Answers for "most efficient binary search code in java"

12

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); 
	} 
}
Posted by: Guest on April-02-2020
2

binary search java

binary search program in java.
public class BinarySearchExample
{
   public static void binarySearch(int[] arrNumbers, int start, int end, int keyElement)
   {
      int middle = (start + end) / 2;
      while(start <= end)
      {
         if(arrNumbers[middle] < keyElement)
         {
            start = middle + 1;
         }
         else if(arrNumbers[middle] == keyElement)
         {
            System.out.println("Element found at index: " + middle);
            break;
         }
         else
         {
            end = middle - 1;
         }
         middle = (start + end) / 2;
      }
      if(start > end)
      {
         System.out.println("Element not found!");
      }
   }
   public static void main(String[] args)
   {
      int[] arrNumbers = {14,15,16,17,18};
      int keyElement = 16;
      int end = arrNumbers.length - 1;
      binarySearch(arrNumbers, 0, end, keyElement);
   }
}
Posted by: Guest on October-23-2020

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