Algorithms Illuminated pdf epub mobi txt 電子書下載 2024

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Tim Roughgarden

Soundlikeyourself Publishing

2017-9-27

218

USD 17.99

Paperback

Algorithms Illuminated

9780999282908

圖書標籤: Algorithm 算法計算機科學計算機 2017 美國工具教程 CS

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发表于2024-11-22

Algorithms Illuminated epub 下載 mobi 下載 pdf 下載 txt 電子書下載 2024

Algorithms Illuminated pdf epub mobi txt 電子書下載 2024

圖書描述

Algorithms are the heart and soul of computer science. Their applications range from network routing and computational genomics to public-key cryptography and machine learning. Studying algorithms can make you a better programmer, a clearer thinker, and a master of technical interviews. Algorithms Illuminated is an accessible introduction to the subject for anyone with at least a little programming experience. The exposition emphasizes the big picture and conceptual understanding over low-level implementation and mathematical details---like a transcript of what an expert algorithms tutor would say over a series of one-on-one lessons. Part 1 covers asymptotic analysis and big-O notation, divide-and-conquer algorithms and the master method, randomized algorithms, and several famous algorithms for sorting and selection.

Algorithms Illuminated 下載 mobi epub pdf txt 電子書

著者簡介

Tim Roughgarden is a Professor of Computer Science and (by courtesy) Management Science and Engineering at Stanford University, where he has taught and researched algorithms since 2004. This book is the first of a four-part series based on his online algorithms courses that have been running regularly since 2012.

圖書目錄

Algorithms Illuminated pdf epub mobi txt 電子書下載

想要找書就要到小哈圖書下載中心

qciss.net

立刻按 ctrl+D收藏本頁

你會得到大驚喜!!

用戶評價

評分☆☆☆☆☆

非常細緻的書，算法證明和靈感來源都有

評分☆☆☆☆☆

Tim老師跟著自己在coursera上algorithms的specialization寫的書。沒有CLRS覆蓋的內容多，但是我覺得講解的更清晰更細緻，尤其是裏麵涉及瞭很多code時候需要注意的細節。不過這個感受也可能存在bias，因為這是第二遍深入復習和dive的時候的感受。

評分☆☆☆☆☆

非常細緻的書，算法證明和靈感來源都有

評分☆☆☆☆☆

非常細緻的書，算法證明和靈感來源都有

讀後感

評分☆☆☆☆☆

原书是这么写的，但是经过反复验证，c=a*a，而不是c=b*b。所以这里应该是作者笔误写错了。顺便贴一下js的实现： /* * 快速n方计算 * */ function fastPower (a, b) { var c, temp; if (b === 1) { return a; } else { c = b * b; temp = fastPower(c, Math.floor(b/2)) } if (...

評分☆☆☆☆☆