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Palindrome Program in Java — A Clear Guide (with Examples)

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Palindrome Program in Java — A Clear Guide (with Examples)

Introduction

A palindrome is a sequence that reads the same forward and backward. Examples: "madam", "racecar", and numeric palindromes like 121 or 1331. Palindromes are a common exercise for beginners learning string handling, loops, and algorithms. In Java, you can check palindromes in several ways: reversing and comparing, two-pointer checking, recursion, or numeric reversal for integers. Below are explanations and practical Java programs you can copy, run, and adapt.

Why learn palindrome checks?

Teaches string manipulation and indexing.
Introduces two-pointer technique (important for many algorithms).
Encourages thinking about edge cases (case, spaces, punctuation, negative numbers).
Useful in interview practice and small utilities (e.g., validating symmetric data).

Key approaches (high level)

Reverse and compare — reverse the string and compare with original. Easy and readable.
Two-pointer — compare characters from both ends moving inward; more memory-efficient.
Numeric reverse — for integers, reverse digits and compare (watch overflow and negatives).
Ignoring non-alphanumeric + case-insensitive checks — common in real problems (like palindrome phrases).

Example 1 — Palindrome Program (String) — two methods

Method A: Using `StringBuilder.reverse()` (simple)

public class PalindromeCheck {

    public static boolean isPalindromeUsingReverse(String s) {

        if (s == null) return false;

        String cleaned = s; // optionally clean: remove spaces/punctuation, and lowercase

        String reversed = new StringBuilder(cleaned).reverse().toString();

        return cleaned.equals(reversed);

    }

public static void main(String[] args) { String[] tests = {"madam", "Racecar", "step on no pets", "hello"}; for (String t : tests) { String cleaned = t.replaceAll("[^A-Za-z0-9]", "").toLowerCase(); System.out.printf("\"%s\" -> %b%n", t, isPalindromeUsingReverse(cleaned)); } } }

Notes:

We cleaned the string to remove non-alphanumeric characters and lowercase it before checking, so "Racecar" and "step on no pets" will be treated as palindromes.
This method uses extra memory for the reversed string (O(n) space).

Method B: Two-pointer (no extra string)

public class PalindromeCheck {

    public static boolean isPalindromeTwoPointer(String s) {

        if (s == null) return false;

        int i = 0, j = s.length() - 1;

        while (i < j) {

            // skip non-alphanumeric if desired

            while (i < j && !Character.isLetterOrDigit(s.charAt(i))) i++;

            while (i < j && !Character.isLetterOrDigit(s.charAt(j))) j--;

            if (Character.toLowerCase(s.charAt(i)) != Character.toLowerCase(s.charAt(j))) return false;

            i++; j--;

        }

        return true;

    }

public static void main(String[] args) { String[] tests = {"Madam", "A man, a plan, a canal: Panama", "hello"}; for (String t : tests) { System.out.printf("\"%s\" -> %b%n", t, isPalindromeTwoPointer(t)); } } }

Notes:

Two-pointer is O(n) time and O(1) extra space.
We skip non-alphanumeric characters and do case-insensitive comparison.

Example 2 — Palindrome Program (Integer)

For integers, you can reverse digits or compare digit-by-digit from ends. Reversing risks overflow for very large integers, so we can reverse only until the halfway point.

public class PalindromeNumber {

    public static boolean isPalindrome(int x) {

        // Negative numbers are not palindromes (because of the leading '-')

        if (x < 0) return false;

        // Numbers ending with 0 are not palindromes unless the number is 0

        if (x != 0 && x % 10 == 0) return false;
        int reverted = 0;

        while (x > reverted) {

            reverted = reverted * 10 + x % 10;

            x /= 10;

        }

        // For odd length, discard the middle digit by reverted/10

        return x == reverted || x == reverted / 10;

    }

public static void main(String[] args) { int[] tests = {121, -121, 10, 12321, 123321}; for (int t : tests) { System.out.printf("%d -> %b%n", t, isPalindrome(t)); } } }

Why this works: The loop builds the reversed lower half of the number. When the original x becomes less than or equal to reverted, we have processed half (or half+1) of the digits. This avoids overflow and is O(log10(n)) time.

Complexity summary

String reverse & compare: Time O(n), Space O(n) (due to reversed string).
Two-pointer string: Time O(n), Space O(1) (ignoring input).
Integer half-reverse: Time O(d) where d = number of digits (≈O(log n)), Space O(1).

Common pitfalls & tips

Case sensitivity: Convert to lower/upper case if you want case-insensitive checks.
Spaces and punctuation: Use replaceAll("[^A-Za-z0-9]", "") or two-pointer skipping to ignore non-alphanumerics.
Negative numbers: Conventionally not palindromes because of - sign.
Leading zeros: 010 as string can be palindrome, but as integer 10 is not.
Integer overflow: Don’t fully reverse huge integers into another int—use the half-reverse trick or use long with care.
Unicode: If handling non-ASCII text, consider using int code points rather than char and be careful with normalization.

Testing suggestions

Try these tests:

Simple: "madam", "hello".
Case: "RaceCar".
Phrases: "A man, a plan, a canal: Panama".
Numbers: 121, -121, 10, 12321.
Edge cases: empty string "" (decide whether to treat as palindrome), single-character strings, very long strings.

Conclusion

A palindrome program in Java is an excellent exercise for learning string handling and algorithmic thinking. For most purposes, the two-pointer string method is efficient and simple; for integers, the half-reverse technique is safe and avoids overflow. Start with the straightforward StringBuilder.reverse() approach to understand the idea, then move to two-pointer or numeric half-reverse for better space efficiency and interview-style solutions.