1.数据 --创建职员表create table tbEmp( eID number primary key, --职员编号 eName varchar2(20) not null, --职员姓名 eSex varchar2(2) not null --职员性别 check(esex in ('男','女')), eAge number not null check(eage>=18), --职员年龄 eAddr varchar2(50) not null, --职员地址 eTel varcha
本文是受网文 <一次非常有意思的SQL优化经历:从30248.271s到0.001s>启发而产生的. 网文没讲创建表的数据过程,我帮他给出. 创建科目表及数据: CREATE TABLE tb_course ( id NUMBER not null primary key, name NVARCHAR2(10) not null ) Insert into tb_course select rownum,dbms_random.string('*',dbms_random.value(6,10
最近被算法虐了一下,刷一下leetcode,找找存在感 如题: Given an array of integers, return indices of the two numbers such that they add up to a specific target. You may assume that each input would have exactly one solution, and you may not use the same element twice. Exam
可以使用如下sql语句: select t1.username,t1.logon_time last_logon_time,t2.account_status,created 账号创建时间 from (select username,max(timestamp) logon_time from dba_audit_session where action_name='LOGON' and username in (select username from dba_users) group by
EMP表是Oracle测试账户SCOTT中的一张雇员表,首先,我们来看看emp表的数据 SQL> select * from emp; EMPNO ENAME JOB MGR HIREDATE SAL COMM DEPTNO ----- ---------- --------- ---------- --------- ---------- ---------- ---------- SMITH CLERK ALLEN SALESMAN WARD SALESMAN JONES MANAGER M
第二课主要介绍第一课余下的BFPRT算法和第二课部分内容 1.BFPRT算法详解与应用 找到第K小或者第K大的数. 普通做法:先通过堆排序然后取,是n*logn的代价. // O(N*logK) public static int[] getMinKNumsByHeap(int[] arr, int k) { if (k < 1 || k > arr.length) { return arr; } int[] kHeap = new int[k];//存放第k小的数 for (int i =