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There Is More To A Baby Bag Than That
Perform this exercise two rounds.These examples of simple exercises that can be performed in a large bag. Always remember to keep your hands on traini...
分类:其他好文   时间:2014-08-08 12:06:45    阅读次数:298
amazon 面经2
http://www.geeksforgeeks.org/amazon-interview-set-108-campus/F2F-1:1) Given a sorted circular link list and a pointer to random node, now insert a new...
分类:其他好文   时间:2014-08-08 04:18:25    阅读次数:248
【LeetCode】Copy List with Random Pointer
A linked list is given such that each node contains an additional random pointer which could point to any node in the list or null. Return a deep copy of the list. 思路:第一遍正常复制链表,同时用哈希表保存链表中原始节点和新...
分类:其他好文   时间:2014-08-07 23:22:45    阅读次数:280
perl C/C++ 扩展(二)
第二讲perl 加载c/c++的库先通过h2xs 创建一个新的工程h2xs -A -n two_test进入目录cd two_test创建一个mylib文件夹,存放静态库mkdir mylibc程序头文件chen.h#includevoid chen(char *);c程序代码chen.c#incl...
分类:编程语言   时间:2014-08-07 22:42:35    阅读次数:484
Light Oj 1214 大数整除
Large Division Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu Submit Status Practice LightOJ 1214 Description Given two integers, a and b, you should check whe...
分类:其他好文   时间:2014-08-07 19:13:30    阅读次数:244
HDU-4930 Fighting the Landlords 多校训练赛斗地主
只需要判断一个回合就可以了,枚举判断可以一次出完所有牌或者大过对面的牌的可能,注意的是4张相同的牌带两张牌的话是可以被炸弹炸的。 #include #include #include #include #include #include #include #include using namespace std; struct node { int k,mk; int two...
分类:其他好文   时间:2014-08-07 19:04:10    阅读次数:258
HDU 4925 Apple Tree
Problem Description I’ve bought an orchard and decide to plant some apple trees on it. The orchard seems like an N * M two-dimensional map. In each grid, I can either plant an apple tree to get one a...
分类:移动开发   时间:2014-08-07 18:59:10    阅读次数:293
AS 7 Internal Architecture Overview--reference
High Level OverviewAt a coarse level, AS 7 consists of two main elements:A core manageable service container based on modular classloadingExtensions t...
分类:其他好文   时间:2014-08-07 18:25:10    阅读次数:262
【leetcode系列】3Sum
这个题我最开始的思路是:先一个数定下来,然后在除这个数之外的集合里面找另外两个数,最后计算和。如此反复,对于N个数,需要进行N-2次循环。 我遇到的问题就是怎么找另外两个数,其实我想过参照Two Sum里面的解法,就是用Hashtable存,键值对的结构是>,但是构造这个Hashtable就需要O(N^2),后面真正解的时候有需要O(N^2)。 参考了大牛的解法后,明白了找两个数还是用两个下标...
分类:其他好文   时间:2014-08-07 13:19:10    阅读次数:171
openebula vm无法获取IP问题解决
http://archives.opennebula.org/documentation:archives:rel2.2:congContextualizing Virtual Machines 2.2There are two contextualization mechanisms availa...
分类:其他好文   时间:2014-08-07 13:03:20    阅读次数:275
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