for..of会遍历可迭代的对象,调用对象上的Symbol.iterator方法(可迭代对象,数组,字符串等) let arr = ["hello", "ts", "test"]; for (let item of arr) { console.log(item) } //"hello", "ts", "test" for...of VS for...in(均可迭代一个列表.但是用于迭代

转:http://www.itmian4.com/thread-6504-1-1.html 最小区间原题 k个有序的数组,找到最小的区间范围使得这k个数组中,每个数组至少有一个数字在这个区间范围内.比如: 数组1:[4, 10, 15, 24, 26] 数组2:[0, 9, 12, 20] 数组3:[5, 18, 22, 30] 最小的区间是[20, 24],这个区间包含了数组1中的24,数组2中的20,数组3中的22 思考时间~~~ 分析 该题看起来还算比较简单,大家通常都会想到:为每一个数组

目录 1.Python计数器Counter 2.Python有序字典OrderredDict 3.Python默认字典default 4.python可命名元组namedtuple 5.Python双向队列deque 6.Python单向队列deque 7.Python深浅拷贝原理 8.Python深浅拷贝应用 9.python函数的基本定义(open.lambda.递归.装饰器) collections系列

一,利用网站浏览器F12键,利用谷歌浏览器插件找到视频的.m3u8文件,并打开. 二,打开m3u8文件后,里面有很多.ts的链接,和key的链接. 三,保存为html文件,下载ts文件,代码如下:可加多线程,可能需要用代理. # 爬虫 123.html就是打开m3u8文件右键保存为html格式. htmlf=open('./123.html','r',encoding="utf-8") htmlcont=htmlf.read() # print(htmlcont) import req

不要让自己的上限成为你的底线 前言 诈尸更新系列,为了跟上脚步,尝试了vue3,在学习vue3的时候顺便学习一手electron和ts,本教程将分别发布,源码会在最后的文章发布.因为还在开发中,目前也是为了防止有些遇到坑不会及时忘掉,所以先整理一部分出来 将分2部分发出教程,因为配置的东西个人感觉有点多并且跟开发内容相关的东西并不是很多,因此单独发出,见谅,第二篇这两天发出,长文警告!. 开发思路: 页面: 列表页index.vue 头部.搜索.内容部分,只能有一个列表页存在 设置页settin

本文版权归博客园和作者吴双本人共同所有,博客园蜗牛NoSql系列分享 http://www.cnblogs.com/tdws/tag/NoSql/ Sorted Set 有序集合—Sorted Set是我最后要分享的一种Redis数据类型,有序集合在集合的基础上,为每一个元素,关联了一个分数的概念(Score).不要忘了,集合中元素是唯一的,但是有序集合中的Score是可以相同的. 下面通过命令的介绍,来帮助大家认识有序集合. 添加命令,ZADD  key score member score

Given a n x n matrix where each of the rows and columns are sorted in ascending order, find the kth smallest element in the matrix. Note that it is the kth smallest element in the sorted order, not the kth distinct element. Example: matrix = [ [ 1, 5

Given an array of integers that is already sorted in ascending order, find two numbers such that they add up to a specific target number.The function twoSum should return indices of the two numbers such that they add up to the target, where index1 mu

Suppose a sorted array is rotated at some pivot unknown to you beforehand. (i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2). Find the minimum element. You may assume no duplicate exists in the array. 这道寻找旋转有序数组的最小值肯定不能通过直接遍历整个数组来寻找,这个方法过于简单粗暴,这样的话,旋不

Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST. 这道题是要求把有序链表转为二叉搜索树,和之前那道Convert Sorted Array to Binary Search Tree 将有序数组转为二叉搜索树思路完全一样,只不过是操作的数据类型有所差别,一个是数组,一个是链表.数组方便就方便在可以通过index直接访问任意一个元

Given an array where elements are sorted in ascending order, convert it to a height balanced BST. 这道题是要将有序数组转为二叉搜索树,所谓二叉搜索树,是一种始终满足左<根<右的特性,如果将二叉搜索树按中序遍历的话,得到的就是一个有序数组了.那么反过来,我们可以得知,根节点应该是有序数组的中间点,从中间点分开为左右两个有序数组,在分别找出其中间点作为原中间点的左右两个子节点,这不就是是二分查找法的核

Given two sorted integer arrays A and B, merge B into A as one sorted array. Note:You may assume that A has enough space (size that is greater or equal to m + n) to hold additional elements from B. The number of elements initialized in A and B are m 

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. 移除有序链表中的重复项需要定义个指针指向该链表的第一个元素,然后第一个元素和第二个元素比较,如果重复了,则删掉第二个元素,

Follow up for "Search in Rotated Sorted Array":What if duplicates are allowed? Would this affect the run-time complexity? How and why? Write a function to determine if a given target is in the array. 这道是之前那道Search in Rotated Sorted Array 在旋转有序数组

Follow up for "Remove Duplicates":What if duplicates are allowed at most twice? For example,Given sorted array A = [1,1,1,2,2,3], Your function should return length = 5, and A is now [1,1,2,2,3]. 这道题是之前那道Remove Duplicates from Sorted Array 有序数组中

Suppose a sorted array is rotated at some pivot unknown to you beforehand. (i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2). You are given a target value to search. If found in the array return its index, otherwise return -1. You may assume no duplic

Given a sorted array, remove the duplicates in place such that each element appear only once and return the new length. Do not allocate extra space for another array, you must do this in place with constant memory. For example,Given input array A = [

Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity. 这道题让我们合并k个有序链表,之前我们做过一道Merge Two Sorted Lists 混合插入有序链表,是混合插入两个有序链表.这道题增加了难度,变成合并k个有序链表了,但是不管合并几个,基本还是要两两合并.那么我们首先考虑的方法是能不能利用之前那道题的解法来解答此题.答案是肯定的,但是需要修改

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. 这道混合插入有序链表和我之前那篇混合插入有序数组非常的相似Merge Sorted Array,仅仅是数据结构由数组换成了链表而已,代码写起来反而更简洁.具体思想就是新建一个链表,然后比较两个链表中的元素值,把较小的

There are two sorted arrays nums1 and nums2 of size m and n respectively. Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)). 这道题让我们求两个有序数组的中位数,而且限制了时间复杂度为O(log (m+n)),看到这个时间复杂度,自然而然的想到了应该使用二分查找法来求解.但是这道题

pes : http://wenku.baidu.com/link?url=KjcA0qXqZ1bWVQTa8i1YOmygofldSQL7Pjj-zGRw1e_6_LFmVLo5DIWF0SNwVns-vao8r02kUwoYXUglzYZTeoIZxa2rc9QmtITJxMeGj4m TS协议解析第一部分(PAT) TS协议解析第二部分(PMT) TS协议解析第三部分(PES)