!!!Interview

《剑指offer》二叉树中和为某一值的路径

rp=2&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking题目
月下键客·2015-09-11 16:12

SAP SD 顾问面试问题 consultant interview questionnaire

以下是我个人目前面试团队SD顾问问题的小结，希望对大家有所帮助，也可能有回答错误的地方，希望同行不吝赐教。也希望各位顾问们讲你们遇到的一些面试问题做个跟贴，方便大家共同进步，谢谢。 面试问题清单： 一、销售订单类型影响：可用性检查，交货类型，输出，信贷，定价，需求，报表，合作伙伴，文本，不完整日志，交货开票，订单开票，公司间开票，统计组，屏幕顺序组，参数文件二、行项目决定与作用：1.影响：允许的计
徐骁俊,您的CIO咨询·2015-09-10 14:00

Leetcode: N-Queens II

insteadoutputtingboardconfigurations,returnthetotalnumberofdistinctsolutions.ShowTagsShowSimilarProblemsHaveyoumetthisquestioninarealinterview
ayst123·2015-09-10 11:00

《剑指offer》从尾到头打印链表

rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking题
libin1105·2015-09-10 09:00

《剑指offer》替换空格

rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking题
libin1105·2015-09-10 09:00

《剑指offer》二维数组中的查找

rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking题
libin1105·2015-09-10 08:00

链表中环的入口结点（剑指offer）Floyd+O(1)

rp=3&ru=/ta/coding-interviews&q
u010579068·2015-09-09 13:00

链表中倒数第k个结点（剑指offer）

rp=1&ru=/ta/coding-interviews&qru=
u010579068·2015-09-07 20:00

反转链表（剑指offer）

rp=1&ru=/ta/coding-interviews&qru=
u010579068·2015-09-07 11:00

hiho一下 第六十一周 题目1 : Combination Lock 线段树 成段更新

时间限制:10000ms单点时限:1000ms内存限制:256MB描述Finally,youcometotheinterviewroom.YouknowthataMicrosoftinterviewerisintheroomthoughthedoorislocked.Thereisacombinationlockonthedoor.ThereareNrotatorsonthelock
u010422038·2015-09-04 09:00

求第n个丑数

rp=2&ru=/ta/coding-interviews已通过测试，第一种超时，所以细节错误不确定。
u010668907·2015-08-31 12:00

2nd Round Google Interview

    1600AmphitheaterPwky.,MountainView,CA94043               backordered                Nexus5         HelloKittyPhoneCase                        delivered       …          …     … …publicabstractclas
virgoboy2004·2015-08-28 05:00

招聘一个靠谱的 iOS(下)

博客地址：https://github.com/ChenYilong/iOSInterviewQuestions/blob/master/01《招聘一个靠谱的iOS》面试题参考答案/《招聘一个靠谱的iOS
sinat_27706697·2015-08-24 14:00

招聘一个靠谱的 iOS(笔试面试)

问题：http://blog.sunnyxx.com/2015/07/04/ios-interview/答案：https://github.com/ChenYilong/iOSInterviewQuestions
jijiji000111·2015-08-24 09:00

CSR公布蓝牙“大杀器” CSR Mesh技术细节，号称要“干掉Zigbee”

原文链接：http://www.eetrend.com/interview/100051279作者张国斌“CSRMesh技术由CSR剑桥大学研发团队打造，利用星型网络和中继技术，每个网络可以连接超过65000
djyos·2015-08-22 10:44

To new is C++; To malloc is C; To mix them is sin （混淆C++中的new和C中的malloc是一种犯罪）

IntroductionOneofthemostcommonquestionsthatgetaskedduringinterviewsforC++programmersistoexplainthedifferencesbetweenusingmallocandusingnew.It
wangshubo1989·2015-08-19 10:00

First Round of Google Phone screen interview backup

qeustion1:Writeaprogramthatreadsinatextfile,andonlyoutputstheuniquelines:(can’tuseunixuniqcommand)Input:ABACOutput:ABCpublicvoidreadFile(Filefile){ if(file==null) return; BufferReaderreader=null; File
virgoboy2004·2015-08-19 04:00

8.14_end

thefirstinterview事件委托（ul、li)jsonp原理实现印象最深的项目each的实现ajax的实现性能优化的方法判断Function和Boolean印象最深的项目thesecondinterview
青青flye·2015-08-14 17:00

Interview questions

阅读更多yx:1、hashmap原理，如何存取，为什么初始化为16，以2的倍数扩容。2、读写锁的原理。jd:1、TreeMap实现原理？（红黑树）2、ConcurrentHashMap原理？3、Mapkey的计算原理？扩展因子？4、实时计算（STORM）？5、线程池场景：coreThread=4,50个max,5个有界队列，启动过程是什么样的？6、线程池的拒绝策略有什么？7、外层设置事务，内层函数
zhb8015·2015-08-14 13:00

Interview questions

yx:1、hashmap原理，如何存取，为什么初始化为16，以2的倍数扩容。2、读写锁的原理。 jd:1、TreeMap实现原理？（红黑树）2、ConcurrentHashMap原理？3、Mapkey的计算原理？扩展因子？4、实时计算（STORM）？5、线程池场景：coreThread=4,50个max,5个有界队列，启动过程是什么样的？6、线程池的拒绝策略有什么？7、外层设置事务，内层函数是不是
zhb8015·2015-08-14 13:00

读《编程之美》励志篇

我读《编程之美》主要关注点不是面试经验，而是想了解一下对于编程，技术公司的interviewer们认为的核心点在哪里？他们是如何利用编程来考察一个人的编程素养的。
LG1259156776·2015-08-12 18:00

Node.js 常见面试题

http://www.liuzhixiang.com/2015/07/01/Node.js-Interview-Questions-and-Answers/By刘志祥前言打开这篇文章时，你要么想要聘请Node.js
salmonellavaccine·2015-08-12 13:41

LintCode - Sort Colors II

sortthemsothatobjectsofthesamecolorareadjacent,withthecolorsintheorder1,2,...k.Haveyoumetthisquestioninarealinterview
yuanhsh·2015-08-08 14:00

LintCode - Sort Colors II

Haveyoumetthisquestioninarealinterview? YesExampleGiven
yuanhsh·2015-08-08 14:00

Linkedin Interview - Paint House with Colors

There are a row of houses, each house can be painted with three colors red, blue and green. The cost of painting each house with a certain color is different. You have to paint all the houses such tha
yuanhsh·2015-08-06 03:00

Linkedin Interview - Isomorphic Strings

Given two strings, determine if they are isomorphic. Two words are called isomorphic if the letters in one word can be remapped to get the second word. Remapping a letter means replacing all occu
yuanhsh·2015-08-06 03:00

Linkedin Interview - Paint House with Colors

阅读更多Therearearowofhouses,eachhousecanbepaintedwiththreecolorsred,blueandgreen.Thecostofpaintingeachhousewithacertaincolorisdifferent.Youhavetopaintallthehousessuchthatnotwoadjacenthouseshavethesamecol
yuanhsh·2015-08-04 00:00

Linkedin Interview - Isomorphic Strings

阅读更多Giventwostrings,determineiftheyareisomorphic.Twowordsarecalledisomorphicifthelettersinonewordcanberemappedtogetthesecondword.Remappingalettermeansreplacingalloccurrencesofitwithanotherletterwhilet
yuanhsh·2015-08-03 23:00

Google Interview - 警察到房间的最短距离

一个 n x n 矩阵，每个房间可能是封闭的房间，可能是警察，可能是开的房间， 封闭的房间不能过，返回一个n x n矩阵，每一个元素是最近的警察到这个房间的最短距离。 初始矩阵中-1代表封闭房间，INT_MAX代表普通房间，0代表有警察的房间。   Solution: 把警察都找出来，然后一起push到BFS的queue里面，同时搜索。复杂度可降为O(n^2)。 Starting
yuanhsh·2015-08-01 12:00

Google Interview - 警察到房间的最短距离

阅读更多一个nxn矩阵，每个房间可能是封闭的房间，可能是警察，可能是开的房间，封闭的房间不能过，返回一个nxn矩阵，每一个元素是最近的警察到这个房间的最短距离。初始矩阵中-1代表封闭房间，INT_MAX代表普通房间，0代表有警察的房间。Solution:把警察都找出来，然后一起push到BFS的queue里面，同时搜索。复杂度可降为O(n^2)。Startingfrompoliceandoverr
yuanhsh·2015-08-01 12:00

Google Interview - Flip Game

算法游戏，给一个只有＋－两种字符的array，两个玩家，轮到某个玩家他可以任选 两个连续的＋＋将他们变成－－，如果某个玩家发现对方无法行动则赢得游戏，要求写 isWin（String s）判断先行动的玩家能否赢。   Followup 如何优化，时间上和空间上。 public boolean canWin(char[] s) { int start = -1;
yuanhsh·2015-07-30 14:00

Google Interview - 判断任意两个人是否有血缘关系

Given a family tree for a few generations for the entire population and two people, write a routine that will find out if they are blood related. Siblings are blood related since they have the same pa
yuanhsh·2015-07-29 13:00

Google Interview - Rearrange a string so that all same characters become d dista

Given a string and a positive integer d. Some characters may be repeated in the given string. Rearrange characters of the given string such that the same characters become d distance away from ea
yuanhsh·2015-07-29 11:00

Google Interview - Perimeter of region that has the same color of that point

given grid of colors, coordinate of a point and its color, find the perimeter of the region that has the same color of that point. BFS或DFS，构成perimeter的条件是只要上下左右有一个不是同颜色或者是out of bound 用一个set记录visit的信
yuanhsh·2015-07-29 09:00

Google Interview - Check String Order

given an order string "abc" check if "aabdccd" maintain the order "aabdccd" -> true; "abbca" -> false; note:order does not contain all chars in s &nbs
yuanhsh·2015-07-29 09:00

Linkedin Interview - Shortest distance between two words

  /* This class will be given a list of words (such as might be tokenized * from a paragraph of text), and will provide a method that takes two * words and returns the shortest distance (in
yuanhsh·2015-07-26 14:00

Airbnb Interview - Nested Integer List Parser

实现一个mini parser, 输入是以下格式的string:"324" or"[123,456,[788,799,833],[[]],10,[]]"要求输出:324 or [123,456,[788,799,833],[[]],10,[]].也就是将字符串转换成对应的格式的数据. 输入一个数组的字符串, 要返回一个数组, 里面每一个元素是要么一个整数,
yuanhsh·2015-07-26 13:00

Hedvig Interview - Count of number with no consecutive 1's

Given integer n, find the number of different binary strings of size n where there are no consecutive 1's.Eg: If n is 2, valid strings are 00, 01, 10. You should return 3.   假设f(n, k)是第n位上为k的数字
yuanhsh·2015-07-24 12:00

史上最全 前端开发面试问题及答案整理 http://segmentfault.com/blog/trigkit4/1190000002562454

https://github.com/hawx1993/Front-end-Interview-questions本文旨在加深对前端知识点的理解，资料来源于网络，由本人（博客：http://segmentfault.com
forlong401·2015-07-22 19:00

Hedvig Interview - Count of number with no consecutive 1's

Given integer n, find the number of different binary strings of size n where there are no consecutive 1's.Eg: If n is 2, valid strings are 00, 01, 10. You should return 3.   假设f(n, k)是第n位上为k的数字
yuanhsh·2015-07-22 14:00

Google Interview - Compress String to i18n

i18n (where 18 stands for the number of letters between the first i and the last n in the word “internationalization,”) Wiki it.    Generate all such possible i18n strings for any given st
yuanhsh·2015-07-20 11:00

Google Interview - Compress String to i18n

i18n(where18standsforthenumberoflettersbetweenthefirstiandthelastnintheword“internationalization,”)Wikiit.  Generateallsuchpossiblei18nstringsforanygivenstring.foreg."careercup"=>"c7p","ca6p","c6up","
yuanhsh·2015-07-20 11:00

Google Interview - Peek Iterator

Suppose you have a Iterator class with has_next() and get_next() methods. Please design and implement a PeekIterator class as a wrapper of Iterator and provide
yuanhsh·2015-07-20 10:00

Google Interview - Peek Iterator

Supposeyouhavea Iterator classwith has_next() and get_next() methods.Pleasedesignandimplementa PeekIterator classasawrapperofIteratorandprovidea peek() method.Whencalling peek(),theuserwillonlygetthec
yuanhsh·2015-07-20 10:00

Google Interview - Zigzag Iterator

Suppose you have a Iterator class with has_next() and get_next() methods. Please design and implement a ZigzagIterator class as a wrapper of two iterators. F
yuanhsh·2015-07-20 10:00

Google Interview - Zigzag Iterator

Supposeyouhavea Iterator classwith has_next() and get_next() methods.Pleasedesignandimplementa ZigzagIterator classasawrapperoftwoiterators.Forexample,giventwoiterators:i0= [1,2,3,4]i1= [5,6]ZigzagIte
yuanhsh·2015-07-20 10:00

程序员面试资源集锦

资源一：《crackthecodeinterview》——谷歌资深技术面试官经典之作本书的中文目录如下，大部分内容由Hawstein君原创翻译，部分缺失的由快课网Jay13补充。
蛀虫·2015-07-19 10:00

程序员面试资源集锦

资源一：《crackthecodeinterview》――谷歌资深技术面试官经典之作本书的中文目录如下，大部分内容由Hawstein君原创翻译，部分缺失的由快课网Jay13补充。
快课教程·2015-07-19 10:17

程序员面试资源集锦

资源一：《crackthecodeinterview》――谷歌资深技术面试官经典之作本书的中文目录如下，大部分内容由Hawstein君原创翻译，部分缺失的由快课网Jay13补充。
快课教程·2015-07-19 10:17

spring面试题

转载自：http://ifeve.com/spring-interview-questions-and-answers/Spring概述1.什么是spring?
竹叶苿·2015-07-17 10:06
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