A given number can be said palindromic in nature if the reverse of the given number is the same as that of a given number. In this article, we will write a Program to check if a number is a Palindrome Number in Java.
Example of Palindrome Number
Input : n = 46355364
Output: Reverse of n = 46355364
Palindrome : YesInput : n = 12345678999999999987654321
Output: Reverse of n = 12345678999999999987654321
Palindrome : Yes
Methods to Check If a Number is Palindrome Number in Java
There are certain methods to check if a Number is a Palindrome Number in Java as mentioned below:
- Using Iterative Approach
- Using Recursive Approach
- Using BigInteger Class
- Using String
1. Check for Palindrome Number using Iteration
We can reverse a number in multiple ways, below is the iterative implementation for the same.
Algorithm
Iterative Algorithm when the given input is an integer number.
- Initialize reversed_number = 0
- Loop while number > 0
- Multiply reversed_number by 10 and add number % 10 to reversed_number
- reversed_number = reversed_number*10 + number %10;
- Divide the number by 10
- Return reversed_number
Program to Check for Palindrome Numbers in Java
Java
// Java program to reverse a number// and find if it is a palindrome or not // Driver Classclass GFG { // Iterative function to // reverse the digits of number static int reversNumber(int n) { int reversed_n = 0; while (n > 0) { reversed_n = reversed_n * 10 + n % 10; n = n / 10; } return reversed_n; } // Main function public static void main(String[] args) { int n = 123464321; int reverseN = reversNumber(n); System.out.println("Reverse of n = " + reverseN); // Checking if n is same // as reverse of n if (n == reverseN) System.out.println("Palindrome = Yes"); else System.out.println("Palindrome = No"); }} |
Reverse of n = 123464321 Palindrome = Yes
The complexity of the above method
Time Complexity: O(log10(n)) where n is the input number.
Auxiliary space: O(1) because constant variables have been used
2. Check for Palindrome Numbers using Recursion
The idea is to solve this problem by using recursion in a different way, Below are the steps:
- Define a recursive method recursive_func().
- It should have two parameters as the given number N and the resultant reverse number as rev.
- Recursively reversing the number until N becomes less than 10 (base condition).
- Finally returning the reversed number.
Program to Check for Palindrome Numbers in Java
Java
// Java program to reverse a number// and find if it is a palindrome or not // Driver Classclass GFG { // Recursive function to reverse // the digits of number static int recursive_func(int n, int rev) { if (n < 10) { return rev * 10 + n; } else { int last_digit = n % 10; rev = rev * 10 + last_digit; return recursive_func(n / 10, rev); } } // Driver Code public static void main(String[] args) { int n = 123464321; int rev = recursive_func(n, 0); System.out.println("Reverse of n = " + rev); // Checking if n is same // as reverse of n if (n == rev) System.out.println("Palindrome = Yes"); else System.out.println("Palindrome = No"); }} |
Reverse of n = 123464321 Palindrome = Yes
The complexity of the above method is mentioned below:
Time Complexity: O(log10N)
Space Complexity: O(n) n-no of digits in a number [stack space]
3. Check for Palindrome Number using BigInteger Class
Big Integer can’t be operated using the same as the normal Integer so we need to use BigInteger Class to solve the issue faced like Overflowing of the values. The length of the number is log10(n), i.e. For BigIntegers using string operations like creation reverse and check palindrome will take log10(n) time.
Approach
- Take input in BigInteger variable.
- Reverse the given BigInteger using the reverse method.
- Compare both BigIntegers using compareTo() method.
Program to Check for Palindrome Numbers in Java
Java
// Java program to reverse a number// and find if it is a palindrome or notimport java.io.*;import java.math.BigInteger; // Driver Classclass GFG { // Reverse Big Integer public static BigInteger reverse(BigInteger n) { String s = n.toString(); StringBuilder sb = new StringBuilder(s); return new BigInteger(sb.reverse().toString()); } // Main Function public static void main(String[] args) { BigInteger n = new BigInteger("12345678999999999987654321"); BigInteger reverseN = reverse(n); System.out.println("Reverse of n = " + reverseN); // Checking if n is same // as reverse of n if (n.compareTo(reverseN) == 0) System.out.println("Palindrome = Yes"); else System.out.println("Palindrome = No"); }} |
Reverse of n = 12345678999999999987654321 Palindrome = Yes
Time Complexity: O(log10(n)) where n is the input number.
Auxiliary Space: O(n) for StringBuilder
4. Using String to Check if a Number is Palindrome
We can convert a number into String, and after conversion, we can use the functions available with Strings for conversions.
Program to Check for Palindrome Numbers in Java
Java
// Java Program to Check if a Number// is Palindrome using String Classimport java.io.*;import java.util.*; // Driver Classclass GFG { // main function public static void main(String[] args) { String s; // Taking input using Scanner Class Scanner in = new Scanner(System.in); System.out.print("Enter Number = "); // Storing the the input value in s s = in.nextLine(); // Length of the String int n = s.length(); String rev = ""; for (int i = n - 1; i >= 0; i--) { rev = rev + s.charAt(i); } // Printing the reversed Number System.out.println("Reverse Number = " + rev); // Checking Palindrome if (s.equals(rev)) System.out.println("Palindrome = Yes"); else System.out.println("Palindrome = No"); }} |
Input:
Enter Number = 12345654321
Output:
Reverse Number = 12345654321
Palindrome = Yes
For more information, please refer to the complete article here.
