Java Source Code: Binary Search in Recursion
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Below is the java source code for a recursive binary search. The program requires to sort the numbers first in an ascending order before the binary search can be implemented successfully. Basically, what the program does is prompt the user to input the size of the array, enter the numbers that the user wants, sort it, then ask for the number he wants to search. If the number is on the array, it outputs the index of the array where it is located but if it is not on the list is simple returns -1.Here how the program works.
How the program works
Important note: Binary Search algorithm would work only if the numbers are sorted.
For example:
User's enters: 2 1 3 5 4
The sorted numbers are: 1 2 3 4 5
Search number: 2
The numbers is on the array index 1, since array index starts from 0. if the search number is 6, then the program returns -1, since 6 is not on the list.
How Binary Search Works?
1. It computes the middle index of the array
From the given numbers above, since there are 5 numbers, we will divide it into 2 to get the middle.
middle = 5/2, returns 2, since they are both integers so any decimal value will not be counted.
meaning on: 1 2 3 4 5, 3 is the middle, Since array index starts from 0.
2. Compare if the search number is equal to the middle if, yes return the index middle.
if(middle == search)
return index middle.
3. If not, it then compare if the search number is greater than or less than the middle. if it is greater than the value of the middle, the search starts at the middle + 1 number till at the last number in the array, however if the search number is less than the middle, it compares next the number located on the middle - 1 index value up to the first value of the array.
The Source code is given below.
Java Source code: How to implement Binary Search Using Recursion.
//Java Class
public class binarySearch //Example of Java Class
{
//biSearch is an example of a method or a function
public int binSearch(int[] arr, int fIndex, int lIndex,int search)
{
int middle = (fIndex + (lIndex - fIndex) / 2);
if(fIndex<lIndex ){
if (search == arr[middle]){
return middle;
}
else if(search < arr[middle]){
if(search == arr[0])
return 0;
return binSearch(arr, fIndex, middle, search);
}
else if(search > arr[middle]){
if(search == arr[middle+1])
return middle + 1;
return binSearch(arr, middle+1, lIndex, search);
}
}
return -1;
}
//this is also a class method
public void sort(int[] arr)
{
for(int i=0; i<arr.length; i++)
{
for(int j=i+1; j<arr.length; j++ )
{
if(arr[i] > arr[j])
{
int temp = arr[j];
arr[j]=arr[i];
arr[i]= temp;
}
}
}
for(int i=0; i<arr.length; i++)
{
System.out.print(arr[i] + " ");
}
}
}
//main class
import java.util.Scanner;
public class Main {
public static void main(String[] args)
{
Scanner input = new Scanner(System.in);
System.out.print("Enter the size of the array: ");
int n = input.nextInt();
int[] x = new int[n];
System.out.print("Enter "+ n +" numbers: ");
int middle;
for(int i=0; i<n; i++)
{
x[i] = input.nextInt();
}
binarySearch access = new binarySearch(); //this is how to instantiate an object to access a class
System.out.println("The sorted numbers are: ");
access.sort(x);//this is how to access a method
System.out.println();
System.out.print("Enter the number you want to search: ");
int value = input.nextInt();
System.out.print("The search number is on the index ");
System.out.print(access.binSearch(x, 0, x.length-1, value)); //how to access a class
}
}
Sample Output 1:
Enter the size of the array: 5
Enter 5 numbers: 2 1 3 5 4
The sorted numbers are:
1 2 3 4 5
Enter the number you want to search: 3
The search number is on the index 2
Sample Output 2:
Enter the size of the array: 5
Enter 5 numbers: 3 2 1 4 5
The sorted numbers are:
1 2 3 4 5
Enter the number you want to search: 1
The search number is on the index 0
Sample Output 3:
Enter the size of the array: 5
Enter 5 numbers: 2 1 3 5 4
The sorted numbers are:
1 2 3 4 5
Enter the number you want to search: 6
The search number is on the index -1
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Comments
AUTHORHi payal;
I'm so sorry for the pause, numbers are sorted in binary search mainly because it is what binary search is. The algorithm of binary search is to sort the numbers first so that it can determine the middle value of the input numbers and the index that holds it. When the value is determined and its index, it would then be use as a comparison to determine the index of the search number. To understand more about what I am talking about, please review again and trace the codes above :D
thanks for sharing your knowledge...
but why the numbers must be sorted for binary search?
please answer me...
- AUTHOR
Hi hillymillydee,
Nah thanks for stopping by and for commenting, You are thoughtful hilly, you belong to those who inspires me...:)
Always have a great day and happy writing...^_^
Aisha,
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