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1 | 1 | package com.codefortomorrow.advanced.chapter13.examples;
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2 | 2 |
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3 | 3 | public class BinarySearch {
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4 |
| - // Java implementation of recursive Binary Search |
5 |
| - // Returns index of x if it is present in arr[l.. |
6 |
| - // r], else return -1 |
7 |
| - public static int binarySearch(int arr[], int l, int r, int x) |
8 |
| - { |
9 |
| - if (r >= l) { |
10 |
| - int mid = l + (r - l) / 2; |
11 |
| - |
12 |
| - // If the element is present at the |
13 |
| - // middle itself |
14 |
| - if (arr[mid] == x) |
15 |
| - return mid; |
16 |
| - |
17 |
| - // If element is smaller than mid, then |
18 |
| - // it can only be present in left subarray |
19 |
| - if (arr[mid] > x) |
20 |
| - return binarySearch(arr, l, mid - 1, x); |
21 |
| - |
22 |
| - // Else the element can only be present |
23 |
| - // in right subarray |
24 |
| - return binarySearch(arr, mid + 1, r, x); |
25 |
| - } |
26 |
| - |
27 |
| - // We reach here when element is not present |
28 |
| - // in array |
29 |
| - return -1; |
30 |
| - } |
31 | 4 |
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32 |
| - // Driver method to test above |
33 |
| - public static void main(String args[]) |
34 |
| - { |
35 |
| - int arr[] = { 2, 3, 4, 10, 40 }; |
36 |
| - int n = arr.length; |
37 |
| - int x = 10; |
38 |
| - int result = binarySearch(arr, 0, n - 1, x); |
39 |
| - if (result == -1) |
40 |
| - System.out.println("Element not present"); |
41 |
| - else |
42 |
| - System.out.println("Element found at index " + result); |
| 5 | + // Java implementation of recursive Binary Search |
| 6 | + // Returns index of x if it is present in arr[l.. |
| 7 | + // r], else return -1 |
| 8 | + public static int binarySearch(int arr[], int l, int r, int x) { |
| 9 | + if (r >= l) { |
| 10 | + int mid = l + (r - l) / 2; |
| 11 | + |
| 12 | + // If the element is present at the |
| 13 | + // middle itself |
| 14 | + if (arr[mid] == x) return mid; |
| 15 | + |
| 16 | + // If element is smaller than mid, then |
| 17 | + // it can only be present in left subarray |
| 18 | + if (arr[mid] > x) return binarySearch(arr, l, mid - 1, x); |
| 19 | + |
| 20 | + // Else the element can only be present |
| 21 | + // in right subarray |
| 22 | + return binarySearch(arr, mid + 1, r, x); |
43 | 23 | }
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| 24 | + |
| 25 | + // We reach here when element is not present |
| 26 | + // in array |
| 27 | + return -1; |
| 28 | + } |
| 29 | + |
| 30 | + // Driver method to test above |
| 31 | + public static void main(String args[]) { |
| 32 | + int arr[] = { 2, 3, 4, 10, 40 }; |
| 33 | + int n = arr.length; |
| 34 | + int x = 10; |
| 35 | + int result = binarySearch(arr, 0, n - 1, x); |
| 36 | + if (result == -1) System.out.println( |
| 37 | + "Element not present" |
| 38 | + ); else System.out.println("Element found at index " + result); |
| 39 | + } |
44 | 40 | /* This code is contributed by Rajat Mishra
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45 | 41 | https://www.geeksforgeeks.org/binary-search/ */
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46 | 42 |
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