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