 4.2 |
|
||
| 2 | 4.2 | 2 | 4.2 |
|
||
| 3 | 4.2 | 3 | 4.2 |
|
||
| 3 | 4.2 | 2 | 4.1 |
|
||
| 1 | 4.1 | 3 | 4.2 |
|
||
| 1 | 4.1 | 2 | 4.2 |
|
||
| 1 | 4.1 | 1 | 4.1 |
|
||
|
||
我们想要的结果是，把分数相同的放在同一个排名，并且相同分数只占一个位置，例如上面的分数，Id=2 和 Id=3 的记录都有相同的分数，并且最高，他们并列第一。而 Id=1 的记录应该排第二名，而不是第三名。所以在进行 COUNT 计数统计时，我们需要使用 COUNT( DISTINCT S2.score ) 从而只统计一次相同的分数。
|
||
|
||
```sql
|
||
SELECT
|
||
S1.score 'Score',
|
||
COUNT( DISTINCT S2.score ) 'Rank'
|
||
FROM
|
||
Scores S1
|
||
INNER JOIN Scores S2
|
||
ON S1.score <= S2.score
|
||
GROUP BY
|
||
S1.id, S1.score
|
||
ORDER BY
|
||
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, 4.1 ),
|
||
( 2, 4.1 ),
|
||
( 3, 4.2 ),
|
||
( 4, 4.2 ),
|
||
( 5, 4.3 ),
|
||
( 6, 4.3 );
|
||
```
|
||
|
||
# 180. Consecutive Numbers
|
||
|
||
https://leetcode.com/problems/consecutive-numbers/description/
|
||
|
||
## Description
|
||
|
||
数字表：
|
||
|
||
```html
|
||
+----+-----+
|
||
| Id | Num |
|
||
+----+-----+
|
||
| 1 | 1 |
|
||
| 2 | 1 |
|
||
| 3 | 1 |
|
||
| 4 | 2 |
|
||
| 5 | 1 |
|
||
| 6 | 2 |
|
||
| 7 | 2 |
|
||
+----+-----+
|
||
```
|
||
|
||
查找连续出现三次的数字。
|
||
|
||
```html
|
||
+-----------------+
|
||
| ConsecutiveNums |
|
||
+-----------------+
|
||
| 1 |
|
||
+-----------------+
|
||
```
|
||
|
||
## 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
|
||
DROP TABLE
|
||
IF
|
||
EXISTS LOGS;
|
||
CREATE TABLE LOGS ( Id INT, Num INT );
|
||
INSERT INTO LOGS ( Id, Num )
|
||
VALUES
|
||
( 1, 1 ),
|
||
( 2, 1 ),
|
||
( 3, 1 ),
|
||
( 4, 2 ),
|
||
( 5, 1 ),
|
||
( 6, 2 ),
|
||
( 7, 2 );
|
||
```
|
||
|
||
# 626. Exchange Seats
|
||
|
||
https://leetcode.com/problems/exchange-seats/description/
|
||
|
||
## Description
|
||
|
||
seat 表存储着座位对应的学生。
|
||
|
||
```html
|
||
+---------+---------+
|
||
| 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。
|
||
|
||
```html
|
||
+---------+---------+
|
||
| id | student |
|
||
+---------+---------+
|
||
| 1 | Doris |
|
||
| 2 | Abbot |
|
||
| 3 | Green |
|
||
| 4 | Emerson |
|
||
| 5 | Jeames |
|
||
+---------+---------+
|
||
```
|
||
|
||
## Solution
|
||
|
||
使用多个 union。
|
||
|
||
```sql
|
||
# 处理偶数 id，让 id 减 1
|
||
# 例如 2,4,6,... 变成 1,3,5,...
|
||
SELECT
|
||
s1.id - 1 AS id,
|
||
s1.student
|
||
FROM
|
||
seat s1
|
||
WHERE
|
||
s1.id MOD 2 = 0 UNION
|
||
# 处理奇数 id，让 id 加 1。但是如果最大的 id 为奇数，则不做处理
|
||
# 例如 1,3,5,... 变成 2,4,6,...
|
||
SELECT
|
||
s2.id + 1 AS id,
|
||
s2.student
|
||
FROM
|
||
seat s2
|
||
WHERE
|
||
s2.id MOD 2 = 1
|
||
AND s2.id != ( SELECT max( s3.id ) FROM seat s3 ) UNION
|
||
# 如果最大的 id 为奇数，单独取出这个数
|
||
SELECT
|
||
s4.id AS id,
|
||
s4.student
|
||
FROM
|
||
seat s4
|
||
WHERE
|
||
s4.id MOD 2 = 1
|
||
AND s4.id = ( SELECT max( s5.id ) FROM seat s5 )
|
||
ORDER BY
|
||
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' );
|
||
```
|
||
|
||
|
||
|
||
|
||
|
||
|
||
<div align="center"><img width="320px" src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/githubio/公众号二维码-2.png"></img></div>
|