I love programming

A  binary search tree  ( BST ), which may sometimes also be called an  ordered  or  sorted binary tree , is a  node-based   binary tree   data structure  which has the following properties: The left  subtree  of a node contains only nodes with keys less than the node's key. The right subtree of a node contains only nodes with keys greater than or equal to the node's key. Both the left and right subtrees must also be binary search trees.  Searching Searching a binary search tree for a specific value can be a  recursive  or  iterative  process.  bool BinarySearchTree :: search ( int val ) { Node * next = this - > root ( ) ; while ( next ! = NULL ) { if ( val == next - > value ( ) ) { return true ; } else if ( val < next - > value ( ) ) { next = next - > left ( ) ; } else { nex...

When using xCode, it is always fun to tap 'ES" button and use the feature of code sense. How about Eclipse? by default, you have to type 'Ctrl+Space‘， which is a very good way to interrupt your coding feeling. As you have to move your left hand back to touch the keys. By using ESC key, you just need to reach your little finger, Done! It's a very nice invention from apple. To do it in Eclipse, you only need to change the hot key to 'ESC' and you then can enjoy this great feature.

 source: http://www.algolist.net/Algorithms/Sorting/Quicksort. The divide-and-conquer strategy is used in quicksort. Below the recursion step is described: 1st: Choose a pivot value. We take the value of the middle element as pivot value, but it can be any value(e.g. some people would like to pick the first element and do the exchange in the end) 2nd: Partition. Rearrange elements in such a way, that all elements which are lesser than the pivot go to the left part of the array and all elements greater than the pivot, go to the right part of the array. Values equal to the pivot can stay in any part of the array. Apply quicksort algorithm recursively to the left and the right parts - the previous pivot element excluded! Partition algorithm in detail: There are two indices i and j and at the very beginning of the partition algorithm i points to the first element in the array and j points to the last one. Then algorithm moves i forward, until an element with value greater or e...

public class MergeArrays { public static void main(String args[]) { int[] a1 = new int[] { 1, 3, 5, 7, 9,23 }; int[] a2 = new int[] { 2, 4, 6,10,21 }; int[] c = mergeArrays(a1, a2); for (int i : c) System.out.println(" " + i); } public static int[] mergeArrays(int[] a1, int[] a2) { int len = a1.length + a2.length; int[] newA = new int[len]; int indexA1 = 0; int indexA2 = 0; int indexNew = 0; while (indexA1 < a1.length && indexA2 < a2.length) { if (a1[indexA1] <= a2[indexA2]) { newA[indexNew] = a1[indexA1]; indexA1++; } else { newA[indexNew] = a2[indexA2]; indexA2++; } indexNew++; } while(indexA1 <a1.length) { newA[indexNew] = a1[indexA1]; indexNew++; indexA1++; } while(indexA2 <a2.length) { newA[indexNew] = a2[indexA2++]; indexNew++; indexA2++; } return newA; }

public class MergeSort { private int[] numbers; private int count; public static void main(String[] args) { int[] a1 = new int[] { 3, 10, 5, 1, 9, 20, 11,8,15 }; MergeSort ms = new MergeSort(a1); ms.mergesort(0, ms.count - 1); for (int i : a1) System.out.println(" " + i); } public MergeSort(int[] a) { numbers = a; count = a.length; } private void mergesort(int low, int high) { if (low < high) { int middle = (low + high) / 2; mergesort(low, middle); mergesort(middle + 1, high); merge(low, middle, high); } } private void merge(int low, int middle, int high) { int helper[] = new int[high+1]; for (int i = low; i <= high; i++) { helper[i] = numbers[i]; } int i = low; int j = middle + 1; int k = low; while (i <= middle && j <= high) { if (helper[i] <= helper[j]) { numbers[k] = helper[i]; i++; } else { numbers[k] = helper[j]; ...

A  binary heap  is a set of nodes with keys arranged in a complete heap-ordered binary tree A binary tree is  heap-ordered  if the key in each node is larger than (or equal to) the keys in that nodes two children (if any). Rule:  In a heap, the parent of the node in position k is in position k/2; and, conversely, the two children of the node in position k are in positions 2k and 2k + 1. We can travel up and down by doing simple arithmetic on array indices: to move up the tree from a[k] we set k to k/2; to move down the tree we set k to 2*k or 2*k+1. Top-down heapify (sink).  If the heap order is violated because a node's key becomes smaller than one or both of that node's children's keys, then we can make progress toward fixing the violation by exchanging the node with the larger of its two children. This switch may cause a violation at the child; we fix that violation in the same way, and so forth, moving down the heap until we reach a node with ...

We did some web service testing based on SQL Server, they all worked well. However, after move to oracle, they all failed. After adding below codes,  they all works. _context.ContextOptions.LazyLoadingEnabled = false; Notice the reason could be entity framework has different behavior. In SQL Server, eager loading by default. In Oracle, Lazy loading by default. Sigh :(

For whatever reason, my Visual Studio.NET 2010 crashed very often. I found the best way of resolving this issue is reset the environment. Steps: step 1: go to folder; c:\Program Files (x86)\Microsoft Visual Studio 10.0\VC step 2: issue below command: Devenv.exe /ResetSettings

After installing the ORACLE ODP.NET, I met THE ORA-12154:tns: cound not resolve the connect identifier specified MESSAGE. To resove this,  need to config the connection details(you can get the dialog through visual studio - > tools -> connect to database) correctly. Data source:  Oracle Database(Oracle ODP.NET) Connection details, Data source name: localhost/xe   comments: the format is machine name or ip/sid  , no space between them user name: oracle user name, should have create session  privilege password: the password for that user

After created user sxu, I got below error when tried to connect the database by using Oracle SQL Developer. command: create user sxu identified by sxu; error:  ORA-01045:user name lacks CREATE SESSION privilege; logon denied The Solution For this Error Is Simple : sqlplus / as sysdba grant create session to "USER-NAME"; Done

When using XML-based configuration metadata, you use the 'id' and/or 'name'attributes to specify the object identifier(s). The 'id' attribute allows you to specify exactly one id, and because it is a real XML element ID attribute, the XML parser is able to do some extra validation when other elements reference the id. As such, it  is the preferred way to specify an object id. However, the XML specification does limit the characters which are legal in XML IDs. This is usually not a constraint, but if you have a need to use one of these special XML  characters, or want to introduce other aliases to the object

Our new project will use Spring.net 2.0, hence, I started learning how to use it. Need to create a prototype implementation for team. How to install it? Like other .net library, you just copy the dll files (Spring.Services.dll, Spring.Aop.dll and Spring.Core, Common.Logging.dll and the counterpart xmls files to  to your project and do the configuration. How to use it? Take Spring Integration with WCF(Windows Communication Foundation) as an example, Step 1: define the Services  namespace Spring.WcfQuickStart {     [Serializable]     public class BinaryOperationArgs     {         private double x;         private double y;         public BinaryOperationArgs(double x, double y)         {             this.x = x;             this.y = y;         }       ...