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CyC2018
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@ -1,4 +1,10 @@
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<!-- 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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@ -10,6 +16,360 @@
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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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## SQL Schema
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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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## 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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# 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 | 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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## SQL Schema
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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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## Solution
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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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# 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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邮件地址表：
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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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## 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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## 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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# 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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| Email |
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+---------+
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| a@b.com |
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+---------+
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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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## Solution
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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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# 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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## 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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## Solution
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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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# 196. Delete Duplicate Emails
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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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## SQL Schema
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与 182 相同。
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## Solution
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连接：
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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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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 ( SELECT id FROM ( SELECT min( id ) AS id FROM Person GROUP BY email ) 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 ( SELECT min( id ) AS id FROM Person GROUP BY email );
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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)
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# 175. Combine Two Tables
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https://leetcode.com/problems/combine-two-tables/description/
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@ -75,7 +435,7 @@ SELECT
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City,
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State
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FROM
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Person AS P
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Person P
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LEFT JOIN Address AS A ON P.PersonId = A.PersonId;
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```
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@ -103,14 +463,16 @@ Employee 表：
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## SQL Schema
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```sql
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DROP TABLE IF EXISTS Employee;
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DROP TABLE
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IF
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EXISTS Employee;
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CREATE TABLE Employee ( Id INT, NAME VARCHAR ( 255 ), Salary INT, ManagerId INT );
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INSERT INTO Employee ( Id, NAME, Salary, ManagerId )
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VALUES
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( '1', 'Joe', '70000', '3' ),
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( '2', 'Henry', '80000', '4' ),
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( '3', 'Sam', '60000', NULL ),
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( '4', 'Max', '90000', NULL );
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( 1, 'Joe', 70000, 3 ),
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( 2, 'Henry', 80000, 4 ),
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( 3, 'Sam', 60000, NULL ),
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( 4, 'Max', 90000, NULL );
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```
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## Solution
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@ -119,8 +481,8 @@ VALUES
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SELECT
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E1.NAME AS Employee
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FROM
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Employee AS E1
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INNER JOIN Employee AS E2 ON E1.ManagerId = E2.Id
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Employee E1
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INNER JOIN Employee E2 ON E1.ManagerId = E2.Id
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AND E1.Salary > E2.Salary;
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```
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@ -168,20 +530,24 @@ Orders 表：
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## SQL Schema
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|
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```sql
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DROP TABLE IF EXISTS Customers;
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DROP TABLE
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IF
|
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EXISTS Customers;
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CREATE TABLE Customers ( Id INT, NAME VARCHAR ( 255 ) );
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DROP TABLE IF EXISTS Orders;
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DROP TABLE
|
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IF
|
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EXISTS Orders;
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CREATE TABLE Orders ( Id INT, CustomerId INT );
|
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INSERT INTO Customers ( Id, NAME )
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VALUES
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( '1', 'Joe' ),
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( '2', 'Henry' ),
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( '3', 'Sam' ),
|
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( '4', 'Max' );
|
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( 1, 'Joe' ),
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( 2, 'Henry' ),
|
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( 3, 'Sam' ),
|
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( 4, 'Max' );
|
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INSERT INTO Orders ( Id, CustomerId )
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VALUES
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( '1', '3' ),
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( '2', '1' );
|
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( 1, 3 ),
|
||||
( 2, 1 );
|
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```
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## Solution
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|
|
@ -192,8 +558,8 @@ VALUES
|
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SELECT
|
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C.NAME AS Customers
|
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FROM
|
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Customers AS C
|
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LEFT JOIN Orders AS O ON C.Id = O.CustomerId
|
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Customers C
|
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LEFT JOIN Orders O ON C.Id = O.CustomerId
|
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WHERE
|
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O.CustomerId IS NULL;
|
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```
|
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|
|
@ -204,7 +570,7 @@ WHERE
|
|||
SELECT
|
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C.NAME AS Customers
|
||||
FROM
|
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Customers AS C
|
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Customers C
|
||||
WHERE
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C.Id NOT IN ( SELECT CustomerId FROM Orders );
|
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```
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|
|
@ -277,16 +643,16 @@ VALUES
|
|||
|
||||
```sql
|
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SELECT
|
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D.NAME AS Department,
|
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E.NAME AS Employee,
|
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D.NAME Department,
|
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E.NAME Employee,
|
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E.Salary
|
||||
FROM
|
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Employee AS E,
|
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Department AS D,
|
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( SELECT DepartmentId, MAX( Salary ) AS Salary FROM Employee GROUP BY DepartmentId ) AS M
|
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Employee E,
|
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Department D,
|
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( SELECT DepartmentId, MAX( Salary ) Salary FROM Employee GROUP BY DepartmentId ) M
|
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WHERE
|
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E.DepartmentId = D.Id
|
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AND E.DepartmentId = M.DepartmentId
|
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E.DepartmentId = D.Id
|
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AND E.DepartmentId = M.DepartmentId
|
||||
AND E.Salary = M.Salary;
|
||||
```
|
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|
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|
|
@ -327,9 +693,9 @@ IF
|
|||
CREATE TABLE Employee ( Id INT, Salary INT );
|
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INSERT INTO Employee ( Id, Salary )
|
||||
VALUES
|
||||
( '1', '100' ),
|
||||
( '2', '200' ),
|
||||
( '3', '300' );
|
||||
( 1, 100 ),
|
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( 2, 200 ),
|
||||
( 3, 300 );
|
||||
```
|
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|
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## Solution
|
||||
|
|
@ -338,7 +704,7 @@ VALUES
|
|||
|
||||
```sql
|
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SELECT
|
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( SELECT DISTINCT Salary FROM Employee ORDER BY Salary DESC LIMIT 1, 1 ) AS SecondHighestSalary;
|
||||
( SELECT DISTINCT Salary FROM Employee ORDER BY Salary DESC LIMIT 1, 1 ) SecondHighestSalary;
|
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```
|
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|
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# 177. Nth Highest Salary
|
||||
|
|
@ -407,12 +773,12 @@ IF
|
|||
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' );
|
||||
( 1, 3.5 ),
|
||||
( 2, 3.65 ),
|
||||
( 3, 4.0 ),
|
||||
( 4, 3.85 ),
|
||||
( 5, 4.0 ),
|
||||
( 6, 3.65 );
|
||||
```
|
||||
|
||||
## Solution
|
||||
|
|
@ -420,10 +786,10 @@ VALUES
|
|||
```sql
|
||||
SELECT
|
||||
S1.score,
|
||||
COUNT( DISTINCT S2.score ) AS Rank
|
||||
COUNT( DISTINCT S2.score ) Rank
|
||||
FROM
|
||||
Scores AS S1
|
||||
INNER JOIN Scores AS S2 ON S1.score <= S2.score
|
||||
Scores S1
|
||||
INNER JOIN Scores S2 ON S1.score <= S2.score
|
||||
GROUP BY
|
||||
S1.id
|
||||
ORDER BY
|
||||
|
|
@ -471,26 +837,26 @@ IF
|
|||
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' );
|
||||
( 1, 1 ),
|
||||
( 2, 1 ),
|
||||
( 3, 1 ),
|
||||
( 4, 2 ),
|
||||
( 5, 1 ),
|
||||
( 6, 2 ),
|
||||
( 7, 2 );
|
||||
```
|
||||
|
||||
## Solution
|
||||
|
||||
```sql
|
||||
SELECT
|
||||
DISTINCT L1.num AS ConsecutiveNums
|
||||
DISTINCT L1.num ConsecutiveNums
|
||||
FROM
|
||||
Logs AS L1,
|
||||
Logs AS L2,
|
||||
Logs AS L3
|
||||
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;
|
||||
AND l2.num = l3.num;
|
||||
```
|
||||
|
|
|
|||
Loading…
Reference in New Issue
Block a user