11 2014 档案

Larbin初试
摘要：前阵子找工作的时候经常会看到epoll多路复用的知识点，无奈自己一点都不懂。慌忙之际也只能去了解个大概。所以最近闲下来之后想要基于epoll机制实现一个比较有用的东西，刚好最近又想爬些东西，希望这次能够写一个高效一点的爬虫。以前只看过一点点的nutch，自己写的就是用python的几个简单功能来爬，... 阅读全文
posted @ 2014-11-30 12:12 linyx 阅读(1743) 评论(0) 推荐(0)
vim的常用命令
摘要：平常最多是用vim来编辑单个文件，看看源码。就是写几k行代码时也没有用一些其他的插件，只是设置了高亮等一些自带的属性。这样的好处是，换到任何一台新机上都能立马使用。网上流传了大量的“vim命令合集”，“把vim打造成一个IDE”之类的文章，我觉得至少对我来说意义不大。我觉得把以下这些命令用熟的话，v... 阅读全文
posted @ 2014-11-29 18:26 linyx 阅读(12811) 评论(0) 推荐(0)
风投？
摘要：天使投资：种子期，可能只有一个概念什么都没开始，或者刚开始运营，还没有出来产品，或者出来了产品却没有大规模开卖。此时浇浇水，种子会长大，你就是一个天使。种子会长大，成长的过程之中还是有许多的风险，需要更多的呵护。于是Venture Capital就有了风险投资的名字，A轮之后就是B轮。如果你愿意还可... 阅读全文
posted @ 2014-11-28 22:40 linyx 阅读(155) 评论(0) 推荐(0)
Cocos2d-x环境搭建
摘要：资源列表官网上下载最新的cocos2d-x-3.3。安装JDK，Eclipse，CDT插件，ADT插件。下载Android SDK，更新。因为国内经常访问不了google，用vpn速度也有点慢。网上搜来的ip还可以用，改hosts文件。210.242.125.108 dl-ssl.google.co... 阅读全文
posted @ 2014-11-15 17:04 linyx 阅读(797) 评论(2) 推荐(0)
结语
摘要：找工作到此为止了。希望有好消息。其实每一轮面都有问题，我也不知道是大是小，希望有机会进入下一轮，希望走得越远。昨天在mitbbs上看到一句话深有感触：“表面上看是差一点点，或者说运气不好，实际上是毫无机会，因为比你优秀的人太多了，如果你不能表现完美无可挑剔，总是能被他们找到拒你的理由．我能做到每次都... 阅读全文
posted @ 2014-11-12 01:00 linyx 阅读(192) 评论(0) 推荐(0)
Print all nodes at distance k from a given node
摘要：Given a binary tree, a target node in the binary tree, and an integer value k, print all the nodes that are at distance k from the given target node. ... 阅读全文
posted @ 2014-11-09 22:02 linyx 阅读(343) 评论(0) 推荐(0)
Construct a tree from Inorder and Level order traversals
摘要：Given inorder and level-order traversals of a Binary Tree, construct the Binary Tree. Following is an example to illustrate the problem.BinaryTreeInpu... 阅读全文
posted @ 2014-11-09 19:59 linyx 阅读(345) 评论(0) 推荐(0)
Transform a BST to greater sum tree
摘要：Given a BST, transform it into greater sum tree where each node contains sum of all nodes greater than that node.自己想的复杂了，其实就是一个反向的inorder。新的值就是前面所有元素的... 阅读全文
posted @ 2014-11-09 14:23 linyx 阅读(163) 评论(0) 推荐(0)
Minimum no. of iterations to pass information to all nodes in the tree
摘要：Given a very large n-ary tree. Where the root node has some information which it wants to pass to all of its children down to the leaves with the cons... 阅读全文
posted @ 2014-11-09 00:47 linyx 阅读(628) 评论(0) 推荐(0)
Print a Binary Tree in Vertical Order
摘要：Given a binary tree, print it vertically. The following example illustrates vertical order traversal. 1 / \ 2 3 / \ / \ 4 5 6 7 \ \ 8 9The output of p... 阅读全文
posted @ 2014-11-08 18:43 linyx 阅读(432) 评论(0) 推荐(0)
Rearrange a string so that all same characters become d distance away
摘要：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 s... 阅读全文
posted @ 2014-11-08 14:54 linyx 阅读(277) 评论(0) 推荐(0)
Suffix array
摘要：A suffix array is a sorted array of all suffixes of a given string. The definition is similar to Suffix Tree which is compressed trie of all suffixes ... 阅读全文
posted @ 2014-11-08 13:04 linyx 阅读(432) 评论(0) 推荐(0)
Graph | Eulerian path
摘要：In graph theory, a Eulerian trail (or Eulerian path) is a trail in a graph which visits every edge exactly once. Similarly, an Eulerian circuit or Eul... 阅读全文
posted @ 2014-11-07 19:53 linyx 阅读(487) 评论(0) 推荐(0)
Careercup | Chapter 8
摘要：8.2Imagine you have a call center with three levels of employees: respondent, manager, and director. An incoming telephone call must be first allocate... 阅读全文
posted @ 2014-11-04 21:38 linyx 阅读(247) 评论(0) 推荐(0)
Leetcode | Pow(x, n)
摘要：Implement pow(x, n).快速求幂，注意n是负数的情况。 1 class Solution { 2 public: 3 double pow(double x, int n) { 4 double ans = 1.0; 5 int symbol ... 阅读全文
posted @ 2014-11-04 19:45 linyx 阅读(130) 评论(0) 推荐(0)
Leetcode | Gray code
摘要：The gray code is a binary numeral system where two successive values differ in only one bit.Given a non-negative integer n representing the total numb... 阅读全文
posted @ 2014-11-04 16:27 linyx 阅读(156) 评论(0) 推荐(0)
分布式缓存
摘要：分布式缓存出于如下考虑，首先是缓存本身的水平线性扩展问题，其次是缓存大并发下的本身的性能问题，再次避免缓存的单点故障问题(多副本和副本一致性)。分布式缓存的核心技术包括首先是内存本身的管理问题，包括了内存的分配，管理和回收机制。其次是分布式管理和分布式算法，其次是缓存键值管理和路由。Memcache... 阅读全文
posted @ 2014-11-03 18:07 linyx 阅读(214) 评论(0) 推荐(0)
Read4096
摘要：Given API: int Read4096(char* buf); It reads data from a file and records the position so that the next time when it is called it read the next 4k cha... 阅读全文
posted @ 2014-11-02 15:50 linyx 阅读(504) 评论(0) 推荐(0)
graph | hungary
摘要：匈牙利算法，求二分图最大匹配。若P是图G中一条连通两个未匹配顶点的路径，并且属于M的边和不属于M的边(即已匹配和待匹配的边)在P上交替出现，则称P为相对于M的一条增广路径。（M为一个匹配）由增广路的定义可以推出下述三个结论：P的路径长度必定为奇数，第一条边和最后一条边都不属于M。所以Line 25-... 阅读全文
posted @ 2014-11-02 13:25 linyx 阅读(209) 评论(0) 推荐(0)
A Great Alchemist
摘要：Time limit : 2sec / Stack limit : 256MB / Memory limit : 256MBProblemCarol is a great alchemist.In her world, each metal has a name of 2N (N is an int... 阅读全文
posted @ 2014-11-01 23:26 linyx 阅读(292) 评论(0) 推荐(0)