void Worker::listen(void) 用于实例化Worker后执行监听。 此方法主要用于在Worker进程启动后动态创建新的Worker实例,能够实现同一个进程监听多个端口,支持多种协议。 例如一个http Worker启动后实例化一个websocket Worker,那么这个进程即能通过http协议访问,又能通过websocket协议访问。由于websocket Worker和ht
List接口概述 有序的 collection(也称为序列/线性表)。此接口的用户可以对列表中每个元素的插入位置进行精确地控制。用户可以根据元素的整数索引(在列表中的位置)访问元素,并搜索列表中的元素。与 set 不同,列表通常允许重复的元素。 List接口是Collection接口的一个子接口,List集合的特性是:有序,可重复,元素有索引,List接口有三个实现类 ArrayList:底层数据
介绍 瀑布流滚动加载,用于展示长列表,当列表即将滚动到底部时,会触发事件并加载更多列表项。 引入 import { createApp } from 'vue'; import { List } from 'vant'; const app = createApp(); app.use(List); 代码演示 基础用法 List 组件通过 loading 和 finished 两个变量控制加载
list|列表标签: 标签名称:list 功能说明:表示列表模板里的分页内容列表 适用范围:仅列表模板 list_*.htm 基本语法: {dede:list col='' titlelen='' infolen='' imgwidth='' imgheight='' listtype='' orderby='' pagesize='' orderway=''}{/dede:list} 参数说明:
lset key index value 成功返回1,key或者下标不存在返回错误
A linked list is given such that each node contains an additional random pointer which could point to any node in the list or null. Return a deep copy of the list. 这题要求深拷贝一个带有random指针的链表random可能指向空,也可
You are given two linked lists representing two non-negative numbers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a link
Given a linked list and a value x, partition it such that all nodes less than x come before nodes greater than or equal to x. You should preserve the original relative order of the nodes in each of th
Given a singly linked list L: L0→L1→…→Ln-1→Ln, reorder it to: L0→Ln→L1→Ln-1→L2→Ln-2→… You must do this in-place without altering the nodes’ values. For example, Given {1,2,3,4}, reorder it to {1,4,2,3
Given a list, rotate the list to the right by k places, where k is non-negative. For example: Given 1->2->3->4->5->NULL and k = 2, return 4->5->1->2->3->NULL. 这题要求把链表后面k个节点轮转到链表前面。 对于这题我们首先需要遍历链表,得到链表
Sort a linked list in O(n log n) time using constant space complexity. 这题要求我们对链表进行排序,我们可以使用divide and conquer的方式,依次递归的对链表左右两半进行排序就可以了。代码如下: class Solution { public: ListNode *sortList(ListNode *hea
Given a linked list, swap every two adjacent nodes and return its head. For example, Given 1->2->3->4, you should return the list as 2->1->4->3. Your algorithm should use only constant space. You may
Reverse a linked list from position m to n. Do it in-place and in one-pass. For example: Given 1->2->3->4->5->NULL, m = 2 and n = 4, return 1->4->3->2->5->NULL. Note: Given m, n satisfy the following
Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists. 这题要求合并两个已经排好序的链表,很简单的题目,直接上代码: class Solution { public:
Given a sorted linked list, delete all duplicates such that each element appear only once. For example, Given 1->1->2, return 1->2. Given 1->1->2->3->3, return 1->2->3. 这题要求在一个有序的链表里面删除重复的元素,只保留一个,也是比