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属性名重名，用表名区分

select dept_manager.dept_no, dept_manager.emp_no, salaries.salary from dept_manager, salaries where salaries.emp_no=dept_manager.emp_no and dept_manag...

2020-06-27

2 813

有歧义的部分，必须指定表名

select salaries.emp_no, salary, salaries.from_date, salaries.to_date, dept_no from salaries, dept_manager where dept_manager.to_date='9999-01-01' and ...

2020-06-26

8 1128

select 表.属性（命名空间：重名管理）

select distinct employees.emp_no, salaries.salary from employees, salaries where employees.emp_no=salaries.emp_no and employees.hire_date between sala...

2020-06-26

1 645

员工的上司的编号

select dept_emp.emp_no as emp_no, dept_manager.emp_no as manager_no from dept_emp, dept_manager where dept_emp.dept_no=dept_manager.dept_no and dept_e...

2020-06-26

10 2219

工资表，项目表，每个项目平均工资

select title, avg(salary) from salaries, titles where salaries.to_date='9999-01-01' and titles.to_date='9999-01-01' and salaries.emp_no=titles.emp_no ...

2020-06-26

6 1881

版本号大小

import java.util.*; public class Main{ public static void main(String[] args) { Scanner sc = new Scanner(System.in); String[] v = ...

2020-06-26

3 663

连线游戏

正n边形有n条边，首个点连除自己和2个邻居以外的n-3个点： import java.util.*; public class Main{ public static void main(String[] args) { Scanner sc = new Scanner(Sy...

2020-06-24

5 575

扑克牌顺子

import java.util.*; public class Solution { public boolean isContinuous(int [] numbers) { if(numbers.length == 0) return false; Li...

2020-06-17

1 714

斐波那契通项公式

O(1)通项公式解法： public class Solution { public int Fibonacci(int n) { double a = Math.sqrt(5.0); double temp1 = (1.0 + a)/2.0; ...

2020-06-12

1 547

矩阵中的路径DFS

class Solution { public boolean hasPath(char[] matrix, int rows, int cols, char[] str){ char[][] mat = new char[rows][cols]; for(i...

2020-06-10

1 793