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) Â Â Â 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
// C program for the above approachÂ
#include <math.h>#include <stdio.h>#include <stdlib.h>Â
// Function to reverse the digits of// the given integerint reverse(int N){    return ((N <= 9))               ? N               : reverse(N / 10)                     + ((N % 10)                        * (pow(10,                               (floor(log10(                                   abs(N)))))));}Â
// Utility function to reverse the// digits of the given integervoid reverseUtil(int N){    // Stores reversed integer    int result = reverse(N);Â
    // Print reversed integer    printf("%d", result);}Â
// Driver Codeint main(){Â Â Â Â // Given integer NÂ Â Â Â int N = 123;Â
    // Function Call    reverseUtil(N);Â
    return 0;} |
Output:
321
Time Complexity: O(log10N)
Auxiliary Space: O(log10N) for call stack
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