implement binary search in java code example

Example 1: binary search java

// Java implementation of iterative Binary Search 
class BinarySearch { 
	// Returns index of x if it is present in arr[], 
	// else return -1 
	int binarySearch(int arr[], int x) 
	{ 
		int l = 0, r = arr.length - 1; 
		while (l <= r) { 
			int m = l + (r - l) / 2; 

			// Check if x is present at mid 
			if (arr[m] == x) 
				return m; 

			// If x greater, ignore left half 
			if (arr[m] < x) 
				l = m + 1; 

			// If x is smaller, ignore right half 
			else
				r = m - 1; 
		} 

		// if we reach here, then element was 
		// not present 
		return -1; 
	} 

	// Driver method to test above 
	public static void main(String args[]) 
	{ 
		BinarySearch ob = new BinarySearch(); 
		int arr[] = { 2, 3, 4, 10, 40 }; 
		int n = arr.length; 
		int x = 10; 
		int result = ob.binarySearch(arr, x); 
		if (result == -1) 
			System.out.println("Element not present"); 
		else
			System.out.println("Element found at "
							+ "index " + result); 
	} 
}

Example 2: binary search iterative

// Binary Search using Iterative Approach

import java.io.*;
class Binary_Search
{
	public static void main(String[] args) throws Exception
	{
		Binary_Search obj = new Binary_Search();
		InputStreamReader isr = new InputStreamReader(System.in);
		BufferedReader br = new BufferedReader(isr);
		System.out.println("Insert the length of the Array : ");
		int n = Integer.parseInt(br.readLine());
		int arr[] = new int[n];
		System.out.println("Insert elements into the array : ");
		for(int i=0;i<n;i++)
		{
			arr[i] = Integer.parseInt(br.readLine());
		}
		System.out.println("Enter the num which you want to Search : ");
		int num = Integer.parseInt(br.readLine());
		obj.logic(arr,num);
	}
	void logic(int arr[],int num)
	{
		int r = arr.length - 1;
		int l = 0;
		int mid;
		while(l<=r)
		{
			mid = l + (r-l)/2;
			if(arr[mid] == num)
			{
				System.out.println("Number found at "+mid+"th index");
				break;
			}
			else if(arr[mid]>num)
			{
				r = mid - 1;
			}
			else
			{
				l = mid + 1;
			}
		}
	}
}