51 lines
1.8 KiB
Java
51 lines
1.8 KiB
Java
// Java implementation of recursive Binary Search
|
|
class BinarySearch {
|
|
// Returns index of x if it is present in arr[l..r],
|
|
// else return -1
|
|
int binarySearch(int arr[], int l, int r, int x)
|
|
{
|
|
if (r >= l) {
|
|
int mid = l + (r - l) / 2;
|
|
//We use (l + (r - l)) rather than using (l - r) to avoid arithmetic overflow.
|
|
//Arithmetic overflow is the situation when the value of a variable increases
|
|
//beyond the maximum value of the memory location, and wraps around.
|
|
|
|
// If the element is present at the
|
|
// middle itself
|
|
if (arr[mid] == x)
|
|
return mid;
|
|
|
|
// If element is smaller than mid, then
|
|
// it can only be present in left subarray
|
|
if (arr[mid] > x)
|
|
return binarySearch(arr, l, mid - 1, x);
|
|
|
|
// Else the element can only be present
|
|
// in right subarray
|
|
return binarySearch(arr, mid + 1, r, x);
|
|
}
|
|
|
|
// We reach here when element is not present
|
|
// in array
|
|
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, 0, n - 1, x);
|
|
if (result == -1)
|
|
System.out.println("Element not present");
|
|
else
|
|
System.out.println("Element found at index " + result);
|
|
}
|
|
}
|
|
|
|
// For running in terminal rename this file to BinarySearch.java
|
|
//then run the commands <javac BinarySearch.java> followed by <java BinarySearch>
|
|
//It will generate and a BinarySearch.class file which is a file containing java bytecode that is executed by JVM.
|