element

一、环境、工具 jdk1.8 maven spring-boot idea VSVode vue 百度网盘(vue+springboot+nginx源码)： 链接：https://pan.baidu.com/s/1m3ciEUmUsjqoQBnIJBR7Zw 提取码：6gi9 二、搭建后台spring-boot框架 步骤： 1、new- project选择Spring Initializr next 2、创建项目文件结构以及jdk版本 3、选择项目需要的依赖 next然后finish 4、创建完毕的文件结构如图 5、对pom.xml更改，信息如下 <?xml version="1.0" encoding="UTF-8"?> <project xmlns="http://maven.apache.org/POM/4.0.0" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation ="http://maven.apache.org/POM/4.0.0 http://maven.apache.org/xsd/maven-4.0.0.xsd"> <modelVersion>4.0.0</modelVersion> <groupId>com.zks</groupId> <artifactId

下载链接： https://files.cnblogs.com/files/gshang/element.rar 效果预览，本人当前博客样式即为element主题。 注意几点： 1.主题需要去除默认css样式，记得在后台勾选； 2.默认会将markdown里的图片 [] 中的文字作为图注信息，显示在博文以及查看图片； 3.博客头像在页脚里修改，相信你会找到的； 4.请尽量使用 Markdown 去写博客，以获得更好的阅读体验。 来源： oschina 链接： https://my.oschina.net/u/4337936/blog/4270263

1 # #if语句 2 arr=[ ' hello ' , ' audi ' , ' subaru ' , ' toyota ' ] 3 4 for element in arr: 5 if element== ' audi ' : 6 print (element.upper()) # #所有字符大写 7 else : 8 print (element.title()) # #首字母大写 9 10 # #and or not 与或非 11 # #检查特定值是否不包含在列表中 12 num_arr=[num for num in range(1,21,2 )] 13 print (num_arr) 14 print ( ' 输入元素并开始判断 ' ) 15 flag= input() 16 print (type(flag)) # #刚刚才知道input()得到的值是str类型的 17 print (type(num_arr[0])) 18 if int(flag) not in num_arr: 19 print ( ' 该列表没有 ' ,flag, ' 这个元素！ ' ) 20 else : 21 print ( ' 该元素在列表中！ ' ) 22 23 # #if-elif-else 24 print ( ' 输入年龄： ' ) 25 age= int(input()) 26

★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★ ➤微信公众号：山青咏芝（shanqingyongzhi） ➤博客园地址：山青咏芝（ https://www.cnblogs.com/strengthen/ ） ➤GitHub地址： https://github.com/strengthen/LeetCode ➤原文地址： https://www.cnblogs.com/strengthen/p/10478921.html ➤如果链接不是山青咏芝的博客园地址，则可能是爬取作者的文章。 ➤原文已修改更新！强烈建议点击原文地址阅读！支持作者！支持原创！ ★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★ You have k lists of sorted integers in ascending order. Find the smallest range that includes at least one number from each of the k lists. We define the range [a,b] is smaller than range [c,d] if b-a < d-c or a < c if b-a == d-c . Example 1: Input:[[4,10,15,24,26],

工作需要做一个带滑块效果的导航栏，初步想法是用element的导航组件去做，后面摸坑结合各位大佬的博客初步实现效果，话不多说，直接上代码，记录一下爬坑之旅 1 <template> 2 <div class="y-nav"> 3 <el-row class="nav"> 4 <el- menu 5 : default -active="$route.path" 6 class="el-menu-demo" 7 mode="horizontal" 8 @select="handleSelect" 9 text-color="#fff" 10 > 11 <el-menu-item index="/index"> 12 <router- link 13 to="/" 14 @mouseover.native="itransition($event )" 15 @mouseout.native="outtransition($event)" 16 >首页</router-link> 17 </el-menu-item> 18 <el-menu-item index="/leaderboard/leaderlist"> 19 <router- link 20 :to="{name:'leaderboard'}" 21 @mouseover.native="itransition($event

问题： Is there a way in jQuery to loop through or assign to an array all of the classes that are assigned to an element? jQuery中是否有一种方法可以循环或将分配给元素的所有类分配给数组？ ex. 例如 <div class="Lorem ipsum dolor_spec sit amet">Hello World!</div> I will be looking for a "special" class as in "dolor_spec" above. 我将在上面的“ dolor_spec”中寻找“特殊”类。 I know that I could use hasClass() but the actual class name may not necessarily be known at the time. 我知道我可以使用hasClass（），但是实际的类名可能在那时不一定是已知的。 解决方案： 参考一： https://stackoom.com/question/59Gw/使用jQuery获取元素的类列表 参考二： https://oldbug.net/q/59Gw/Get-class-list-for-element-with-jQuery 来源：

Implement the following operations of a stack using queues. push(x) -- Push element x onto stack. pop() -- Removes the element on top of the stack. top() -- Get the top element. empty() -- Return whether the stack is empty. Notes: You may assume that all operations are valid (for example, no pop or top operations will be called on an empty stack). Depending on your language, queue may not be supported natively. You may simulate a queue by using a list or deque (double-ended queue), as long as you use only standard operations of a queue -- which means only push to back , pop from front , size ,

A quack is a data structure combining properties of both stacks and queues. It can be viewed as a list of elements written left to right such that three operations are possible: push(x) : add a new item x to the left end of the list pop() : remove and return the item on the left end of the list pull() : remove the item on the right end of the list. Implement a quack using three stacks and O(1) additional memory, so that the amortized time for any push, pop, or pull operation is O(1) . Your solution. 1. stack s1 for push/pop; stack s3 for pull; stack s2 is a copy of s1; 2. instance variable sz

Implement the following operations of a queue using stacks. push(x) -- Push element x to the back of queue. pop() -- Removes the element from in front of queue. peek() -- Get the front element. empty() -- Return whether the queue is empty. Notes: You must use only standard operations of a stack -- which means only push to top , peek/pop from top , size , and is empty operations are valid. Depending on your language, stack may not be supported natively. You may simulate a stack by using a list or deque (double-ended queue), as long as you use only standard operations of a stack. You may assume

实现一个单链表及相关注释 1 function LinkList() { // 单链表 2 // 节点 3 let Node = function (element) { 4 this .element = element 5 this .next = null 6 } 7 8 // 初始头节点 9 let head = null 10 11 // 链表长度 12 let length = 0 13 14 // 操作 operation 15 this .getList = function () { // 获取列表 16 return head 17 } 18 // 思路：遍历单链表，判断节点值是否等于待查找值，相等则返回 true ，否则继续遍历下一个节点，直到遍历完整个链表还未找到，则返回 false 19 this .search = function (element) { // 查找节点 20 let p = head 21 if (!p) return false // 链表为null 直接返回false 22 while (p) { // 此时p为head首节点 23 if (p.element === element) return true // 如果相等直接返回 true 24 p = p.next // 把p赋值为下一节点 25 } 26 return