Given an integer N, the task is to reverse the digits of given integer using recursion.
Examples:
Input: N = 123
Output: 321
Explanation:
The reverse of the given number is 321.Input: N = 12532
Output: 23521
Explanation:
The reverse of the given number is 23521.
Approach: Follow the steps below to solve the problem:
- Recursively iterate every digit of N.
- If the current value of N passed is less than 10, return N.
if(num < 10) return N;
- Otherwise, after each recursive call (except the base case), return the recursive function for next iteration:
return reverse(N/10) + ((N%10)*(pow(10, (floor(log10(abs(N))))))) where, floor(log10(abs(x))) gives the count of digits of x ((x%10)*(pow(10, (floor(log10(abs(x))))))) places the extracted unit place digits (x%10) to their desired positions
Below is the implementation of the above approach:
// C program for the above approach #include <math.h> #include <stdio.h> #include <stdlib.h> // Function to reverse the digits of // the given integer int reverse( int N)
{ return ((N <= 9))
? N
: reverse(N / 10)
+ ((N % 10)
* ( pow (10,
( floor ( log10 (
abs (N)))))));
} // Utility function to reverse the // digits of the given integer void reverseUtil( int N)
{ // Stores reversed integer
int result = reverse(N);
// Print reversed integer
printf ( "%d" , result);
} // Driver Code int main()
{ // Given integer N
int N = 123;
// Function Call
reverseUtil(N);
return 0;
} |
321
Time Complexity: O(log10N)
Auxiliary Space: O(log10N) for call stack