Linear and Binary Searching of Parallel Arrays

I'm an AP student who used the "Fundamentals of Java" by Lambert/Osborne 3rd Edition. This text book has code that doesn't match anything else I've found in other how-to's, guides, or teach yourself books. Not even online can I find code that matches up with the format used in this book!
I've got an assignment that wants me to read in a 4 digit account number of N number of customers, places them in two parallel arrays. Data found in two separate txt files. Create a prompt that ask's for customer's account number then displays account balance. Same program for both linear and binary search methods (2 programs).
I know the search method and how to read the files with a scanner. It is the body of the program that is stumping me. How to call and search the arrays themselves. Any help would be great.

First of all, you have posted this question in the wrong place. Please post these kinds of general questions in the New to Java forum.
Second, if you're in an AP class, don't you have a teacher you can ask?
But anyway, here's the idea
For a linear search, you go thru the array element by element, and on each element you call equals(x) to see if that element is equal to what you're searching for (note that primitives use == rather than equals)
For binary search, note first of all that your data MUST BE COMPARABLE (primative or implement the comparable interface) and MUST BE SORTED.
Then what you can do if go to the middle of the list, and if what you are searching for is less than that element, go to the middle of the first half of the list (if it's greater, go the the middle of the upper half of the list) and keep breaking the list in half until you've found the element or you know its not there.

Similar Messages

  • Linear search and binary search

    can any one tell me what is linear and binary search in detail.
    and what is the difference between them .
    which one is useful in coding.

    If you read entries from standard tables using a key other than the default key, you can use a binary search instead of the normal linear search. To do this, include the addition BINARY SEARCH in the corresponding READ statements.
    READ TABLE <itab> WITH KEY <k1> = <f1>... <kn> = <fn> <result>  BINARY SEARCH.
    The standard table must be sorted in ascending order by the specified search key. The BINARY SEARCH addition means that you can access an entry in a standard table by its key as quickly as you would be able to in a sorted table.
    REPORT demo_int_tables_read_index_bin.
    DATA: BEGIN OF line,
            col1 TYPE i,
            col2 TYPE i,
          END OF line.
    DO 4 TIMES.
      line-col1 = sy-index.
      line-col2 = sy-index ** 2.
      APPEND line TO itab.
    SORT itab BY col2.
    WRITE: 'SY-SUBRC =', sy-subrc.
    The output is:
    SY-SUBRC =    0
    The program fills a standard table with a list of square numbers and sorts them into ascending order by field COL2. The READ statement uses a binary search to look for and find the line in the table where COL2 has the value 16.
    Linear search use sequential search means each and every reord will be searched to find. so it is slow.
    Binary search uses logrim for searching. Itab MUST be sorted on KEY fields fro binary search. so it is very fast.
    The search takes place as follows for the individual table types :
    standard tables are subject to a linear search. If the addition BINARY SEARCH is specified, the search is binary instead of linear. This considerably reduces the runtime of the search for larger tables (from approximately 100 entries upwards). For the binary search, the table must be sorted by the specified search key in ascending order. Otherwise the search will not find the correct row.
    sorted tables are subject to a binary search if the specified search key is or includes a starting field of the table key. Otherwise it is linear. The addition BINARY SEARCH can be specified for sorted tables, but has no effect.
    For hashed tables, the hash algorithm is used if the specified search key includes the table key. Otherwise the search is linear. The addition BINARY SEARCH is not permitted for hashed tables.
    Binary search must be preffered over linear sarch.
    Hope this is helpful, Do reward.

  • Arrays,bubblesort, and binary search

    Hi I need help I have been working on this homework assignment for the past week and I can't seem to get it so if anyone can help me to figure out what is wrong I would reall grateful. Thanks ahead of time for the help.
    Here is what is required for the assignment and also the errors I am getting.
    Write a program that sorts an integer array in ascending order and checks whether an integer entered by user is in the array or not. Please follow the following steps to complete the assignment:
    1. Declare and create a one-dimensional array consisting of 20 integers.
    2. Read 20 integers from the user to initialize the array. Use input dialog box and repetition statement.
    3. Build an output string containing the content of the array.
    4. Sort the array in ascending order using bubbleSort( ) and swap( ) methods. Then, append the content of the sorted array to the output string.
    5. Read an integer search key from the user;
    6. Use binarySearch( ) method to check whether the search key is in the array or not. Then, append the search result to the output string.
    7. Display the output string in a message box.
    Here is my code
    import javax.swing.*;
    import java.awt.*;
    import java.awt.event.*;
    public class SortSearch {
    public static void main (String args[])
    String input, ouput;
    int key;
    int index;
    int[] array=new int [20];
    input=JOptionPane.showInputDialog("Enter 20 Numbers");
    for(int counter=0; counter<array.length; counter++)
    JTextArea inputArea=new JTextArea();
    public void bubblesort(int array2[] )
    for(int pass=1; pass<array2.length; pass++){
    for(int element=0; element<array2.length-1; element++){
    swap(array2, element, element+1);
    public void swap(int array3[], int first, int second)
    int hold;
    public void actionPerformed(ActionEvent actionEvent)
    String searchKey=actionEvent.getActionCommand();
    int element=binarySearch(array, Integer.parseInt(searchKey) );
    output.setText("Found value in element " + element);
    output.setText("Value not found ");
    public int binary search(iny array2[], int key)
    int low=0;
    int high=array2.length-1;
    int middle;
    middle=(low + high)/2;
    buildOutput(array2, low, middle, high);
    if(key==array[middle] )
    return middle;
    else if(key<array[middle] )
    JOptionPane.showMessageDialog(null, outputArea);
    } //end main
    } //end class
    Here is my errors
    C:\java>javac illegal start of expression
    public void bubblesort(int array2[] )
    ^ cannot resolve symbol
    symbol : variable output
    location: class SortSearch
    ^ cannot resolve symbol
    symbol : variable counter
    location: class SortSearch
    ^ cannot resolve symbol
    symbol : variable output
    location: class SortSearch
    ^ cannot resolve symbol
    symbol : variable outputArea
    location: class SortSearch
    5 errors

    I am still having problems.
    Here is my code
    import javax.swing.*;
    import java.awt.*;
    import java.awt.event.*;
    public class SortSearch {
          public static void main (String args[]){
          String input, output;
          int key;
          int index;
          int[] array=new int [20];
          input=JOptionPane.showInputDialog("Enter 20 Numbers");
          for(int counter=0; counter<array.length; counter++)
          JTextArea outputArea=new JTextArea();
            public void bubblesort(int array2[] )
              for(int pass=1; pass<array2.length; pass++){
                for(int element=0; element<array2.length-1; element++){
                    swap(array2, element, element+1);
                }  //end inner for
              }  //end outer for
            }  //end bubblesort 
            public void swap(int array3[], int first, int second)
              int hold;
            }  //end swap
            public void actionPerformed(ActionEvent actionEvent)
              String searchKey=actionEvent.getActionCommand();
              int element=binarySearch(array, Integer.parseInt(searchKey) );
                outputArea.setText("Found value in element " + element);
                outputArea.setText("Value not found ");
            }  //end actionperformed
            JOptionPane.showMessageDialog(null, outputArea,"Comparisons");       
    }  //end classHere is my errors
    C:\java>javac <identifier> expected
    JOptionPane.showMessageDialog(null, outputArea,"Comparisons");
    ^ cannot resolve symbol
    symbol : class showMessageDialog
    location: class javax.swing.JOptionPane
    JOptionPane.showMessageDialog(null, outputArea,"Comparisons");
    ^ cannot resolve symbol
    symbol : method array (int)
    location: class SortSearch
    ^ cannot resolve symbol
    symbol : variable array
    location: class SortSearch
    int element=binarySearch(array, Integer.parseInt(searchKey) );
    ^ cannot resolve symbol
    symbol : variable outputArea
    location: class SortSearch
    outputArea.setText("Found value in element " + element);
    ^ cannot resolve symbol
    symbol : variable outputArea
    location: class SortSearch
    outputArea.setText("Value not found ");
    6 errors
    Thanks ahead of time I still don't understand the stuff so I sometime don't understand what my errors are telling me so that is why I ask for your help so that maybe it will help make sense.

  • Read table and binary search

    Hi all,
    I have a simple query regarding read int_tab with binary search.
    Why reading  internal table with binary search fails if it is sorted in descending order table must be sorted in ascending order?
    I check fo the algorithm of binary search, it does not talk about sort order. As far as my understanding goes binary search only require sorted table but while reading table in SAP it has to be sorted in ascending order!!

    By default binary search assumes that the sort order is ASCENDING.
    If you sort the list in descending and then try to binary search your quires will fail. Look at an example:
    Let the descending order internal table as:
    Field1      Field2
    Sam        50000
    John       34786
    Boob      54321
    Alice       12345
    When you do binary search with key = 'Sam' then it will directly go to 2nd and 3rd records for comparision. The binary search algorithm compares 'Sam' with 'John' and it concludes that 'Sam' is greater than 'John' and it will continue to look downward into the internal table. And when it reaches the end of the internal table then the value of SY-TABIX = 5 and SY-SUBRC = 8 (Key is greater than the all).
    And if you do binary search with key = 'Alice' then the binary search algorithm compares 'Alice' with 'John' and it concludes that 'Alice' is lower than 'John' and it will continue to look upward into the internal table. And when it reaches above the first record in internal table then the value of SY-TABIX = 1 and SY-SUBRC = 4 (points to the next largest entry).
    The only correct result you will get is when you execute statement with key='John' (In this particular case) . SY-TABIX = 2 and SY-SUBRC = 0. I think you got this binary search algorithm.

  • Search in object array

    I am working on a program in which one of the problems is to search an object array to allow a user to edit an item in the array. I have examples on how to do linear and binary searches, but they are all simple strings of numeric data. How do I apply this algorithm to a more complex array?I have a class file called Application that contains my constructor, and my get & set methods for creating the objects in the array. And I have a driver class to allow a user to view the array, add items to it , and edit items in it.
    This is an example of my declaration of the initial items:
    Application inventory[] = new Application[100];
      inventory[0] = new Application("Jumpstart Toddlers", 14.99, 500);
      inventory[1] = new Application("Norton Antivirus 2003", 49.99, 1200);
    I would like to set it up so that the user is prompted for the name of the software item they would like to edit, the array is searched using a method which returns the index the name was found in.

    Sounds like a homework assignment to me but...
    Appliction apps[] = new Application[100];
    ...Add Items to the apps array...
    String software_to_search_for = .. user entered
    for (int i = 0; i < apps.length; ++i)
    edit(apps[i], software_to_search_for);
    public void edit(Application app, String newName)
    // You can do anything here
    } // edit
    True enough, this is an assignment. But..These assignments were given to us without all the necessary tools. They are set up by the dept. and we have to finish them even if the instructor does not finish covering all the relevant material.In this case, we have not covered the equalsIgnoreCase(something obviously necessary for my problem). Now that I know what I am looking for, I have found it in my book and can apply it to my code appropriately. I was not necessarily looking for code, just the tools I need for my own design. This is something that the Computer dept. allows for in that they have assistants in the open labs that will help in this manner. Problem is, everyone has been out for Thanksgiving since Tues. Please dont feel as though you assisted someone in cheating!! I can assure you, that is not the case here. Thank you for your help.

  • Difference between Linear & Binary search.

           Could Anybody describe me, What is the difference between Linear Search and Binary Search in ABAP ?
    Moderator Message: Please search before posting your question. Therad locked.
    Edited by: Suhas Saha on Oct 19, 2011 11:21 AM

    In case of linear search system will search from begining.
    that means Example : z table contains single field with values 
    1 2 3 4 5 6 7 8 9 
    if u r searching for a value then system will starts from
    first position. if required value is founded then execution
    will comes out from z table.
    In case of binary search system will starts from mid point.
    if value is not founded then it will search for upper half.
    in  that upper half it will check mid that search
    will takes place.

  • Java Linear, Quicksort, Binary Time Calc Problems

    Objective of program: Simple program demonstrating the time it takes to do each algorithm.
    My approach:
    1) Prompt a number
    2) Make array of linear ints
    3) Calculate time it takes for a linear search
    4) Repeat for quick sort, and Binary search
    1) Linear search time is always 0
    2) Quick sort never ends.
    3) Don't know if binary works yet since it never goes beyond quicksort.
    Any help or suggestions are greatly appreciated.
    import java.util.Date;
    import java.util.ArrayList;
    import java.util.Random;
    import java.util.Arrays;
    import java.util.Scanner;
    public class Driver {
         public static void main(String[] args) {
              Random generator = new Random();               
              int[] linear = new int[1000000];                           // Create Linear Array
              for(int i = 0; i < linear.length; i++)                 // filling up the array
                   linear[i] = i;
              Scanner input = new Scanner(; 
              System.out.print("Please enter number to search: ");       // Asks
              int search = input.nextInt();                              // Stores number
              Date end = new Date();                                   // Create Timer          
              long startTime1 = end.getTime();                           // Start Time
              for(int i = 0; i < linear.length; i++) {               // Linear Search
                   if (linear[i] == search) {
                        long endTime1 = end.getTime();                 // If found, end time
                        System.out.println("Time of Linear search: " + (endTime1 - startTime1));          // Prints elapsed time
              int[] quicksort = new int[1000000];                    // Creates quicksort array
              for(int i = 0; i < quicksort.length; i++)                // Initializes the array
                   quicksort[i] = generator.nextInt(100000);
              long startTime2 = end.getTime();                         // Starts the time
              for(int i = 0; i < 1000000; i++)                         // Sorts...
              long endTime2 = end.getTime();                         // Ends time
              System.out.println("Time of QuickSort: " + (startTime2 - endTime2));               // Prints elapsed time
              int[] binary = new int[1000000];                         // Creates binary array
              for(int i = 0; i < binary.length; i++)                    // Initializes binary array
                   binary[i] = generator.nextInt();
              long startTime3 = end.getTime();                         // Start time
              Arrays.binarySearch(binary,search);                      // Binary Search
              long endTime3 = end.getTime();                         // Ends time
              System.out.println("Time of Binary Search: " + (endTime3 - startTime3));     // Prints out time
    }Edited by: onguy3n on Mar 26, 2009 4:39 AM

    ibanezplayer85 wrote:
    Any help or suggestions are greatly appreciated.
    Suggestion: Break your code up into different methods and even separate classes, if necessary; don't just use main for everything. Maybe you posted it this way to have it all in one class for the forum to read, but it's very confusing to look at it and understand it this way. I know that this isn't an answer to your question, but you did ask for suggestions :)Thanks, it was just a demonstration program in class so he didn't really care about readability, but yes I should have separated it in a different class.
    Linear search time is always 0I'm not sure what the convention is, but whenever I needed to measure time for an algorithm, I used System.currentTimeMillis(); rather than the Date class.
    long startTime = System.currentTimeMillis();
    long endTime = System.currenTimeMillis();
    long totalTime = endTime - startTime;Although, I think if you're not printing anything out to the console, it will most likely print out 0 as the time (because most of the processing is working on printing out the data). That is, unless you're doing some heavy processing.Thanks! I tried System.currentTimeMillis() and it now works. I still don't understand your explanation why it prints out 0 though :(
    Quick sort never ends.I think it's just taking a while. It's not an efficient algorithm with a worst case time complexity of O(n^2) and you gave it a very large array of random values to work with. I wouldn't be surprised if you ran out of heap space before it finished. If you knock off some zero's on the array size, you'll see that it does, in fact, finish.Ok, thanks! In class we didn't talk much about the heap. How do I calculate how much heap space my program will use and how to allocate more? I vaguely remember something like xmx512m or something as the parameter, but every time I get an error:
    Unrecognized option: -xmx512m
    Could not create the Java virtual machine.

  • Sequential vs binary search

    i am trying to know how to find the theoretical running time for both the sequential and binary search algorithms. i have written a program that determines the timing when running on my pc, however, i would like to compare these results against the ones ran via the program i have written.
    thanks for any tips,

    I'm not sure what you're asking either.
    It sounds like you're asking how to determine the time required to run the search. You can't determine that precisely. What you can do is determine the "big O" time, which doesn't tell you how fast it will run for a given input set, but does give you an idea of how fast the running time grows as the input size grows.
    As for how you determine the big-O time, as already mentioned, that's well documented in textbooks, google, wikipedia...

  • Difference between Binary search and Linear search

    Dear all,
    If anyone helps me to get the basic difference between binary and Linear search in SAP environment.

    In case of linear search system will search from begining.
    that means Example : z table contains single field with values 
    1 2 3 4 5 6 7 8 9 
    if u r searching for a value then system will starts from
    first position. if required value is founded then execution
    will comes out from z table.
    In case of binary search system will starts from mid point.
    if value is not founded then it will search for upper half.
    in  that upper half it will check mid that search
    will takes place.

  • Java binary search and insert

    I'm currently writing a program which is an appointment book. I currently have 4 classes and at the minute it can sort the array and print it out. I'm stuck at binary search and inserting a new appointment record. I will include the classes which i have got.
       import java.util.*;
       import java.util.Scanner;
        class Appointment implements Comparable
          private String description;
          private int day;
          private int month;
          private int year;
          private String startTime;
          private String endTime;
          protected static Scanner keyboard = new Scanner(;     
           public Appointment()
             description = "";
             day = 0;
             month = 0;;
             year = 0;;
             startTime = "";
             endTime = "";
           public Appointment(String appDesc, int appDay, int appMonth, int appYear, String appStartTime, String appEndTime)
             description = appDesc;
             day = appDay;
             month = appMonth;
             year = appYear;
             startTime = appStartTime;
             endTime = appEndTime;
           public void display()      
             System.out.print("  Description: " + description);
             System.out.print(", Date: " + day + "/" +month+ "/" +year);
             System.out.println(", Start Time: " + startTime);
             System.out.println(", End Time: " + endTime);
           public void setDay(int day)
          { = day; }
           public int getDay()     
             return day; }
           public void setMonth(int month)
          { this.month = month; }
           public int getMonth()     
             return month; }
           public void setYear(int year)
          { this.year = year; }
           public int getYear()
             return year; }
           public int compareTo(Object obj)
             if (obj instanceof Appointment)
                Appointment appt = (Appointment) obj;
                if ( > appt.getDay())
                   return 1;
                else if ( < appt.getDay());
                return -1;
             return 0;
           public String toString() {
             StringBuffer buffer = new StringBuffer();
             buffer.append("Description: " + description);
             buffer.append(", Date: " + day + "/" +month+ "/" +year);
             buffer.append(", Start Time: " + startTime);
             buffer.append(", End Time: " + endTime);
             return buffer.toString();
           public void read(){
             System.out.print("Description : ");String;
             System.out.print("Day : ");int dayIn=keyboard.nextInt();
             System.out.print("Month : ");int monthIn=keyboard.nextInt();
             System.out.print("Year : ");int yearIn=keyboard.nextInt();
             System.out.print("Start Time : ");String;
             System.out.print("End Time : ");String;
             boolean goodInput = false;
                   goodInput = true;
                    catch(IllegalArgumentException e){
                      System.out.println("INVALID ARGUMENT PASSED FOR day or month or year");
                      System.out.print("RE-ENTER VALID ARGUMENT FOR DAY : ");dayIn=keyboard.nextInt();
                      System.out.print("RE-ENTER VALID ARGUMENT FOR MONTH : ");monthIn=keyboard.nextInt();
                      System.out.print("RE-ENTER VALID ARGUMENT FOR YEAR : ");yearIn=keyboard.nextInt();                                        

    import java.util.*;
    class Array
         private Appointment[] app;
         private int nElems;
         Appointment tempApp;
         public Array(int max)
              app = new Appointment[max];
              nElems = 0;
         public Array(String desc, int day, int month, int year, String sTime, String eTime)
              app = new Appointment[100];
              nElems = 0;
       public int size()
          { return nElems; }
         void add(){
              Appointment appointmentToAdd = new Appointment();
              // Read its details
              // And add it to the studentList
         public void add(String desc, int day, int month, int year, String sTime, String eTime)
            app[nElems] = new Appointment(desc, day, month, year, sTime, eTime);
            nElems++;             // increment size
              Appointment appointmentToAdd = new Appointment(desc, day, month, year, sTime, eTime);
              // And add it to the studentList
          public void insert(Appointment tempApp) {
        int j;
        for (j = 0; j < nElems; j++)
          // find where it goes
          if (app[j] > tempApp) // (linear search)
        for (int k = nElems; k > j; k--)
          // move bigger ones up
          app[k] = app[k - 1];
        app[j] = tempApp; // insert it
        nElems++; // increment size
       public void display()       // displays array contents
            for(int j=0; j<nElems; j++)    // for each element,
              app[j].display();     // display it
         public void insertionSort()
       int in, out;
       for(out=1; out<nElems; out++) // out is dividing line
         Appointment temp = app[out];    // remove marked person
         in = out;          // start shifting at out
         while(in>0 &&        // until smaller one found,
          app[in] = app[in-1];     // shift item to the right
          --in;          // go left one position
         app[in] = temp;        // insert marked item
         } // end for
       } // end insertionSort()
    import java.util.*;
    class Menu{
       private static Scanner keyboard = new Scanner(;
       int option;
         void display(){
              // Clear the screen
            System.out.println("\n1 Display");
              System.out.println("\n2 Insert");          
              System.out.println("3 Quit");          
         int readOption(){
            System.out.print("Enter Option [1|2|3] : ");
              return option;
       import java.util.*;
       import java.util.Arrays;
        class ObjectSortApp
           public static void main(String[] args)
                   int maxSize = 100;
                   Array arr;
                   arr = new Array(maxSize)
                   Appointment app1 = new Appointment("College Closed", 30, 4, 2009, "09:30", "05:30");;
                   Appointment app2 = new Appointment("Assignment Due", 25, 4, 2009, "09:30", "05:30");
             Appointment app3 = new Appointment("College Closed", 17, 4, 2009, "09:30", "05:30");
             Appointment app4 = new Appointment("Easter Break", 9, 4, 2009, "01:30", "05:30");
             Appointment app5 = new Appointment("College Opens", 15, 4, 2009, "09:30", "05:30");
             Appointment app6 = new Appointment("Assignment Due", 12, 4, 2009, "09:30", "05:30");
             Appointment app7 = new Appointment("Exams Begin", 11, 4, 2009, "09:30", "05:30");
                //To sort them we create an array which is passed to the Arrays.sort()
            Appointment[] appArray = new Appointment[] {app1, app2, app3, app4, app5, app6, app7};
             System.out.println("Before sorting:");
            //Print out the unsorted array
            for (Appointment app : appArray)
             //arr.insertionSort();      // insertion-sort them
             System.out.println("\n\nAfter sorting:");               
            //Print out the sorted array
            for (Appointment app : appArray)
              Menu appMenu = new Menu();
              int chosenOption;
                   for (Appointment app : appArray)
                      case 1 : app.display(); break;
                        case 2 : arr.add(); break;
              }while(chosenOption != 3);    
          } // end main()
       } // end class ObjectSortApp

  • Binary search tree in java using a 2-d array

    Good day, i have been wrestling with this here question.
    i think it does not get any harder than this. What i have done so far is shown at the bottom.
    We want to use both Binary Search Tree and Single Linked lists data structures to store a text. Chaining
    techniques are used to store the lines of the text in which a word appears. Each node of the binary search
    tree contains four fields :
    (i) Word
    (ii) A pointer pointing to all the lines word appears in
    (iii) A pointer pointing to the subtree(left) containing all the words that appears in the text and are
    predecessors of word in lexicographic order.
    (iv) A pointer pointing to the subtree(right) containing all the words that appears in the text and are
    successors of word in lexicographic order.
    Given the following incomplete Java classes BinSrchTreeWordNode, TreeText, you are asked to complete
    three methods, InsertionBinSrchTree, CreateBinSrchTree and LinesWordInBinSrchTree. For
    simplicity we assume that the text is stored in a 2D array, a row of the array represents a line of the text.
    Each element in the single linked list is represented by a LineNode that contains a field Line which represents a line in which the word appears, a field next which contains the address of a LineNode representing the next line in which the word appears.
    public class TreeText{
    BinSrchTreeWordNode RootText = null;// pointer to the root of the tree
    String TextID; // Text Identification
    TreeText(String tID){TextID = tID;}
    void CreateBinSrchTree (TEXT text){...}
    void LinesWordInBinSrchTree(BinSrchTreeWordNode Node){...}
    public static void main(String[] args)
    TEXT univ = new TEXT(6,4);
    univ.textcont[0][0] = "Ukzn"; univ.textcont[0][1] ="Uct";
    univ.textcont[0][2] ="Wits";univ.textcont[0][3] ="Rhodes";
    univ.textcont[1][0] = "stellenbosch";
    univ.textcont[1][1] ="FreeState";
    univ.textcont[1][2] ="Johannesburg";
    univ.textcont[1][3] = "Pretoria" ;
    univ.textcont[2][0] ="Zululand";univ.textcont[2][1] ="NorthWest";
    univ.textcont[2][2] ="Limpopo";univ.textcont[2][3] ="Wsu";
    univ.textcont[3][0] ="NorthWest";univ.textcont[3][1] ="Limpopo";
    univ.textcont[3][2] ="Uct";univ.textcont[3][3] ="Ukzn";
    univ.textcont[4][0] ="Mit";univ.textcont[4][1] ="Havard";
    univ.textcont[4][2] ="Michigan";univ.textcont[4][3] ="Juissieu";
    univ.textcont[5][0] ="Cut";univ.textcont[5][1] ="Nmmu";
    univ.textcont[5][2] ="ManTech";univ.textcont[5][3] ="Oxford";
    // create a binary search tree (universities)
    // and insert words of text univ in it
    TreeText universities = new TreeText("Universities");
    // List words Universities trees with their lines of appearance
    public class BinSrchTreeWordNode {
    BinSrchTreeWordNode LeftTree = null; // precedent words
    String word;
    LineNode NextLineNode = null; // next line in
    // which word appears
    BinSrchTreeWordNode RightTree = null; // following words
    BinSrchTreeWordNode(String WordValue)
    {word = WordValue;} // creates a new node
    BinSrchTreeWordNode InsertionBinSrchTree
    (String w, int line, BinSrchTreeWordNode bst)
    public class LineNode{
    int Line; // line in which the word appears
    LineNode next = null;
    public class TEXT{
    int NBRLINES ; // number of lines
    int NBRCOLS; // number of columns
    String [][] textcont; // text content
    TEXT(int nl, int nc){textcont = new String[nl][nc];}
    The method InsertionBinSrchTree inserts a word (w) in the Binary search tree. The method Create-
    BinSrchTree creates a binary search tree by repeated calls to InsertionBinSrchTree to insert elements
    of text. The method LinesWordInBinSrchTree traverses the Binary search tree inorder and displays the
    words with the lines in which each appears.
    //InsertionBinTree is of type BinSearchTreeWordNode
    BinSrchTreeWordNode InsertionBinSrchTree(String w, int line,                                             BinSrchTreeWordNode bst)
    //First a check must be made to make sure that we are not trying to //insert a word into an empty tree. If tree is empty we just create a //new node.
         If (bst == NULL)
                   System.out.println(“Tree was empty”)
         For (int rows =0; rows <= 6; rows++)
                   For (int cols = 0; cols <= 4; cols++)
                        Textcont[i][j] = w

    What is the purpose of this thread? You are yet to ask a question... Such a waste of time...
    For future reference use CODE TAGS when posting code in a thread.
    But again have a think about how to convey a question to others instead of blabbering on about nothing.
    i think it does not get any harder than this.What is so difficult to understand. Google an implementation of a binary tree using a single array. Then you can integrate this into the required 2-dimension array for your linked list implemented as an array in your 2-d array.

  • Java Binary Search and sorting in Java

    My program is suppose to search news articles and alphabetize all the words article individually of the text file. Right now the program alphabetizes all the words of the articles including the numbers. The text file will be located below the code. So basically i need to know how to alphabetize every articles words individually.
    //This program reads an input line from the reader put the worda into an array with a count and increases
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    //then prints out the array. There are 4 different articles like this one in the text file
    <BODY>Assets of money market mutual funds
    increased 720.4 mln dlrs in the week ended yesterday to 236.90
    billion dlrs, the Investment Company Institute said.
        Assets of 91 institutional funds rose 356 mln dlrs to 66.19
    billion dlrs, 198 general purpose funds rose 212.5 mln dlrs to
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    import java.util.StringTokenizer;
    import java.util.ArrayList;
    public class WordsFrequency
       public static void main(String[] args)
          // Initializations
          FileReader reader = null;
          FileWriter writer = null;
          // Open input and output files
              reader = new FileReader("Reuters00.txt");
              writer = new FileWriter("WordsReport.txt");
          catch(FileNotFoundException e)
             System.err.println("Cannot find input file");
          catch(IOException e)
           System.err.println("Cannot open input/output file");
          // Set up to read a line and write a line
          BufferedReader in = new BufferedReader(reader);
          PrintWriter out = new PrintWriter(writer);
          out.println("Copied file is: Words followed by frequency");
          int count = 0;
           wordCount[] wordsArray = new wordCount[100000];
          boolean done = false;
             String inputLine;
                  inputLine= in.readLine();
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                System.err.println("Problem reading input, program terminates. " );
                inputLine = null;
             if (inputLine == null)
               done = true;
               for(int i = 0; i < count;i++)
                   StringTokenizer tokenizer = new StringTokenizer(inputLine);
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                        int lengthofString = token.length();
                        char ch = token.charAt(lengthofString-1);
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                                  wordsArray[count] = wordAndCount;
    // Close files
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    System.err.println("Error closing file.");
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         for (int j = 0; j < size-1; j++)
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    Edited by: IronManNY on Sep 25, 2008 3:24 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                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                             

    IronManNY wrote:
    My program is suppose to search news articles and alphabetize all the words article individually of the text file. Right now the program alphabetizes all the words of the articles including the numbers. The text file will be located below the code. So basically i need to know how to alphabetize every articles words individually.You want to strip out the numbers?

  • Search given string array and replace with another string array using Regex

    Hi All,
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    As stated, no, it can't be done. But that doesn't mean you have to make a separate pass over the input for each word you want to replace. You can employ a regex that matches any word, then use the lower-level Matcher methods to replace the word or not depending on what was matched. Here's an example: import java.util.*;
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      public static void main(String[] args) throws Exception
        String str = "If you wish to search for any of these characters, "
            + "they must be preceded by the character to be interpreted";
      public static String doReplace(String str)
        Pattern p = Pattern.compile("\\b\\w+\\b");
        Matcher m = p.matcher(str);
        StringBuffer sb = new StringBuffer();
        while (m.find())
          int pos = oldWords.indexOf(;
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            m.appendReplacement(sb, "");
        return sb.toString();
    } This is just a demonstration of the technique; a real-world solution would require a more complicated regex, and I would probably use a Map instead of the two Lists (or arrays).

  • Search for Emails and place them on a array

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  • Binary search in java

    I keep getting a java.lang.ClassCastException with these two classes when I try to perform a binary search. Any tips?
    Phonebook class:
    import java.util.*;
    public class Phonebook
    private static long comparisons = 0;
    private static long exchanges = 0;
         public static void main (String[] args)
         // Create an array of Phonebook records
         PhoneRecord[] records = new PhoneRecord[10];
         records[0] = new PhoneRecord("Smith","Bob", 1234367);
         records[1] = new PhoneRecord("Jones","Will", 1234548);
         records[2] = new PhoneRecord("Johnson","Frank", 1234569);
         records[3] = new PhoneRecord("Mc John","Pete", 1234560);
         records[4] = new PhoneRecord("OBrien","Frank", 1234571);
         records[5] = new PhoneRecord("OConnor","Joe", 1234572);
         records[6] = new PhoneRecord("Bloggs","Ricky", 1233570);
         records[7] = new PhoneRecord("empty","empty", 8888888);
         records[8] = new PhoneRecord("empty","empty", 9999999);
         records[9] = new PhoneRecord("Van Vliet","Margreet", 1244570);
         // call menu
         } // end main
    // menu
    static void Menu(PhoneRecord[] records)
         int option;
         // menu options
         System.out.println(" ");
         System.out.println("1. Find record (Advanced Search) ");
         System.out.println("2. Quick Search ");
         System.out.println("3. Add a record ");
         System.out.println("4. Show database ");
         System.out.println("5. Sort database ");
         System.out.println("6. Exit     ");
         System.out.println(" ");
         System.out.println(" ");
         System.out.println("Choose a number ");
         option = Console.readInt();
         // every menu option has its own method
         if (option == 1)
         if (option == 2)
         else if (option == 3)
         else if (option == 4)
         else if (option == 5)
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         // if 6 then terminate the program
    } // end of menu
    // menu option 1: Find a record - using linear search
    static void FindRecord(PhoneRecord[] records)
    int option;
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    System.out.println("1. First Name");
    System.out.println("2. Last Name");
    System.out.println("3. End search");
    option = Console.readInt();
         // option 1 is search based on first name
         if (option == 1)
              System.out.println("Enter First Name");
              String first = Console.readString();
              int notthere = -1;
              for (int i=0; i < 10; i++)
                   if (first.equals(records.first_Name))
                        System.out.println(records[i].last_Name + ", " + records[i].first_Name);
                   // if a record is found, the variable notthere will be > -1
                   notthere = i;
              } // end search array
                   // if notthere is -1, then there is no record available
                   if (notthere < 0)
                   System.out.println("No record available");
         } // end option 1 First Name
         // option 2 allows the user to search based on last name
         else if (option == 2)
              System.out.println("Enter Last Name");
              String last = Console.readString();
              int notthere = -1;
              for (int i=0; i < 10; i++)
                   if (last.equals(records[i].last_Name))
                        System.out.println(records[i].last_Name + ", " + records[i].first_Name);
                   notthere = i;
                   // if notthere is -1 then there is no record available
                   // if notthere is > -1 then there is a record available
                   } // end search array
                   if (notthere < 0)
                   System.out.println("No record available");
         } // end option 2 Last Name
              // if the user types in a wrong number, he or she returns to the menu
    while (option != 3);
    } // end FindRecord
    // menu option 2: Quick Search - using binary search
    static void QuickSearch(PhoneRecord[] records)
         // Sort array - Using Quicksort
         quickSort(records, 0, 9);
         // allow user to enter the last name
         System.out.println("Enter Last Name");
         String last = Console.readString();
         // use binary search to find the target
         int index = binarySearch(records, last);
         // -1 means that there are no records
         if (index == -1)
         System.out.println("No record available");
         // print out the record
         System.out.println(records[index].last_Name + ", " + records[index].first_Name);
         // return to menu
    } // end QuickSearch
    public static int binarySearch( Comparable [ ] a, Comparable x )
    int low = 0;
    int high = 9;
    int mid;
    while( low <= high )
    mid = ( low + high ) / 2;
    if( a[ mid ].compareTo( x ) < 0 )
    low = mid + 1;
    else if( a[ mid ].compareTo( x ) > 0 )
    high = mid - 1;
    return mid;
    return -1; // not found
    // menu option 3: Add a record
    static void AddRecord(PhoneRecord[] records)
    int option;
    int index = 0;
    // enter details
         // to say that the array is not full yet, I use the variable filled
         int filled = 0;
    System.out.println("Enter the First Name");
    String frst = Console.readString();
    System.out.println("Enter the Last Name");
    String lst = Console.readString();
    System.out.println("Enter the phone number");
    int phn = Console.readInt();
    // search the array for the empty slot
         for (int i=0; i < 10; i++)
              if (records[i].first_Name.equals("empty") && filled == 0)
              records[i].first_Name = frst;
              records[i].last_Name = lst;
              records[i].phonenumber = phn;
              filled = 1;
         // Sort array - Using Quicksort
         quickSort(records, 0, 9);
         // Print out sorted values
         for(int i = 0; i < records.length; i++)
              System.out.println(records[i].last_Name + ", " + records[i].first_Name);
         System.out.println("Do you want to add more records?");
         System.out.println("1. Yes");
         System.out.println("2. No");
         option = Console.readInt();
         if (option == 2)
         // sets the database to full
         int empty = 0;
         for (int i=0; i < 10; i++)
              // empty = 1 means that there is an empty slot
              if (records[i].first_Name.equals("empty"))
                   empty = 1;
         // if the system didn't find an empty slot, the database must be full
         if (empty == 0)
         System.out.println("Database is full");
         option = 2;
    while (option != 2);
    } // end AddRecord
    // menu option 4: Show database
    static void ShowDatabase(PhoneRecord[] records)
              // shows the entire database
              for (int i=0; i < 10; i++)
                        System.out.println(records[i].last_Name + ", " + records[i].first_Name);
    // Sort array
    public static void quickSort (Comparable[] a, int left, int right)
         // Sort a[left?right] into ascending order.
         if (left < right) {
         int p = partition(a, left, right);
         quickSort(a, left, p-1);
         quickSort(a, p+1, right);
    static int partition (Comparable[] a, int left, int right)
         // Partition a[left?right] such that
         // a[left?p-1] are all less than or equal to a[p], and
         // a[p+1?right] are all greater than or equal to a[p].
         // Return p.
         Comparable pivot = a[left];
         int p = left;
         for (int r = left+1; r <= right; r++) {
         int comp = a[r].compareTo(pivot);
         if (comp < 0) {
         a[p] = a[r]; a[r] = a[p+1];
    a[p+1] = pivot; p++;          }
         return p;
    } // end class PhoneBook
    PhoneRecord class:
    public class PhoneRecord implements Comparable
    public int phonenumber;
    public String last_Name;
    public String first_Name;
    public PhoneRecord(String last_Name, String first_Name, int phonenumber)
    this.last_Name = last_Name;
    this.phonenumber = phonenumber;
    this.first_Name = first_Name;
    /* Overload compareTo method */
    public int compareTo(Object obj)
    PhoneRecord tmp = (PhoneRecord)obj;
    // sorting based on last name
    String string1 = this.last_Name;
    String string2 = tmp.last_Name;
    int result = string1.compareTo(string2);
    if(result < 0)
    /* instance lt received */
    return -1;
    else if(result > 0)
    /* instance gt received */
    return 1;
    /* instance == received */
    return 0;

    JosAH wrote:
    prometheuzz wrote:
    bats wrote:
    I keep getting a java.lang.ClassCastException with these two classes when I try to perform a binary search. Any tips?
    ...Looking at your binary search method:
    public static int binarySearch( Comparable [ ] a, Comparable x)I see it expects to be fed Comparable objects. So, whatever you're feeding it, it's not a Comparable (ie: it doesn't implement Comparable), hence the CCE.It's even worse: if an A is a B it doesn't make an A[] a B[].
    kind regards,
    JosMy post didn't make much sense: if there were no Comparables provided as an argument, it would have thrown a compile time error. The problem lies in the compareTo(...) method.

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