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Thread: Binary Search for an array of random ints - Can't find the target value

  1. #1
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    Default Binary Search for an array of random ints - Can't find the target value

    Writing a binary search for an array of random ints. When in comes time to pick a target value and search for it, the program will not identify the number. If you run it you'll see what I mean

    /*
    	Exercise4.java
     
    	Binary Search
     
    */
     
     
    import java.io.*;
    import java.util.*;
     
    public class Exercise4
    {
    	private static final int NOT_FOUND = -1; // this value GLOBAL to file - visible in ALL methods in this file
     
    	public static void main( String args[] )
    	{
    		Scanner kbd = new Scanner(System.in);
    		Random rand = new Random();
     
    		if (args.length < 1 )
    		{
    			System.out.println("\n\n      !! You must enter a desired array length on the cmd line !!\n");
    			System.exit(0);
    		}
     
    		int array[] = new int[ Integer.parseInt(args[0]) ];
    		int count=0;
     
            for (int i=0 ; i < array.length ; ++i )
            {
                int r = 1 + rand.nextInt(100);
                if ( binarySearch( array, r ) == NOT_FOUND )
                {
                    insertInOrder( array, count, r  );
                    ++count;
                }
                // else ignore this value its already in the array - no dupes to be inserted
            }
     
    		printArray( array );
     
    		do
    		{
    			System.out.print("number to search for (negative to quit)? ");
    			int target = kbd.nextInt();
    			if (target < 0) break;
     
    			int index=binarySearch( array, target );
     
    			if (index >= 0)
    				System.out.printf("%d FOUND in array at index position %d\n",target,index );
    			else
    				System.out.printf("%d NOT found in array\n",target );
    		} while (true);
     
    	} // END MAIN
     
     
    	private static void printArray( int array[] )
    	{
    		System.out.printf("\nArray has %d values:\n",array.length );
    		for( int i=0 ; i<array.length ;++i )
    			System.out.print(array[i] + " " );
    		System.out.println("\n");
    	}
     
    	// puts the new value into the array at its proper sorted position by shuffline higher values up one place
    	//	and opening up a hole where the new val goes. All values are kept contiguous from the front
     
    	private static void insertInOrder( int array[], int count, int newVal )
    	{
    		int i;
    		for (i=count-1; i>=0 && newVal<array[i] ; --i)
    			array[i+1] = array[i];
     
    		array[i+1] = newVal;
    	}
     
    	// if target not found returns -1
    	// else returns index position where target found
     
    	public static int binarySearch( int array[], int target )
    	{
    	int low=0;
    	int high=array.length;
    	int spot=(low+high)/2;
    	int i=0;
     
    	while (i==0)
    	{
    	if (array[spot]<target)
    	{
    	low=spot+1;
    	break;
    	}
     
    	else if(array[spot]>target)
    	{
    	high=spot-1;
    	break;
    	}
     
    	else if (array[spot]==target)
    	return spot;
     
    	}
    	return NOT_FOUND;
     
    	}
     
    } //END class


  2. #2
    Crazy Cat Lady KevinWorkman's Avatar
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    Default Re: Binary Search for an array of random ints - Can't find the target value

    What exactly do you mean "when in comes time to pick a target value and search for it"? Are you having trouble with the actual search, or reading in a value, or something else?
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  3. #3
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    Default Re: Binary Search for an array of random ints - Can't find the target value

    I mean that the user picks a value to search for in the array. My problem is in the search method

  4. #4
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    Default Re: Binary Search for an array of random ints - Can't find the target value

    Well, what have you done to debug it? Have you stepped through it with a debugger, or at least added some print statements to figure out what's going on? When does the flow of the program differ from what you expect?

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