leetcode SQL题目

文章目录

组合两个表

sql 复制代码
SELECT firstName,lastName,city,state FROM Person LEFT JOIN Address ON Person.PersonId=Address.PersonId

第二高的薪水

sql 复制代码
SELECT
IFNULL(
(SELECT DISTINCT salary FROM Employee Order by salary desc LIMIT 1,1),null
) AS SecondHighestSalary

第N高的薪水

sql 复制代码
CREATE FUNCTION getNthHighestSalary(N INT) RETURNS INT
BEGIN
SET N:=N-1;
  RETURN (
      # Write your MySQL query statement below.
      SELECT IFNULL((SELECT DISTINCT salary FROM Employee ORDER BY salary DESC LIMIT N,1),NULL) 
  );
END

分数排名

sql 复制代码
SELECT score, (SELECT count(DISTINCT score)+1 FROM Scores S2 Where S2.score>Scores.score) AS 'rank' FROM Scores ORDER BY score DESC

连续出现的数字

sql 复制代码
SELECT DISTINCT l1.num AS '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

超过经理收入的员工

sql 复制代码
SELECT e.name AS Employee FROM Employee e,Employee e2 WHERE e2.id=e.managerId AND e.salary > e2.salary

查找重复的电子邮件

sql 复制代码
SELECT DISTINCT p1.email AS Email FROM Person p1,Person p2 WHERE p1.email=p2.email AND p1.id !=p2.id

从不订购的客户

sql 复制代码
SELECT name AS Customers FROM Customers WHERE id NOT IN (SELECT DISTINCT customerId FROM Orders)

部门工资最高的员工

sql 复制代码
SELECT Department.name AS Department,e.name AS Employee,e.salary AS Salary FROM Department,Employee e WHERE Department.id=e.departmentId AND e.salary >= ALL(SELECT salary FROM Employee e2 WHERE e.departmentId = e2.departmentId)

部门工资前三高的所有员工

sql 复制代码
SELECT Department.name AS Department, e.name AS Employee,e.salary AS Salary FROM Employee e,Department WHERE e.departmentId=Department.id AND
(SELECT COUNT(Distinct e2.salary) FROM Employee e2 WHERE e2.departmentId=e.departmentId AND e2.salary>e.salary)<3 ORDER BY e.departmentId ASC

删除重复的电子邮箱

sql 复制代码
DELETE p2 FROM Person p1,Person p2 WHERE p1.email=p2.email AND p1.id<p2.id

上升的温度

sql 复制代码
SELECT w2.id AS Id FROM Weather w1,Weather w2 WHERE DATEDIFF(w2.recordDate,w1.recordDate)=1 AND w1.Temperature < w2.Temperature

DATEDIFF()函数用于计算两个日期的天数差

游戏玩法分析Ⅰ

sql 复制代码
SELECT player_id,MIN(event_date) AS first_login FROM Activity GROUP BY player_id

游戏玩法Ⅳ

sql 复制代码
SELECT ROUND(
(SELECT COUNT(DISTINCT b.player_id) FROM Activity b WHERE (b.player_id,b.event_date) IN
(
SELECT a.player_id,DATE(MIN(a.event_date)+INTERVAL 1 DAY) FROM Activity a GROUP BY a.player_id
))
/
(SELECT COUNT(DISTINCT player_id) FROM Activity),2
) AS fraction
相关推荐
skyang.12 分钟前
LeetCode 85. 最大矩形
算法·leetcode·职场和发展
滋滋不吱吱1 小时前
枚举中间位置基础篇
考研·算法·leetcode
阳光不锈@1 小时前
算法:最长递增子序列解法记录
算法·最长递增子序列·超详细分析·java实现
遇见尚硅谷1 小时前
C语言:20250728学习(指针)
c语言·开发语言·数据结构·c++·笔记·学习·算法
十八岁讨厌编程2 小时前
【算法训练营Day17】二叉树part7
算法
YouQian7722 小时前
(AC)Playlist
算法
算法_小学生3 小时前
决策树(Decision Tree)完整解析:原理 + 数学推导 + 剪枝 + 实战
算法·决策树·剪枝
岁忧3 小时前
(LeetCode 面试经典 150 题 ) 155. 最小栈 (栈)
java·c++·算法·leetcode·面试·go
明明如月学长4 小时前
什么你不知道 Cherry Studio 有快捷助手?
算法
Vegetable_Dragon4 小时前
数论1.01
算法