随笔分类 -  算法

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What the hell is Rotate?
摘要： 阅读全文
posted @ 2014-09-21 08:51 门对夕阳 阅读(108) 评论(0) 推荐(0)
堆和堆排序
摘要：堆有两种: max-heap 和 min-heap. Max-heap 一般用来排序，Min-heap 用来实现 priority queue.max-heap的定义是：for each i: A[parent(i)] >= A[i]min-heap: for each i: A[parent(i)... 阅读全文
posted @ 2014-07-18 02:09 门对夕阳 阅读(204) 评论(0) 推荐(0)
[cc150] check palindrome of a singly linked list
摘要：Problem: Implement a function to check if a singly linked list is a palindrome.思路：最简单的方法是 Reverse and compare.另外一种非常经典的办法是用 Recursive 的思路，把一个list看成这种形... 阅读全文
posted @ 2014-07-16 00:02 门对夕阳 阅读(441) 评论(0) 推荐(0)
Detect loop in a singly linked list
摘要：去Twitter面试的被问到这个问题，当时只想到了用HashMap的办法，这种办法时间复杂度O(n)，空间复杂度是O(n), 更好的办法是用 FastRunner / SlowRunner approach。用两个pointer遍历链表，fast的速度是slow的两倍，如果有loop，二者一定会co... 阅读全文
posted @ 2014-07-15 05:45 门对夕阳 阅读(311) 评论(0) 推荐(0)
Partition an array around an interger
摘要：Partition an array of integers around a value such taht all elements less than x come before elements greater than or equal to x.Idea: Just use of sub... 阅读全文
posted @ 2014-07-15 01:53 门对夕阳 阅读(174) 评论(0) 推荐(0)
Binary search for the first element greater than target
摘要：We all know how to search through an array for an element whose value equals the target value, but how to search for the element that has value greate... 阅读全文
posted @ 2014-07-10 02:24 门对夕阳 阅读(341) 评论(0) 推荐(0)
Searching in a rotated and sorted array
摘要：Given a sorted array that has been rotated serveral times. Write code to find an element in this array. You may assume that the array was originally s... 阅读全文
posted @ 2014-07-09 09:31 门对夕阳 阅读(265) 评论(0) 推荐(0)
[DP] The 0-1 knapsack problem
摘要：Give a dynamic-programming solution to the 0-1 knapsack problem that runs inO(nW) time, where n is the number of items and W is the maximum weight ofi... 阅读全文
posted @ 2014-06-10 12:17 门对夕阳 阅读(365) 评论(0) 推荐(0)
[DP] Rod-cutting problem
摘要：给一个长度为 n 的杆子，切成小段卖出去，价格根据小段的长度不同而不同。下面是一个例子我们要通过切成小段卖出尽可能高的总价钱。问题是：How to decompose the problem?Decomposition 的第一步是：第一刀切在哪？可以切在最左边（等于整根卖出去）；可以切在位置1，位置... 阅读全文
posted @ 2014-06-10 12:13 门对夕阳 阅读(415) 评论(0) 推荐(0)
Dynamic Programming (DP) 问题总结
摘要：所有的 DP 问题都可以简单得用 Recursion 的方式实现。这通常是最容易想到的思路。问题是这种实现不够 efficient，存在 subproblem 被重复计算的情况。有两种解决这个问题的方法：1. 很直观的，在 naive recursion 里加入 一个 save 的环境，把每个 su... 阅读全文
posted @ 2014-06-10 11:06 门对夕阳 阅读(307) 评论(0) 推荐(0)
慎用递归！
摘要：其一，所有的递归实现都可以 iteratively 实现，虽然很多时候递归的代码更简洁。其二，递归会产生多余的开销，包括空间和时间。其三，如果一定要递归，不要在一个递归函数里做多件事，最好只做一件。像下面的代码（未完成，因为写不下去了），试图在递归函数里做三件事，(1) 检测路径是否存在(2) 构造... 阅读全文
posted @ 2014-06-01 06:29 门对夕阳 阅读(398) 评论(0) 推荐(0)
[Leetcode] Balanced Binary Tree
摘要：问题：给一个二叉树，写一个算法判断这个树是不是balanced。Solution #1.第一次遇到这个问题时我的解法，如下：public class Solution { public boolean isBalanced(TreeNode root) { if(root == ... 阅读全文
posted @ 2014-05-04 00:58 门对夕阳 阅读(6718) 评论(2) 推荐(1)
The Bellman-Ford algorithm
摘要：This algorithm deals with the general case, where G is a directed, weight graph, and it can contains negative edge weigths, cycles and even negative-w... 阅读全文
posted @ 2014-04-14 06:05 门对夕阳 阅读(591) 评论(0) 推荐(0)
编程之美 2014资格赛 格格取数
摘要：描述给你一个m x n (1 <= m, n <= 100)的矩阵A (0<=aij<=10000)，要求在矩阵中选择一些数，要求每一行，每一列都至少选到了一个数，使得选出的数的和尽量的小。思路：这个题用DP做不了，因为Solution的结构不具备Optimal Substructure, 比如1 100 1001 100 100100 1 1Solution是1+1+1+1 = 4，但是该矩阵的子矩阵100 1001 1的Solution是 100+1 = 101. 看来只能用DFS做了。 阅读全文
posted @ 2014-04-13 10:26 门对夕阳 阅读(1545) 评论(0) 推荐(0)
Solve Longest Path Problem in linear time
摘要：We know that the longest path problem for general case belongs to the NP-hard category, so there is no known polynomial time solution for it. However, for a special case which is directed acyclic graph, we can solve this problem in linear time.First, take a look at an algorithm that sloves the short 阅读全文
posted @ 2014-04-12 07:20 门对夕阳 阅读(350) 评论(0) 推荐(0)
Why longest path problem doesn't have optimal substructure?
摘要：We all know that the shortest path problem has optimal substructure. The reasoning is like below:Supppose we have a path p from node u to v, another node t lies on path p: u->t->v ("->" means a path).We claim that u->t is also a shortest path from u to t, and t->v is a short 阅读全文
posted @ 2014-04-12 06:45 门对夕阳 阅读(256) 评论(0) 推荐(0)
[leetcode] Reorder List
摘要：Problem:此题的做法和 “检查单链表回文” 那道题的骨架是一样的，用 Recursive 的方法，简洁而优雅。参考回文那道题：http://www.cnblogs.com/Antech/p/3847752.htmlCode in Java:public class Solution { ... 阅读全文
posted @ 2014-04-12 03:08 门对夕阳 阅读(367) 评论(0) 推荐(0)
逆序访问单链表
摘要：This is a simple but important algorithm when dealing with singly linked list.This algorithms can reversely access a singly linked list using O(n) time. Below is my code in Java.class Node{ Node next; int val;}class SinglyLinkedList{ Node head; Node tail; //append a new Node t... 阅读全文
posted @ 2014-04-11 05:28 门对夕阳 阅读(213) 评论(0) 推荐(0)
Binary Tree Traversal
摘要：Binary Tree的三种Traversal方法老是记混，故于此总结下。Preorder: Node -> Left subtree -> Right subtreeInorder: Left subtree -> Node -> Right subtreePostorder: Left subtree -> Right subtree -> NodePre, In, Post指的是Node本身在[Node, Left, Right]中被访问的顺序。Pre is 1st, In is 2nd, Post is 3rd.Note: Inorder trave 阅读全文
posted @ 2014-04-10 03:08 门对夕阳 阅读(117) 评论(0) 推荐(0)
“中文分词 关键词提取”的资料收集
摘要：Fudannlp：开源中文自然语言处理工具包|中文分词|词性标注|依存句法分析|指代消解。支持JAVA调用，和WebServices调用。ictclas:中科院研发的一个分词系统，支持的编程语言较丰富。 阅读全文
posted @ 2013-04-19 23:10 门对夕阳 阅读(233) 评论(0) 推荐(0)

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