Binary Search implementation in java?

Where you use primitive[] in your post, I assume you mean one of the overloads of Arrays.binarySearch that takes an array of one of the primitive types (byte, short, int, long, float, double or char) as a parameter rather than an Object array.

In both cases, the array needs to be sorted first before applying the binarySearch method otherwise the results will be undefined. However, this may not be possible with an Object[] as it could contain both strings and numbers which are not comparable with one another. So, in that event, you'd get an undefined result.

Here's an example:

import java.util.Arrays;

public class BinarySearchDemo

{

public static void main(String[] args)

{

// create unsorted array of ints

int ia[] = {10, 2, 7, 8, 4};

// sort it

Arrays.sort(ia);

// print its elements

System.out.print("The sorted int array is : ");

for (int i : ia)

{

System.out.print(" " + i);

}

System.out.println();

// enter a value to look for

int search = 7;

int index = Arrays.binarySearch(ia, search);

System.out.println("The index of " + search + " is : " + index);

// now create an Object array which includes a string

Object[] oa = {"hello", 10, 2, 7, 8, 4};

// can't sort this without getting an exception so don't bother

// print its elements

System.out.print("The unsorted object array is : ");

for (Object o : oa)

{

System.out.print(" " + o);

}

System.out.println();

// look for 'search' again

index = Arrays.binarySearch(ia, search);

System.out.println("The index of " + search + " is now : " + index);

}

}

When I compiled and ran this, the output was:

The sorted int array is : 2 4 7 8 10

The index of 7 is : 2

The unsorted object array is : hello 10 2 7 8 4

The index of 7 is now : 2

The last line is, of course, wrong since 7 is at index 3 in the (unsorted) array.

2 Replies

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Vulpes12y ago
Accepted answer
Where you use primitive[] in your post, I assume you mean one of the overloads of Arrays.binarySearch that takes an array of one of the primitive types (byte, short, int, long, float, double or char) as a parameter rather than an Object array.

In both cases, the array needs to be sorted first before applying the binarySearch method otherwise the results will be undefined. However, this may not be possible with an Object[] as it could contain both strings and numbers which are not comparable with one another. So, in that event, you'd get an undefined result.

Here's an example:

import java.util.Arrays;

public class BinarySearchDemo
{

public static void main(String[] args)
{

// create unsorted array of ints
int ia[] = {10, 2, 7, 8, 4};

// sort it
Arrays.sort(ia);

// print its elements
System.out.print("The sorted int array is : ");
for (int i : ia)
{
System.out.print(" " + i);
}

System.out.println();

// enter a value to look for
int search = 7;

int index = Arrays.binarySearch(ia, search);
System.out.println("The index of " + search + " is : " + index);

// now create an Object array which includes a string
Object[] oa = {"hello", 10, 2, 7, 8, 4};

// can't sort this without getting an exception so don't bother

// print its elements
System.out.print("The unsorted object array is : ");
for (Object o : oa)
{
System.out.print(" " + o);
}

System.out.println();

// look for 'search' again
index = Arrays.binarySearch(ia, search);
System.out.println("The index of " + search + " is now : " + index);
}
}

When I compiled and ran this, the output was:

The sorted int array is : 2 4 7 8 10
The index of 7 is : 2
The unsorted object array is : hello 10 2 7 8 4
The index of 7 is now : 2

The last line is, of course, wrong since 7 is at index 3 in the (unsorted) array.
Abhijeet Singh12y ago
Thank you so much Sir ... :)