2020年4月14日

今日算法题 445. 两数相加 II

给你两个非空链表来代表两个非负整数。数字最高位位于链表开始位置。它们的每个节点只存储一位数字。将这两数相加会返回一个新的链表。

你可以假设除了数字 0 之外，这两个数字都不会以零开头。

进阶：

如果输入链表不能修改该如何处理？换句话说，你不能对列表中的节点进行翻转。

示例：

输入：(7 -> 2 -> 4 -> 3) + (5 -> 6 -> 4)
输出：7 -> 8 -> 0 -> 7

//官方题解给出的使用栈先进后出的原则，自己解题思路为先将两个链表反转，相加后再将链表反转回去的方式
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {ListNode pro1 = null;ListNode temp = l1;while (temp != null){ListNode next = temp.next;temp.next = pro1;pro1 = temp;temp = next;}ListNode pro2 = null;ListNode temp2 = l2;while (temp2 != null){ListNode next = temp2.next;temp2.next = pro2;pro2 = temp2;temp2 = next;}ListNode result = new ListNode(-1);ListNode tempResult = result ;int flag = 0;while (pro1 != null ||  pro2 != null){int i = (pro1 == null ? 0 : pro1.val) + (pro2 == null ? 0 : pro2.val)+flag;tempResult.next = new ListNode(i%10) ;flag = 0 ;if(i>=10){flag  = 1;}if(pro1 != null){pro1 = pro1.next;}if(pro2 != null){pro2 = pro2.next;}tempResult = tempResult.next;}if(flag == 1){tempResult.next = new ListNode(1) ;}ListNode pro3 = null;ListNode tempResult3 = result.next ;while (tempResult3 != null){ListNode next = tempResult3.next;tempResult3.next = pro3;pro3 = tempResult3;tempResult3 = next;}return pro3;}

182. 查找重复的电子邮箱

//建表语句
Create table If Not Exists Person (Id int, Email varchar(255))
Truncate table Person
insert into Person (Id, Email) values ('1', 'a@b.com')
insert into Person (Id, Email) values ('2', 'c@d.com')
insert into Person (Id, Email) values ('3', 'a@b.com')

编写一个 SQL 查询，查找 Person 表中所有重复的电子邮箱。

示例：
±—±--------+
| Id | Email |
±—±--------+
| 1 | a@b.com |
| 2 | c@d.com |
| 3 | a@b.com |
±—±--------+
根据以上输入，你的查询应返回以下结果：

±--------+
| Email |
±--------+
| a@b.com |
±--------+
说明：所有电子邮箱都是小写字母。

SELECT Email FROM person GROUP BY email HAVING COUNT(email) > 1

183. 从不订购的客户

//建表语句
Create table If Not Exists Customers (Id int, Name varchar(255))
Create table If Not Exists Orders (Id int, CustomerId int)
Truncate table Customers
insert into Customers (Id, Name) values ('1', 'Joe')
insert into Customers (Id, Name) values ('2', 'Henry')
insert into Customers (Id, Name) values ('3', 'Sam')
insert into Customers (Id, Name) values ('4', 'Max')
Truncate table Orders
insert into Orders (Id, CustomerId) values ('1', '3')
insert into Orders (Id, CustomerId) values ('2', '1')

某网站包含两个表，Customers 表和 Orders 表。编写一个 SQL 查询，找出所有从不订购任何东西的客户。
Customers 表：
±—±------+
| Id | Name |
±—±------+
| 1 | Joe |
| 2 | Henry |
| 3 | Sam |
| 4 | Max |
±—±------+
Orders 表：
±—±-----------+
| Id | CustomerId |
±—±-----------+
| 1 | 3 |
| 2 | 1 |
±—±-----------+
例如给定上述表格，你的查询应返回：
±----------+
| Customers |
±----------+
| Henry |
| Max |
±----------+

SELECT NAME AS Customers FROM customers WHERE id NOT in (SELECT customerId FROM orders GROUP BY customerId)

184. 部门工资最高的员工

Create table If Not Exists Employee (Id int, Name varchar(255), Salary int, DepartmentId int);
Create table If Not Exists Department (Id int, Name varchar(255));
Truncate table Employee;
insert into Employee (Id, Name, Salary, DepartmentId) values ('1', 'Joe', '70000', '1');
insert into Employee (Id, Name, Salary, DepartmentId) values ('2', 'Jim', '90000', '1');
insert into Employee (Id, Name, Salary, DepartmentId) values ('3', 'Henry', '80000', '2');
insert into Employee (Id, Name, Salary, DepartmentId) values ('4', 'Sam', '60000', '2');
insert into Employee (Id, Name, Salary, DepartmentId) values ('5', 'Max', '90000', '1');
Truncate table Department;
insert into Department (Id, Name) values ('1', 'IT');
insert into Department (Id, Name) values ('2', 'Sales');

Employee 表包含所有员工信息，每个员工有其对应的 Id, salary 和 department Id。
±—±------±-------±-------------+
| Id | Name | Salary | DepartmentId |
±—±------±-------±-------------+
| 1 | Joe | 70000 | 1 |
| 2 | Henry | 80000 | 2 |
| 3 | Sam | 60000 | 2 |
| 4 | Max | 90000 | 1 |
±—±------±-------±-------------+
Department 表包含公司所有部门的信息。
±—±---------+
| Id | Name |
±—±---------+
| 1 | IT |
| 2 | Sales |
±—±---------+
编写一个 SQL 查询，找出每个部门工资最高的员工。例如，根据上述给定的表格，Max 在 IT 部门有最高工资，Henry 在 Sales 部门有最高工资。
±-----------±---------±-------+
| Department | Employee | Salary |
±-----------±---------±-------+
| IT | Max | 90000 |
| Sales | Henry | 80000 |
±-----------±---------±-------+

SELECTdep1. NAME AS Department,emp1. NAME AS Employee,emp1.Salary AS Salary
FROMEmployee emp1
LEFT JOIN Department dep1 ON emp1.DepartmentId = dep1.id
WHERE (emp1.Salary ,emp1.DepartmentId)in (SELECT  MAX(emp.Salary),emp.DepartmentId FROM Employee emp  GROUP BY emp.DepartmentId) AND dep1.id != '' ;

185. 部门工资前三高的所有员工

Create table If Not Exists Employee (Id int, Name varchar(255), Salary int, DepartmentId int);
Create table If Not Exists Department (Id int, Name varchar(255));
Truncate table Employee;
insert into Employee (Id, Name, Salary, DepartmentId) values ('1', 'Joe', '85000', '1');
insert into Employee (Id, Name, Salary, DepartmentId) values ('2', 'Henry', '80000', '2');
insert into Employee (Id, Name, Salary, DepartmentId) values ('3', 'Sam', '60000', '2');
insert into Employee (Id, Name, Salary, DepartmentId) values ('4', 'Max', '90000', '1');
insert into Employee (Id, Name, Salary, DepartmentId) values ('5', 'Janet', '69000', '1');
insert into Employee (Id, Name, Salary, DepartmentId) values ('6', 'Randy', '85000', '1');
insert into Employee (Id, Name, Salary, DepartmentId) values ('7', 'Will', '70000', '1');
Truncate table Department;
insert into Department (Id, Name) values ('1', 'IT');
insert into Department (Id, Name) values ('2', 'Sales');

Employee 表包含所有员工信息，每个员工有其对应的工号 Id，姓名 Name，工资 Salary 和部门编号 DepartmentId 。
±—±------±-------±-------------+
| Id | Name | Salary | DepartmentId |
±—±------±-------±-------------+
| 1 | Joe | 85000 | 1 |
| 2 | Henry | 80000 | 2 |
| 3 | Sam | 60000 | 2 |
| 4 | Max | 90000 | 1 |
| 5 | Janet | 69000 | 1 |
| 6 | Randy | 85000 | 1 |
| 7 | Will | 70000 | 1 |
±—±------±-------±-------------+
Department 表包含公司所有部门的信息。
±—±---------+
| Id | Name |
±—±---------+
| 1 | IT |
| 2 | Sales |
±—±---------+
编写一个 SQL 查询，找出每个部门获得前三高工资的所有员工。例如，根据上述给定的表，查询结果应返回：
±-----------±---------±-------+
| Department | Employee | Salary |
±-----------±---------±-------+
| IT | Max | 90000 |
| IT | Randy | 85000 |
| IT | Joe | 85000 |
| IT | Will | 70000 |
| Sales | Henry | 80000 |
| Sales | Sam | 60000 |
±-----------±---------±-------+
解释：
IT 部门中，Max 获得了最高的工资，Randy 和 Joe 都拿到了第二高的工资，Will 的工资排第三。销售部门（Sales）只有两名员工，Henry 的工资最高，Sam 的工资排第二。