摘要： url: http://uva.onlinejudge.org/external/9/910.htmlhelpful discussion: http://acm.uva.es/board/viewtopic.php?f=34&t=16714&p=61697&hilit=910&sid=5e08f43bf5aa383425db1a424d7b794d#p61697helpful test Case:5A B E xB D C -C D A -D D B -E E E -55A B E -B D C xC D A -D D B -E E E -55A B E -B 阅读全文

摘要： 题目总体上还是不太难……只是，一个晚上都卡在第四题，做Java实现去了……up date: 最后Java实现了一个上午，终于把第四题做好了。基本思想和Solution Manual是一样的，只是我允许相同key的情况——因为我认为，即使是带着相同的key，data也有可能不同。11.2-1:11.2-2:11.2-3:For everynode, it's a linked list; therefore, by keeping the list sorted, the runningtime would change like this:OperationOriginalNew: k 阅读全文

摘要： 感觉挺难的。尤其是3.2-4， 我花了近2h做出来……还是参考了别人的解法的情况下。3.2-1：3.2-2：3.2-3：3.2-4：3.2-5：3.2-6：3.2-7:Reference:Stirling's Approximation, in Wolfman World, retrieved on 2011年3月5日14:35:08, from:http://mathworld.wolfram.com/StirlingsApproximation.htmlCLRS 2e: Exercise 3.2-4, in My Answers. BlogSpot, retrieved at 20 阅读全文

摘要： 3.1-1：3.1-2:3.1-3：It's useless because the best case don'tnormally happen.3.1-4：3.1-5:3.1-6: Same as 3.1-5;3.1-7:3.1-8: 阅读全文

摘要： 11.1-1:Perform a linear search on T, worst case is O(m)11.1-2:0 stands for no elements in the corresponding key, while 1 means the key correspond to some data.11.1-3:Assumption: besides of the key, the satellite data can also have data to be mapped to a secondary key.Thus we can consider a case of 2 阅读全文

摘要： Normal 0 false 7.8 磅 0 2 false false false EN-US ZH-CN X-NONE 最近上课讲了一个挺陌生的概念，叫做Skip List。搜索了一下，中文名称作“跳表”。写这个题目的原因：不过中文blog讲得都很浅：http://hideto.javaeye.com/blog/297625http://www.kernelchina.org/?q=node/239(刚刚baidu搜索了一下，百度文库里面有很好的介绍SkipList的文章：【1】http://wenku.baidu.com/view/7285945f804d2b160b4ec0a8.ht. 阅读全文

摘要： 5.1-1:Since we can compare between any two elements, we know the total order on the ranks of the candidates.5.1-2:RANDOM(a,b)%range = b - a;%d = the number of binary bits %range hasdofor(%i = 1 to %d):%i th bit of integer %res = rand(0,1);while(%res > %range);return (a + %res);Each time of while 阅读全文

摘要： 6.1 - 1:Define n to be the number of elements in a heap of height h, then:6.1-2:6.1-3:For any of the sub-tree of one max-heap, it maintains the basic heap property, i.e., A[parent(i)] >= A[i]. and the root of one sub-tree is the ancestor of all its nodes, therefore, the root is the largest by rec 阅读全文

摘要： (今天主要解决了昨天遗留的问题。在上课的论坛上讨论还是很有效的。12.2未结)12.2-1：[代码]12.2-5:If one BST node has two children, then its predecessor is the maximum of its leftsub-tree, and its successor is the minimum of its right sub-tree. The smallestelement has no left child, and the largest one has no right child.12.2-6:Think from 阅读全文

摘要： 我打算发一下自己对于算法导论书里面的个人答案。欢迎大家指正。因为读的是英文版，这里也用英文来发吧：）12.1 What is a binary search tree12.1-1:12.1-2:BSP Property: Let x be a node in BST, if y is a node in the left subtree of x, then:key[x]>= key[y]; if y is a node on the right subtree of x, then key[x]<=key[y].Min-Heap Property: for any node x, 阅读全文