MySQL-刷题记录

Posted 2021-01-16 AugusKong

1. Second Highest Salary

SELECT MAX(Salary)
FROM Employee
WHERE Salary < (SELECT max(Salary) FROM Employee)

1. Nth Highest Salary

CREATE FUNCTION getNthHightestSalary(N INT) RETURNS INT
BEGIN
DECLARE M INT;
SET M=N-1;
    RETURN (
        SELECT DISTINCT Salary FROM Employee ORDER BY Salary DESC LIMIT M, 1(LIMIT M, 1 是LIMIT 1 OFFSET M的简写) OFFSET M表示从表的M+1行(0-based开始查找
    );
END

1. Shortest Distance in a Line
   每个点与其他所有的点都做差比较，最后返回最小值
   Self Join: table joined with itself
   SELECT column_name(s) FROM table T1, table T2 WHERE condition

SELECT MIN(ABS(P1.x - P2.x)) AS shortest FROM point AS P1
JOIN point AS P2 ON P1.x > P2.x(等价于P1.x <> P2.x <>符号表示不相等)

SELECT MIN(ABS(P1.x - P2.x)) AS shortest FROM 
point  P1, point P2 
WHERE P1.x > P2.x

KEYWORD LIST

LIMIT: Specify the number of records to return in the result set
LEFT JOIN: Returns all rows from the left table, and the matching rows from the right table
NOT NULL: enforces a column to not accept NULL values CREATE TABLE Persons( ID int NOT NULL, Name varchar(25) NOT NULL);
ROUND(number, decimals):: number: Required The number to be rounded decimals: Optional, the number of decimal places to round number it. 默认是整型
CASE FUNCTION
语法:

CASE
    WHEN condition1 THEN result1
    WHEN condition2 THEN result2
    WHEN conditionN THEN resultN
    ELSE result
END;

IF FUNCTION

WHERE OR条件 与 UNION表的效率差别

1. Friend Requests I: Overall Acceptance Rate
   必须找到unique 的acceptance 以及 request

SELECT
    IFNULL(
        (SELECT ROUND(COUNT(DISTINCT  requester_id, accepter_id)/COUNT(DISTINCT sender_id, send_to_id), 2)
FROM request_accepted, friend_request), 0) 
AS accept_rate; 

1. Tree Node

id	p_id
1	null
2	1
3	1
4	2
5	2

Each node in the tree can be one of three types:
Leaf / Root / Inner

通过Query找出每个node的Type

Query之间的 连接组合可以使用的SQL语法:

• CASE WHEN关键字
• IF关键字
• UNION关键字

SELECT id, ‘Root‘ AS Type
FROM tree 
WHERE p_id IS NULL
UNION

SELECT id, ‘Leaf‘ AS Type
FROM tree
WHERE id NOT IN (SELECT DISTINCT p_id FROM tree WHERE p_id IS NOT NULL)
AND p_id IS NOT NULL
UNION

SELECT id, ‘Inner‘ AS Type
FROM tree
WHERE id IN (SELECT DISTINCT p_id FROM tree WHERE p_id IS NOT NULL) 
AND p_id IS NOT NULL
ORDER BY id;

--使用CASE关键词
SELECT id AS `Id`, 
(CASE 
    WHEN p_id is null THEN ‘ROOT‘
    WHEN id IN (SELECT p_id FROM tree WHERE p_id IS NOT NULL) THEN ‘Inner‘
    ELSE ‘Leaf‘ 
END) as Type
FROM tree
GROUP BY id; 

1. Rank Scores