[Help] Binary Search
I want to do a binary search on a sorted array list.
My question is,
In the java.util package, there is a pre-written methods, Arrays.binarySearch() which I can use it to do the searching straight forward. What I concern here, if I write the binary search algorithm myself, will it be fastest?
Let's assume, there are only integers in the array, and the binary search algorithm that I am going to write in traditional way that widely used today.
Whenever there is a method available in the API we need not write the method once again. All the way whatever we write may contain some pit-falls and it may fail some where at the same it won't be that much efficient also.
Similar Messages
-
Creating A Binary search algorithm !!!! URGENT HELP
hi ..
i;m currently tryin to create a binary search algorithm ..
the user should be able to input the size of the array
and also the key that he would like to find.
it also has to have to ability to measure the run time of the algorithm.. it how long it too the algorithm to search through the array and find they key..
i have created 3 classes
the first class is the binary search class
which is the mathamatical side of things
the second class is the Array
this creates an array selection a random first number
and then incrementing from there, so that its a sorted array
the third class is the binary search class
which is my main class.
this class should take the users input
and pass it to the array
and the binary seach accordingly
it should also measure the running time, from when it passes the array
to the binary search class
i am having a really hard time creating this last class.
i have created the other 2 successfully
the codes for the binary search class is as follows
public class BinarySearch
static int binSearch(int[] array, int val)
// setting the start and the end of the array
int low = 0, high = array.length;
//While loop
while(low <= high) {
// How to find the mid point
int mid = (low + high)/2;
// if the mid point is the value return the value
if(array[mid] == val) {
return mid;
// if the value is smaller than the mid point
// go search the left half
} else if(array[mid] > val) {
high = mid - 1;
//if the value is greater then the mid point
// go search the right half
} else if(array[mid] < val) {
low = mid + 1;
// if value is not found return nothing
return -1;
}and the code for the Array class is as follows
import java.util.Random;
public class RandomSortedArray
public int[] createArray(int length)
// construct array of given length
int[] ary = new int[length];
// create random number generator
Random r = new Random();
// current element of the array; used in the loop below. Starts at
// -1 so that the first element of the array CAN be a 0
int val = -1;
for( int i = 0; i < length; i++)
val += 1 + r.nextInt(10);
ary[i] = val;
return ary;
}can some pne please help me create my binarysearchTest class.
as i mentioned before
it has to take the users input for the array size
and the users input for the value that they want to find
also needs to measure the running time
thanks for all ur help in advanceimport java.util.*;
public class AlgorithmTest
public static void main(String args[])
long StartTime, EndTime, ElapsedTime;
System.out.println ("Testing algorithm");
// Save the time before the algorithm run
StartTime = System.nanoTime();
// Run the algorithm
SelectionSortTest1();
// Save the time after the run
EndTime = System.nanoTime();
// Calculate the difference
ElapsedTime = EndTime- StartTime;
// Print it out
System.out.println("The algorithm took " + ElapsedTime + "nanoseconds to run.");
}this is the code i managed to work up for measuring the time..
how would i include it into the main BinarysearchTest Class -
Ok, maybe someone can help me with this one. Here's the code that I'm having trouble with:
public Object findValue(Comparable aKey)
Node result = (Node) findByKey(aKey);
if(result != null && result.getNodeKey() == aKey)
return result.getNodeData();
else
return null;
}The problem is in the condition for the if statement, particularly, "result.getNodeKey() == aKey". I've checked both values for result.getNodeKey() and aKey, and they are both the same, but the program gives ma a false result, which sends it straight to the "return null;" statement. Any thoughts as to why it might do this?I just found another trouble spot. Here's the code this time:
public boolean delete(Comparable aKey)
Node result = (Node) findByKey(aKey);
if(result != null && result.getNodeKey().equals(aKey))
Node currentNode = rootNode;
Node parentNode = null;
boolean found = false;
while(!found && currentNode != null)
if(aKey.compareTo(currentNode.getNodeKey()) < 0)
parentNode = currentNode;
currentNode = (Node)currentNode.getLeftChild();
else
if(aKey.compareTo(currentNode.getNodeKey()) > 0)
parentNode = currentNode;
currentNode = (Node) currentNode.getRightChild();
else
found = true;
if (parentNode == null)
//here's where I need assistance
return true;
else
parentNode.setLeftChild((Node) currentNode.getLeftChild());
Node holderNode = (Node) currentNode.getLeftChild();
holderNode.setRightChild((Node) currentNode.getRightChild());
return true;
else
return false;
}I'm trying to delete from the binary search tree, and I'm not sure how to adjust for deleting the top of the tree (rootNode, in this case) before the rest of the tree is gone. Any ideas? -
Below is the coding for the binary search:
public class Binary_Search_
public int binarySearch(int arr[],int key)
int low=0;
int high=arr.length-1;
while(low<=high)
int middle=(low+high)/2;
if(key==arr[middle])
return middle;
else
if(key<arr[middle]) high=middle-1;
else low=middle+1;
return -1;//not found
The only thing is that for the binary search the array must be sorted so what could I add to the start of the coding to quickly sort the array.
You must think this is a simple question but im new to java!Another thing..
Try rethinking your algorithm through.
int middle=(low+high)/2will not always hit the center/middle/sweet spot (or what you would call it) of your array.
If low was 2 and high was 5, your middle would be 3 (yes, 3), and if you are unlucky to have an array like this
1, 2, 3, 5, 7, 11, 13, 17, 19your code would return -1 if you were searching for 5.
And that's just one of a jillion possibilities. -
Java question help (applying binary search)
please show me how do i apply a binary search inside this program where i can use the binary serach to search for the book author or book title
// objectSort.java
// demonstrates sorting objects (uses bubble sort)
// to run this program: C>java libmainsys
class libary
private String booktitle;
private String bookauthor;
private String publisher;
private int yearpublished;
private int edition;
private int nofcop;
public libary(String title, String author, String pub, int yrpub, int edt, int nfcp)
{ // constructor
booktitle = title;
bookauthor = author;
publisher = pub;
yearpublished = yrpub;
edition = edt;
nofcop = nfcp;
public void displaylibary()
System.out.print(" Book Title: " + booktitle);
System.out.print(", Book Author: " + bookauthor);
System.out.print(", Book Publisher: " + publisher);
System.out.print(", Year Published: " + yearpublished);
System.out.print(", Edition: " + edition);
System.out.println(",No Of Copies : " + nofcop);
public String getLast() // get title
{ return booktitle; }
} // end class libary
class ArrayInOb
private libary[] nfcp; // ref to array ypub
private int nElems; // number of data items
public ArrayInOb(int max) // constructor
nfcp = new libary[max]; // create the array
nElems = 0; // no items yet
// put libary into array
public void insert(String title, String author, String pub, int yrpub, int edt, int nofcop)
nfcp[nElems] = new libary(title, author, pub, yrpub, edt, nofcop);
nElems++; // increment size
public void display() // displays array contents
for(int j=0; j<nElems; j++) // for each element,
nfcp[j].displaylibary(); // display it
System.out.println("");
public void bubbleSort()
int i, j;
libary temp;
for (i = nElems-1; i >= 0; i--)
for (j = 1; j <= i; j++)
if (nfcp[j-1].getLast().compareTo(nfcp[j].getLast())>0)
temp = nfcp[j-1];
nfcp[j-1] = nfcp[j];
nfcp[j] = temp;
} // end class ArrayInOb
class libmainsys
public static void main(String[] args)
int maxSize = 1000; // array size
ArrayInOb arr; // reference to array
arr = new ArrayInOb(maxSize); // create the array
arr.insert("Java_how__to_program", "Patty_John", "Deitel", 2001, 1, 430);
arr.insert("System_Design", "Dexter_Smith", "Thomson", 2002, 3, 450);
arr.insert("Program_Design", "Lorraine_Paul", "About", 1996, 2, 196);
arr.insert("Computer_Architecture", "Paul_Andrew","Dzone", 2006, 5, 199);
arr.insert("Visual_Basic_How_To_ Program", "Tom_Jones", "Jeffereson_publication", 2007, 4, 207);
arr.insert("Information_ Management", "William_Peter", "Mcgraw_Hill", 1995, 3, 204);
arr.insert("Sofware_ Application", "Henry_Sam", "Pearson", 2001, 6, 278);
arr.insert("English", "Samantha_Julia", "James_Hill", 2005, 1, 200);
arr.insert("Web_Publishing", "Audrey_Cynthia", "Surg", 2004, 3, 201);
arr.insert("Human_Computer_Interaction", "Tony_Edward", "Telde", 2003, 3, 199);
System.out.println("Before sorting:");
arr.display(); // display items
arr.bubbleSort(); // insertion-sort them
System.out.println("After sorting:");
arr.display(); // display them again
} // end main()
} // end class libmainsysI see you haven't worked out bubbleSort either. Since binary search only works on sorted arrays, I suggest you start there. Do you have the algorithms somewhere?
If you really need to be able to search using either author or title, I suggest you create 2 Comparators that compare using the desired property. Comparator is just an interface that defines methods to compare 2 objects. You have to write your own implementation of it to compare library objects. You'll always have to sort and search using the same Comparator. -
How does sy-tabix affect by binary search transforming no fields?
1 report ztx1110.
2 data: begin of it occurs 3,
3 f1(2) type n,
4 f2 type i,
5 f3(2) type c,
6 f4 type p,
7 end of it.
8
9 it-f1 = '40'. it-f3 = 'DD'. it-f2 = it-f4 = 4. append it.
10 it-f1 = '20'. it-f3 = 'BB'. it-f2 = it-f4 = 2. append it.
11
12 sort it by f1.
13 do 5 times.
14 it-f1 = sy-index * 10.
15 it-f3 = 'XX'.
16 it-f2 = it-f4 = sy-index.
17 read table it
18 with key f1 = it-f1
19 binary search
20 transporting no fields.
21 if sy-subrc <> 0.
22 insert it index sy-tabix.
23 endif.
24 enddo.
25
26 loop at it.
27 write: / it-f1, it-f2, it-f3, it-f4.
28 endloop.
in this program changing value of sy-tabix as such sy-index.
means in it's first loop it's value is 1, on second loop it's 2, and so on up to 5.
okay that is due to binary search.
but why binary search affecting this system variable?Hi Vinay,
It is simple.
This depends on your select query.
CASE 1: If you use select ..into table....., then there wil be nothing in your header ls_vbap. Now if you execute READ stmt..you will get nothing.
CASE 2: If you use Select..... into ls_vbap.....append ls_vbap....clear ls_vbap.....endselect. In this case also the output of READ will be nothing because you are clearing header. So while READ stmt you are comparing ls_vbap-vbeln and ls_vbap-posnr , as nothing in it you will get nothing
CASE 3: If you use case 2 without CLEAR stmt then always you will get the index of the last fetched record...i.e., the last record of the internal table before sorting..
I think it is helpful for you....if you are not clear just reply me -
Binary Search need clarification
Hi All,
I am facing a problem here in binary search. I will try to explain the situation with an example.
Internal table : it_vbap
Fields : vbeln posnr qty
Index 1 100 10 5
index 2 100 10 5
index 3 100 20 5
index 4 100 20 5
index 5 100 30 5
index 6 100 30 5
the abap code is this.
sort it_vbap by vbeln posnr.
read table it_vbap into ls_vbap with ket
vbeln = ls_vbap-vbeln
psonr = ls_vbap-posnr
binary search.
If sy-subrc = 0.
write : 'index :' sy-tabix.
endif.
My question here is does it always write index 1,3,5 (means in the all the case it should return the first record ), otherwise it can also return 1, 3, 6.
Pay attendtion the posnr are the same for every two records.
The internal table not only contains 6 six records , my question was however not depends on the number of records in the internal table, the binary search will return the first sorted key record in the table or not.
Please post your views.
Thanks & regards,
VijayHi Vinay,
It is simple.
This depends on your select query.
CASE 1: If you use select ..into table....., then there wil be nothing in your header ls_vbap. Now if you execute READ stmt..you will get nothing.
CASE 2: If you use Select..... into ls_vbap.....append ls_vbap....clear ls_vbap.....endselect. In this case also the output of READ will be nothing because you are clearing header. So while READ stmt you are comparing ls_vbap-vbeln and ls_vbap-posnr , as nothing in it you will get nothing
CASE 3: If you use case 2 without CLEAR stmt then always you will get the index of the last fetched record...i.e., the last record of the internal table before sorting..
I think it is helpful for you....if you are not clear just reply me -
Here is the code
read table <fs_it_crv> assigning <fs_wa_crv> with key
('objty') = wa_plfh-objty
('objid') = wa_plfh-objid
binary search.
If i use binary search system giving error while activating as
You cannot use explicit or implicit index operations on tables with
types "HASHED TABLE" or "ANY TABLE". "" has the type "ANY
TABLE". It is possible that .
How could i use Binary search in dynamic read
a®Hello,
how do you create your internal table "<FS_IT_CRV>" ?
i found this thread [here|Need help regarding Dynamic Read Table statement;.
You can try the code lines of David Klotz.
If you insert a sort statement before read and change read adding binary search it might work.
SORT <gt_itab> BY (ge_key_field1) (ge_key_field2) (ge_key_field3) .
* Finally, perform the search
READ TABLE <gt_itab> ASSIGNING <gs_result>
WITH KEY (ge_key_field1) = pa_cond1
(ge_key_field2) = pa_cond2
(ge_key_field3) = pa_cond3 BINARY SEARCH.
I tried the following code in my own program and it works:
SORT <wlf_t_data> BY (c_trkorr).
read table <wlf_t_data> assigning <wlf_header> with key
(c_trkorr) = 'DRR' binary search.
Where <wlf_t_data> is a dynamic table created by method create_dynamic_table of class cl_alv_table_create.
Cordialement,
Chaouki -
Searching a binary search tree written to a file
hello
I have a binary search tree which has 10 levels. Now I want to search a node
after writing all the nodes to a file.
Can anyone help me in doing this. A sample code would be great ...
Thanks
KYou suggest that I do away with the inner class
then?Absolutely. In fact I strongly suggest this. You are learning how to code and need to do things cleanly and in small steps. That means first creating your Node class and making sure it works. Then creating your Tree class, and making sure it works. In fact I would load the Strings into the Tree class first directly and testing things before even thinking about reading to and from files. Only then should you implement the file input and output steps.
The key here is that you don't go on to the next step until you're reasonably sure that your current code works. Remember, it's MUCH easier to code than to debug. -
Binary search tree - writing to a file in alphabetic order words from tree
Hi
I have written a program that will read a list of words from a file, insert these into a binary search tree, write words from the tree to another file, so that the resulting list contains words in ascending order. My input file Alpha1.txt contains the following contents in the order and format given (one word per line):
Dawn
Dave
Mike
Beth
David
Gina
Pat
Cindy
Sue
My program is supposed to be producing an alphabetical list of these words in another file "final.txt".
Instead it gives me the following list:
Dave Beth David Gina Cindy Sue Pat Mike Dawn
This is obviously wrong, right? My correct list in "final.txt" should be
Beth Cindy Dave David Dawn Gina Mike Pat Sue
I am not sure what is wrong with my code which I reproduce below:
import java.io.*;
import java.util.*;
//read Java Developer's Almanac from exampledepot.com
//Read this: http://en.wikipedia.org/wiki/Tree_traversal
/**preorder(node)
print node.value
if node.left ? null then preorder(node.left)
if node.right ? null then preorder(node.right)
public class AlphabeticBinarySortTree
private static TreeNode root;
private static TreeNode runner;
static String[] alphaArray;
static int alphaCounter;
private static TreeNode alphaRunner;
//Inner class
private static class TreeNode
String word;
TreeNode left;
TreeNode right;
int count;
public TreeNode(String word)
this.word = word;
left = null;
right = null;
public void insertAll(TreeNode newNode)
if(newNode.word.compareTo(runner.word) < 1)
System.out.println("newNode.word = " + newNode.word);
if(runner.left == null)
runner.left = newNode;
runner = runner.left;
else
insertAll(newNode);
else if(newNode.word.compareTo(runner.word) > 1)
System.out.println("newNode.word = " + newNode.word);
if(runner.right == null)
runner.right = newNode;
runner = runner.right;
else
insertAll(newNode);
else
count++;
}// end method insertAll
// Recursively print words (with counts) in sorted order
public static void printInPreOrder(TreeNode root)
System.out.println(root.word + " ");
if(root.left != null)
printInPreOrder(root.left);
if(root.right != null)
printInPreOrder(root.right);
} //end method printInPreOrder()
//called from inside main
public static void arrangeInAscendingOrder(TreeNode root, PrintWriter pWriter)
if(root.left != null)
arrangeInAscendingOrder(root.left, pWriter);
System.out.println();
System.out.println();
System.out.println(root.word + " ");
pWriter.write(root.word + " ");
if(root.right != null)
arrangeInAscendingOrder(root.right, pWriter);
}//end inner class TreeNode
public AlphabeticBinarySortTree()
root = null;
//belong to the outer class
public static void main(String[] args)
System.out.println("This program reads text from a file that it will parse. ");
System.out.println("In doing so, it will eliminate duplicate strings and ");
System.out.println("pick up only unique strings.These strings will be in a ");
System.out.println("stored in alphabetical order in a binary Search tree before they are ");
System.out.println("written out to another text file in alphabetic order");
//open the file for reading
try
BufferedReader bReader = new BufferedReader(new FileReader("Alpha1.txt"));
String words;
int count;
//System.out.println("A test to inspect the contents of words: " + words);
//System.out.println("Words =" + words);
count = 0;
//why is there an endless loop when
//I use "while(str != null)
StringTokenizer st;
st = null;
//based on http://www.exampledepot.com/egs/java.io/ReadLinesFromFile.html
while ((words = bReader.readLine()) != null)
st = new StringTokenizer(words);
while(st.hasMoreTokens())
//shiffman.net/teaching/a2z/concordance
String token = st.nextToken();
System.out.println("Token = " +token);
AlphabeticBinarySortTree.initiateInsert(token);
//count the number of tokens in the string
count++;
}//end inner while
}//end outer while
System.out.println("Here are the contents of your tree:");
//System.out.println("before the call to print()");
print();
System.out.println("the no of words in the file is: " + count);
bReader.close();
}//end of try
catch(IOException exception)
exception.printStackTrace();
/**try
FileWriter fWriter = new FileWriter("final.txt");
BufferedWriter bWriter = new BufferedWriter(fWriter);
PrintWriter pWriter = new PrintWriter(bWriter);
catch(IOExcepion exception)
exception.printStackTrace();
} // end main here
//this method belongs to the outer class
static void initiateInsert(String word)
//TreeNode is also static by the way
TreeNode newNode = new TreeNode(word);
if(root == null)
root = newNode;
System.out.println("root.word = " + root.word);
runner = root;
else
runner.insertAll(newNode);
// Start the recursive traversing of the tree
//without the access specifier 'static'
//I would get the following error message
//AlphabeticBinarySortTree.java:119: non-static method print() cannot be referenced from a static context
public static void print()
//System.out.println("**********AM I INSIDE THE PRINT() METHOD? ********");
if (root != null)
//System.out.println("++++++++ AM I INSIDE THE IF BLOCK OF THE PRINT() METHOD? +++++++");
//System.out.println("Inside THE IF BLOCK OF print() BUT BEFORE THE CALL TO printInPreOrder(),root.word = " + root.word);
AlphabeticBinarySortTree.TreeNode.printInPreOrder(root);
//open the file for writing
try
FileWriter fWriter = new FileWriter("final.txt");
BufferedWriter bWriter = new BufferedWriter(fWriter);
PrintWriter pWriter = new PrintWriter(bWriter);
AlphabeticBinarySortTree.TreeNode.arrangeInAscendingOrder(root, pWriter);
pWriter.close();
catch(IOException eException)
eException.printStackTrace();
}//end of if block
} // end of method print
}//end outer enclosing class here--------
All help is highly appreciated. Thanks for your time and consideration.You suggest that I do away with the inner class
then?Absolutely. In fact I strongly suggest this. You are learning how to code and need to do things cleanly and in small steps. That means first creating your Node class and making sure it works. Then creating your Tree class, and making sure it works. In fact I would load the Strings into the Tree class first directly and testing things before even thinking about reading to and from files. Only then should you implement the file input and output steps.
The key here is that you don't go on to the next step until you're reasonably sure that your current code works. Remember, it's MUCH easier to code than to debug. -
Problem with binary search tree
Hi all, just having troubles with a program im writing.
My program is based on a binary search tree full of items which are passed in via an input file and then saved to an output file.
I have written a sellItem method which gets passed in the item and quantity that has been sold which then needs to be changed in the binary tree.
Here if my sell Item method:
public void sellItem(Item item, int quantity){
stockItem = item;
mQuantity = quantity;
if (tree.includes(stockItem)){
int tempQuantity = stockItem.getQuantityInStock();
tempQuantity -= mQuantity;
else{
throw new IllegalArgumentException("The Barcode " + mBarCode + " does NOT exist.");
}and here is where i am calling it in a test class :
number1.sellItem(item1, 40);Each item is in this format :
ABCD, PENCIL, 1, 0.35, 200, 100, 200
where 200 is the quantity.
Therefore if i pass 40 into the method the binary search tree should then store the quantity as 160.
below is a copy of my binary tree :
public class BSTree extends Object {
private class TreeNode extends Object{
public Comparable data;
public TreeNode left;
public TreeNode right;
public TreeNode() {
this(null);
public TreeNode(Comparable barCode){
super();
data = barCode;
left = null;
right = null;
private TreeNode root;
private int nodeCount;
public BSTree(){
super();
root = null;
nodeCount = 0;
public boolean isEmpty() {
return root == null;
public int size() {
return nodeCount;
private TreeNode attach(TreeNode newPointer, TreeNode pointer){
if (pointer == null){
nodeCount++;
return newPointer;
else {
Comparable obj1 = (Comparable) newPointer.data;
Comparable obj2 = (Comparable) pointer.data;
if (obj1.compareTo(obj2) < 0) //Go left
pointer.left = attach(newPointer, pointer.left);
else //Go right
pointer.right = attach(newPointer, pointer.right);
return pointer;
public void insert(Comparable item){
//Create a new node and initialize
TreeNode newPointer = new TreeNode(item);
//Attach it to the tree
root = attach(newPointer, root);
public Comparable remove(Comparable key) {
TreeNode pointer;
TreeNode parent;
//Find the node to be removed
parent = null;
pointer = root;
while ((pointer != null) && !key.equals(pointer.data)) {
parent = pointer;
if (key.compareTo(pointer.data) <0)
pointer = pointer.left;
else
pointer = pointer.right;
if (pointer == null)
return null;
//Orphans
TreeNode leftSubtree = pointer.left;
TreeNode rightSubtree = pointer.right;
if (parent == null)
root = null;
else if (key.compareTo(parent.data) < 0)
parent.left = null;
else
parent.right = null;
//Reattaching any orphans in the left subtree
if (leftSubtree != null) {
root = attach(leftSubtree, root);
nodeCount--;
//Reattaching any orphans in the right subtree
if (rightSubtree != null) {
root = attach(rightSubtree, root);
nodeCount--;
nodeCount--;
return pointer.data;
private TreeNode search(TreeNode pointer, Comparable key) {
if (pointer == null)
return null;
else if (pointer.data.compareTo(key) == 0)
return pointer;
else if (key.compareTo(pointer.data) < 0)
return search(pointer.left, key);
else
return search(pointer.right, key);
public boolean includes(Comparable key) {
return (search(root, key) != null);
public Comparable retrieve(Comparable key) {
TreeNode pointer;
pointer = search(root, key);
if (pointer == null)
return null;
else
return pointer.data;
public Comparable[] getAllInOrder() {
Comparable[] list = new Comparable[nodeCount];
inOrderVisit(root,list,0);
return list;
private int inOrderVisit(TreeNode pointer, Comparable[] list, int count) {
if (pointer != null) {
count = inOrderVisit(pointer.left, list, count);
list[count++] = pointer.data;
count = inOrderVisit(pointer.right, list, count);
return count;
public String toString() {
StringBuffer result = new StringBuffer(100);
inOrderString(root, result);
return result.toString();
private void inOrderString(TreeNode pointer, StringBuffer result) {
if (pointer != null) {
inOrderString(pointer.left, result);
result.append(pointer.data.toString() + "\n");
inOrderString(pointer.right, result);
}Thanks for everyones help. Keep in mind i'm very new to java.Hi all, just having troubles with a program im writing.
My program is based on a binary search tree full of items which are passed in via an input file and then saved to an output file.
I have written a sellItem method which gets passed in the item and quantity that has been sold which then needs to be changed in the binary tree.
Here if my sell Item method:
public void sellItem(Item item, int quantity){
stockItem = item;
mQuantity = quantity;
if (tree.includes(stockItem)){
int tempQuantity = stockItem.getQuantityInStock();
tempQuantity -= mQuantity;
else{
throw new IllegalArgumentException("The Barcode " + mBarCode + " does NOT exist.");
}and here is where i am calling it in a test class :
number1.sellItem(item1, 40);Each item is in this format :
ABCD, PENCIL, 1, 0.35, 200, 100, 200
where 200 is the quantity.
Therefore if i pass 40 into the method the binary search tree should then store the quantity as 160.
below is a copy of my binary tree :
public class BSTree extends Object {
private class TreeNode extends Object{
public Comparable data;
public TreeNode left;
public TreeNode right;
public TreeNode() {
this(null);
public TreeNode(Comparable barCode){
super();
data = barCode;
left = null;
right = null;
private TreeNode root;
private int nodeCount;
public BSTree(){
super();
root = null;
nodeCount = 0;
public boolean isEmpty() {
return root == null;
public int size() {
return nodeCount;
private TreeNode attach(TreeNode newPointer, TreeNode pointer){
if (pointer == null){
nodeCount++;
return newPointer;
else {
Comparable obj1 = (Comparable) newPointer.data;
Comparable obj2 = (Comparable) pointer.data;
if (obj1.compareTo(obj2) < 0) //Go left
pointer.left = attach(newPointer, pointer.left);
else //Go right
pointer.right = attach(newPointer, pointer.right);
return pointer;
public void insert(Comparable item){
//Create a new node and initialize
TreeNode newPointer = new TreeNode(item);
//Attach it to the tree
root = attach(newPointer, root);
public Comparable remove(Comparable key) {
TreeNode pointer;
TreeNode parent;
//Find the node to be removed
parent = null;
pointer = root;
while ((pointer != null) && !key.equals(pointer.data)) {
parent = pointer;
if (key.compareTo(pointer.data) <0)
pointer = pointer.left;
else
pointer = pointer.right;
if (pointer == null)
return null;
//Orphans
TreeNode leftSubtree = pointer.left;
TreeNode rightSubtree = pointer.right;
if (parent == null)
root = null;
else if (key.compareTo(parent.data) < 0)
parent.left = null;
else
parent.right = null;
//Reattaching any orphans in the left subtree
if (leftSubtree != null) {
root = attach(leftSubtree, root);
nodeCount--;
//Reattaching any orphans in the right subtree
if (rightSubtree != null) {
root = attach(rightSubtree, root);
nodeCount--;
nodeCount--;
return pointer.data;
private TreeNode search(TreeNode pointer, Comparable key) {
if (pointer == null)
return null;
else if (pointer.data.compareTo(key) == 0)
return pointer;
else if (key.compareTo(pointer.data) < 0)
return search(pointer.left, key);
else
return search(pointer.right, key);
public boolean includes(Comparable key) {
return (search(root, key) != null);
public Comparable retrieve(Comparable key) {
TreeNode pointer;
pointer = search(root, key);
if (pointer == null)
return null;
else
return pointer.data;
public Comparable[] getAllInOrder() {
Comparable[] list = new Comparable[nodeCount];
inOrderVisit(root,list,0);
return list;
private int inOrderVisit(TreeNode pointer, Comparable[] list, int count) {
if (pointer != null) {
count = inOrderVisit(pointer.left, list, count);
list[count++] = pointer.data;
count = inOrderVisit(pointer.right, list, count);
return count;
public String toString() {
StringBuffer result = new StringBuffer(100);
inOrderString(root, result);
return result.toString();
private void inOrderString(TreeNode pointer, StringBuffer result) {
if (pointer != null) {
inOrderString(pointer.left, result);
result.append(pointer.data.toString() + "\n");
inOrderString(pointer.right, result);
}Thanks for everyones help. Keep in mind i'm very new to java. -
Hi Gurus.
I am using binary search to read data from internal table with approx more than 3lacs records. Also my table is sorted by BUKRS BELNR. But the problem is that i am getting records from lower half of table only and first half of table is ignoring, so i am not able to get all records. If i use nornal search, i am geting 7856 records and if i use binary search, i am geting only 1786 records....Can anyone help me why this is happening????Hi PV
Actually my table contain line items. I am reading header data from bkpf for a single day. and for that document nos i am fetching line items. But if i use join or for all entries, i will get time exceed errror. so i get Period of that single day and taking data for that period from BSIS and then reading BKPF.. and only taking that line items for which document no in BKPF...
my code is like
SELECT-OPTIONS: cpudt FOR sy-datum DEFAULT '20030102'.
SELECT bukrs belnr gjahr monat FROM bkpf
INTO CORRESPONDING FIELDS OF TABLE ibkpf
WHERE ( cpudt IN cpudt OR aedat IN cpudt ).
ibkpf1[] = ibkpf[].
SORT ibkpf1 BY monat.
DELETE ADJACENT DUPLICATES FROM ibkpf1 COMPARING monat.
IF ibkpf1[] IS NOT INITIAL.
SELECT * FROM bsis INTO TABLE ibsis
FOR ALL ENTRIES IN ibkpf1 WHERE gjahr = ibkpf1-gjahr
AND monat = ibkpf1-monat.
SORT ibsis BY bukrs belnr.
LOOP AT ibsis.
READ TABLE ibkpf WITH KEY bukrs = ibsis-bukrs
belnr = ibsis-belnr
BINARY SEARCH.
IF sy-subrc NE 0.
DELETE ibsis.
ENDIF.
ENDLOOP.
ENDIF. -
Hi all,
I'm thinking about how to implement a binary sort to get,given a key, a value from a text file with a list of words in order. I saw randomaccessfile for j2se and I know that this work it's usually done by seek functions but how to do the seek work without the File class , with only the InputStream and Reader? I'm talking about the j2me midp 2 cldc 1.1 api, is there some methods or class to use for this thing? How could be implemented a binary search in a text file resident in the /res dir or the JAR or which could be the problems that cannot make possible this staff? If there was something like seek one way could be that one of decide a common length of words to read so going to the half of the text file it's possible with simply multiplication, and the do the binary search things.But for now I haven't found any resources or help.
I hope you can help me to understand, thank youHallo all,
it is NOT the problem to locate the text I need to find in the fmb, or fmx.
The problem is that since I want to change not only the text but also some more code around I want to do this in Form Builder - Object Navigator. And I'm not able to discover in which trigger or procedure the text appears.
I do not want to edit fmx, or fmb in a text editor because I suupose theer are some checksums hidden for the format.
-duro
BTW in FMT there is no code seen (only some binary) for v6i as it was in v4.5. -
READ...BINARY SEARCH for more than 1 row.
Hi,
i have an internal table that contains several same entries. now i want to search the table with READ ... BINARY SEARCH ( or in another efficient way ) and get all those entries and not just one/the first one.
how could i do that ? thanks in advance!hi leider,
plz c the below sample code.
loop at i_head into workarea.
READ TABLE i_zvfx_gts008_itm INTO wa_zvfx_gts008_itm
WITH KEY
bukrs = wa_zvfx_gts008_head-bukrs
zzinvoice = wa_zvfx_gts008_head-zzinvoice
BINARY SEARCH.
IF sy-subrc EQ 0.
LOOP AT i_zvfx_gts008_itm INTO wa_zvfx_gts008_itm FROM sy-tabix.
IF ( wa_zvfx_gts008_itm-bukrs NE wa_zvfx_gts008_head-bukrs ) OR
( wa_zvfx_gts008_itm-zzinvoice NE wa_zvfx_gts008_head-zzinvoice ).
EXIT.
ENDIF.
endloop.
this is something like avoiding loop at where condition.
plz reward points if found helpful. -
Hi SDN Experts
Is it must to use a sort statement before we use a binary search in a read statement. because in some cases its failing as SY-SUBRC = 4 if my internal table is not sorted.
Please confirm me
Regards
PratyushaHi,
Yes, it is mandatory to sort the table before doing a binary search.
The principle of binary search only work in case of sorted tables.
A binary search algorithm is a technique for finding a particular value in a sorted list. A binary search finds the median element in a list, compares its value to the one you are searching for, and determines if its greater than, less than, or equal to the one you want. A guess that turns out to be too high becomes the new top of the list, and one too low the new bottom of the list. The binary search's next guess is halfway between the new list's top and bottom. Pursuing this strategy iteratively, it narrows the search by a factor 2 each time, and finds your value. A binary search is an example of a divide and conquer algorithm.
More on binary search.
The idea is to eliminate half of the search space with each comparison.
First, the middle element of the sequence is compared to the value we are searching for. If this element matches the value we are searching for, we are done. If, however, the middle element is less than the value we are chosen for (as specified by the relation used to specify a total order over the set of elements), then we know that, if the value exists in the sequence, it must exist somewhere after the middle element. Therefore we can eliminate the first half of the sequence from our search and simply repeat the search in the exact same manner on the remaining half of the sequence. If, however, the value we are searching for comes before the middle element, then we repeat the search on the first half of the sequence.
Hope this helps.
Regards,
Kate
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