Recursive program to check if a number is a palindrome in C++

We get an integer as input. The goal is to use recursion to determine whether the input number Num is a palindrome.

To check if a number is a palindrome, reverse the number and check if both numbers are the same. If the reversed number is equal to the original number, it is a palindrome.

Example

Input− Num = 34212;

Output− 34212 is not a palindrome!

Explanation− If we reverse 34212, we get 21243. 34212 != 21243 Therefore the entered number is not a palindrome.

Input− Num = 32123;

Output− 32123 is a palindrome!

Explanation - If we reverse 32123, we get 32132. 32123!= 32123, so the input number is a palindrome.

The method used in the program below is as follows

For base case -: If num1 is 0, return num2.

p>

Else-: Use recursion to calculate the reverse order of num1. Returns the reciprocal of the calculation.

If both are the same, the input number is a palindrome.

• Get the input number Num.

  • Get the input number Num. p>

  • Get Num2 = revrsNum(Num,0)

  • The function revrsNum(int num1, int num2) recursively generates the inverse value of num1, and Returns the reversed number.

  • If num1 is 0, the inverted calculation result returns num2.

  • Otherwise multiply num2 by 10 and add num1 .

  • Use num1=num1/10 to reduce num1 by 10.

  • Use revrsNum(recursive num1, num2);

  • to return the result.

  • Print the results obtained inside main.

  Example

  #include 
  using namespace std;
  int revrsNum(int num1, int num2){
     if (num1 == 0){
        return num2;
     }
     num2 *= 10;
     num2 += (num1 % 10);
     num1 = num1/10;
     return revrsNum(num1, num2);
  }
  int main(){
     int Num = 1345431;
     int Num2 = revrsNum(Num,0);
     if (Num == Num2){
        cout <
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  Output

  If we run the above code it will generate the following output

  1345431 is Palindrome!

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