A complete computer science study plan to become a software engineer.
-
Updated
Dec 3, 2019
A complete computer science study plan to become a software engineer.
Learn how to design large-scale systems. Prep for the system design interview. Includes Anki flashcards.
All Algorithms implemented in Python
📝 Algorithms and data structures implemented in JavaScript with explanations and links to further readings
Everything you need to know to get the job.
The Patterns of Scalable, Reliable, and Performant Large-Scale Systems
120+ interactive Python coding interview challenges (algorithms and data structures). Includes Anki flashcards.
我是木易杨,公众号「高级前端进阶」作者,每天搞定一道前端大厂面试题,祝大家天天进步,一年后会看到不一样的自己。
⭐️ Companies that don't have a broken hiring process
System design interview for IT companies
Everything you need to prepare for your technical interview
前端面试每日 3+1,以面试题来驱动学习,提倡每日学习与思考,每天进步一点!每天早上5点纯手工发布面试题(死磕自己,愉悦大家)
📚 C/C++ 技术面试基础知识总结,包括语言、程序库、数据结构、算法、系统、网络、链接装载库等知识及面试经验、招聘、内推等信息。
The top Internet companies android interview questions and answers
A curated collection of common interview questions to help you prepare for your next interview.
🚀 fullstack tutorial 2019,后台技术栈/架构师之路/全栈开发社区,春招/秋招/校招/面试
A collection of Linux Sysadmin Test Questions and Answers. Test your knowledge and skills in different fields with these Q/A.
Adding leading zero in the question number will be an awesome feature for file explorer to sort files by the name.
Interview = 简历指南 + LeetCode + Kaggle
Data Structures and Algorithms explained and implemented in JavaScript
前端开发面试题大收集,前端面试集锦 :heart: :gift_heart: :cupid:
Preparation links and resources for system design questions
Add a description, image, and links to the interview topic page so that developers can more easily learn about it.
To associate your repository with the interview topic, visit your repo's landing page and select "manage topics."
数轴上放置了一些筹码,每个筹码的位置存在数组 chips 当中。
你可以对 任何筹码 执行下面两种操作之一(不限操作次数,0 次也可以):
将第 i 个筹码向左或者右移动 2 个单位,代价为 0。
将第 i 个筹码向左或者右移动 1 个单位,代价为 1。
最开始的时候,同一位置上也可能放着两个或者更多的筹码。
返回将所有筹码移动到同一位置(任意位置)上所需要的最小代价。
示例 1:
输入:chips = [1,2,3]
输出:1
解释:第二个筹码移动到位置三的代价是 1,第一个筹码移动到位置三的代价是 0,总代价为 1。
示例 2:
输入:chips = [2,2,2,3,3]
输出:2
解释:第四和第五个筹码移动到位置二的代价都是 1,所以最小总代价为 2。
提示:
1 <= chips.length <= 1