recursive binary search in c++ code example
Example 1: binary search program c++
#include <iostream>
using namespace std;
// This program performs a binary search through an array, must be sorted to work
int binarySearch(int array[], int size, int value)
{
int first = 0, // First array element
last = size - 1, // Last array element
middle, // Mid point of search
position = -1; // Position of search value
bool found = false; // Flag
while (!found && first <= last)
{
middle = (first + last) / 2; // Calculate mid point
if (array[middle] == value) // If value is found at mid
{
found = true;
position = middle;
}
else if (array[middle] > value) // If value is in lower half
last = middle - 1;
else
first = middle + 1; // If value is in upper half
}
return position;
}
int main ()
{
const int size = 5; // size initialization
int array[size] = {1, 2, 3, 4, 5}; // declare array of size 10
int value; // declare value to be searched for
int result; // declare variable that will be returned after binary search
cout << "What value would you like to search for? "; // prompt user to enter value
cin >> value;
result = binarySearch(array, size, value);
if (result == -1) // if value isn't found display this message
cout << "Not found\n";
else // If value is found, displays message
cout << "Your value is in the array.\n";
return 0;
}
Example 2: dichotomic search c++
using namespace std;
// A recursive binary search function. It returns
// location of x in given array arr[l..r] is present,
// otherwise -1
int binarySearch(int arr[], int l, int r, int x)
{
if (r >= l) {
int mid = l + (r - l) / 2;
// 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;
}
int main(void)
{
int arr[] = { 2, 3, 4, 10, 40 };
int x = 10;
int n = sizeof(arr) / sizeof(arr[0]);
int result = binarySearch(arr, 0, n - 1, x);
(result == -1) ? cout << "Element is not present in array"
: cout << "Element is present at index " << result;
return 0;
}