Check string for palindrome

You can check if a string is a palindrome by comparing it to the reverse of itself:

public static boolean isPalindrome(String str) { return str.equals(new StringBuilder(str).reverse().toString()); } 

or for versions of Java earlier than 1.5,

public static boolean isPalindrome(String str) { return str.equals(new StringBuffer().append(str).reverse().toString()); } 

EDIT: @FernandoPelliccioni provided a very thorough analysis of the efficiency (or lack thereof) of this solution, both in terms of time and space. If you're interested in the computational complexity of this and other possible solutions to this question, please read it!

A concise version, that doesn't involve (inefficiently) initializing a bunch of objects:

boolean isPalindrome(String str) { int n = str.length(); for( int i = 0; i < n/2; i++ ) if (str.charAt(i) != str.charAt(n-i-1)) return false; return true; } 

Alternatively, recursion.

For anybody who is looking for a shorter recursive solution, to check if a given string satisfies as a palindrome:

private boolean isPalindrome(String s) { int length = s.length(); if (length < 2) // If the string only has 1 char or is empty return true; else { // Check opposite ends of the string for equality if (s.charAt(0) != s.charAt(length - 1)) return false; // Function call for string with the two ends snipped off else return isPalindrome(s.substring(1, length - 1)); } } 

OR even shorter, if you'd like:

private boolean isPalindrome(String s) { int length = s.length(); if (length < 2) return true; return s.charAt(0) != s.charAt(length - 1) ? false : isPalindrome(s.substring(1, length - 1)); } 

Go, Java:

public boolean isPalindrome (String word) { String myWord = word.replaceAll("\\s+",""); String reverse = new StringBuffer(myWord).reverse().toString(); return reverse.equalsIgnoreCase(myWord); } isPalindrome("Never Odd or Even"); // True isPalindrome("Never Odd or Even1"); // False 

also a different looking solution:

public static boolean isPalindrome(String s) { for (int i=0 , j=s.length()-1 ; i<j ; i++ , j-- ) { if ( s.charAt(i) != s.charAt(j) ) { return false; } } return true; } 

And here a complete Java 8 streaming solution. An IntStream provides all indexes til strings half length and then a comparision from the start and from the end is done.

public static void main(String[] args) { for (String testStr : Arrays.asList("testset", "none", "andna", "haah", "habh", "haaah")) { System.out.println("testing " + testStr + " is palindrome=" + isPalindrome(testStr)); } } public static boolean isPalindrome(String str) { return IntStream.range(0, str.length() / 2) .noneMatch(i -> str.charAt(i) != str.charAt(str.length() - i - 1)); } 

Output is:

testing testset is palindrome=true testing none is palindrome=false testing andna is palindrome=true testing haah is palindrome=true testing habh is palindrome=false testing haaah is palindrome=true 

public class Palindromes { public static void main(String[] args) { String word = "reliefpfpfeiller"; char[] warray = word.toCharArray(); System.out.println(isPalindrome(warray)); } public static boolean isPalindrome(char[] word){ if(word.length%2 == 0){ for(int i = 0; i < word.length/2-1; i++){ if(word[i] != word[word.length-i-1]){ return false; } } }else{ for(int i = 0; i < (word.length-1)/2-1; i++){ if(word[i] != word[word.length-i-1]){ return false; } } } return true; } } 

public class palindrome { public static void main(String[] args) { StringBuffer strBuf1 = new StringBuffer("malayalam"); StringBuffer strBuf2 = new StringBuffer("malayalam"); strBuf2.reverse(); System.out.println(strBuf2); System.out.println((strBuf1.toString()).equals(strBuf2.toString())); if ((strBuf1.toString()).equals(strBuf2.toString())) System.out.println("palindrome"); else System.out.println("not a palindrome"); } 

}

I worked on a solution for a question that was marked as duplicate of this one. Might as well throw it here...

The question requested a single line to solve this, and I took it more as the literary palindrome - so spaces, punctuation and upper/lower case can throw off the result.

Here's the ugly solution with a small test class:

public class Palindrome { public static boolean isPalendrome(String arg) { return arg.replaceAll("[^A-Za-z]", "").equalsIgnoreCase(new StringBuilder(arg).reverse().toString().replaceAll("[^A-Za-z]", "")); } public static void main(String[] args) { System.out.println(isPalendrome("hiya")); System.out.println(isPalendrome("star buttons not tub rats")); System.out.println(isPalendrome("stab nail at ill Italian bats!")); return; } } 

Sorry that it is kind of nasty - but the other question specified a one-liner.

Checking palindrome for first half of the string with the rest, this case assumes removal of any white spaces.

public int isPalindrome(String a) { //Remove all spaces and non alpha characters String ab = a.replaceAll("[^A-Za-z0-9]", "").toLowerCase(); //System.out.println(ab); for (int i=0; i<ab.length()/2; i++) { if(ab.charAt(i) != ab.charAt((ab.length()-1)-i)) { return 0; } } return 1; } 

I'm new to java and I'm taking up your question as a challenge to improve my knowledge.

import java.util.ArrayList; import java.util.List; public class PalindromeRecursiveBoolean { public static boolean isPalindrome(String str) { str = str.toUpperCase(); char[] strChars = str.toCharArray(); List<Character> word = new ArrayList<>(); for (char c : strChars) { word.add(c); } while (true) { if ((word.size() == 1) || (word.size() == 0)) { return true; } if (word.get(0) == word.get(word.size() - 1)) { word.remove(0); word.remove(word.size() - 1); } else { return false; } } } } 

1. If the string is made of no letters or just one letter, it is a palindrome.
2. Otherwise, compare the first and last letters of the string.
   • If the first and last letters differ, then the string is not a palindrome
   • Otherwise, the first and last letters are the same. Strip them from the string, and determine whether the string that remains is a palindrome. Take the answer for this smaller string and use it as the answer for the original string then repeat from 1.

Try this out :

import java.util.*; public class str { public static void main(String args[]) { Scanner in=new Scanner(System.in); System.out.println("ENTER YOUR STRING: "); String a=in.nextLine(); System.out.println("GIVEN STRING IS: "+a); StringBuffer str=new StringBuffer(a); StringBuffer str2=new StringBuffer(str.reverse()); String s2=new String(str2); System.out.println("THE REVERSED STRING IS: "+str2); if(a.equals(s2)) System.out.println("ITS A PALINDROME"); else System.out.println("ITS NOT A PALINDROME"); } } 

public boolean isPalindrome(String abc){ if(abc != null && abc.length() > 0){ char[] arr = abc.toCharArray(); for (int i = 0; i < arr.length/2; i++) { if(arr[i] != arr[arr.length - 1 - i]){ return false; } } return true; } return false; } 

Another way is using char Array

public class Palindrome { public static void main(String[] args) { String str = "madam"; if(isPalindrome(str)) { System.out.println("Palindrome"); } else { System.out.println("Not a Palindrome"); } } private static boolean isPalindrome(String str) { // Convert String to char array char[] charArray = str.toCharArray(); for(int i=0; i < str.length(); i++) { if(charArray[i] != charArray[(str.length()-1) - i]) { return false; } } return true; } 

}

Here my analysis of the @Greg answer: componentsprogramming.com/palindromes

Sidenote: But, for me it is important to do it in a Generic way. The requirements are that the sequence is bidirectionally iterable and the elements of the sequence are comparables using equality. I don't know how to do it in Java, but, here is a C++ version, I don't know a better way to do it for bidirectional sequences.

template <BidirectionalIterator I> requires( EqualityComparable< ValueType<I> > ) bool palindrome( I first, I last ) { I m = middle(first, last); auto rfirst = boost::make_reverse_iterator(last); return std::equal(first, m, rfirst); } 

Complexity: linear-time,

• If I is RandomAccessIterator: floor(n/2) comparissons and floor(n/2)*2 iterations

• If I is BidirectionalIterator: floor(n/2) comparissons and floor(n/2)*2 iterations plus (3/2)*n iterations to find the middle ( middle function )

• storage: O(1)

• No dymamic allocated memory

Recently I wrote a palindrome program which doesn't use StringBuilder. A late answer but this might come in handy to some people.

public boolean isPalindrome(String value) { boolean isPalindrome = true; for (int i = 0 , j = value.length() - 1 ; i < j ; i ++ , j --) { if (value.charAt(i) != value.charAt(j)) { isPalindrome = false; } } return isPalindrome; } 

Using stack, it can be done like this

import java.io.*; import java.util.*; import java.text.*; import java.math.*; import java.util.regex.*; import java.util.*; public class Solution { public static void main(String[] args) { Scanner in = new Scanner(System.in); String str=in.nextLine(); str.replaceAll("\\s+",""); //System.out.println(str); Stack<String> stack=new Stack<String>(); stack.push(str); String str_rev=stack.pop(); if(str.equals(str_rev)){ System.out.println("Palindrome"); }else{ System.out.println("Not Palindrome"); } } } 

 public static boolean isPalindrome(String word) { String str = ""; for (int i=word.length()-1; i>=0; i--){ str = str + word.charAt(i); } if(str.equalsIgnoreCase(word)){ return true; }else{ return false; } } 

• This implementation works for numbers and strings.
• Since we are not writing anything, so there is no need to convert the string into the character array.

public static boolean isPalindrome(Object obj) { String s = String.valueOf(obj); for(int left=0, right=s.length()-1; left < right; left++,right--) { if(s.charAt(left++) != s.charAt(right--)) return false; } return true; } 

Why not just :

boolean isPalindrom(String s) { char[] myChars = s.toCharArray(); for (int i = 0; i < myChars.length/2; i++) { if (myChars[i] != myChars[myChars.length - 1 - i]) { return false; } } return true; } 

import java.util.Scanner; public class Palindrom { public static void main(String []args) { Scanner in = new Scanner(System.in); String str= in.nextLine(); int x= str.length(); if(x%2!=0) { for(int i=0;i<x/2;i++) { if(str.charAt(i)==str.charAt(x-1-i)) { continue; } else { System.out.println("String is not a palindrom"); break; } } } else { for(int i=0;i<=x/2;i++) { if(str.charAt(i)==str.charAt(x-1-i)) { continue; } else { System.out.println("String is not a palindrom"); break; } } } } } 

private static boolean isPalindrome(String word) { int z = word.length(); boolean isPalindrome = false; for (int i = 0; i <= word.length() / 2; i++) { if (word.charAt(i) == word.charAt(--z)) { isPalindrome = true; } } return isPalindrome; } 

I was looking for a solution that not only worked for palindromes like...

• "Kayak"
• "Madam"

...but as well for...

• "A man, a plan, a canal, Panama!"
• "Was it a car or a cat I saw?"
• "No 'x' in Nixon"

Iterative: This has be proven as a good solution.

private boolean isPalindromeIterative(final String string) { final char[] characters = string.replaceAll("[\\W]", "").toLowerCase().toCharArray(); int iteratorLeft = 0; int iteratorEnd = characters.length - 1; while (iteratorEnd > iteratorLeft) { if (characters[iteratorLeft++] != characters[iteratorEnd--]) { return false; } } return true; } 

Recursive. I think this solution shouldn't be much worse than the iterative one. Is a little bit crapy we need to extract the cleaning step out of the method to avoid unnecesary procesing.

private boolean isPalindromeRecursive(final String string) { final String cleanString = string.replaceAll("[\\W]", "").toLowerCase(); return isPalindromeRecursiveRecursion(cleanString); } private boolean isPalindromeRecursiveRecursion(final String cleanString) { final int cleanStringLength = cleanString.length(); return cleanStringLength <= 1 || cleanString.charAt(0) == cleanString.charAt(cleanStringLength - 1) && isPalindromeRecursiveRecursion (cleanString.substring(1, cleanStringLength - 1)); } 

Reversing: This has been proved as a expensive solution.

private boolean isPalindromeReversing(final String string) { final String cleanString = string.replaceAll("[\\W]", "").toLowerCase(); return cleanString.equals(new StringBuilder(cleanString).reverse().toString()); } 

All the credits to the guys answering in this post and bringing light to the topic.

Considering not letters in the words

public static boolean palindromeWords(String s ){ int left=0; int right=s.length()-1; while(left<=right){ while(left<right && !Character.isLetter(s.charAt(left))){ left++; } while(right>0 && !Character.isLetter(s.charAt(right))){ right--; } if((s.charAt(left++))!=(s.charAt(right--))){ return false; } } return true; } 

———

@Test public void testPalindromeWords(){ assertTrue(StringExercise.palindromeWords("ece")); assertTrue(StringExercise.palindromeWords("kavak")); assertFalse(StringExercise.palindromeWords("kavakdf")); assertTrue(StringExercise.palindromeWords("akka")); assertTrue(StringExercise.palindromeWords("??e@@c_--e")); } 

Here you can check palindrome a number of String dynamically

import java.util.Scanner; public class Checkpalindrome { public static void main(String args[]) { String original, reverse = ""; Scanner in = new Scanner(System.in); System.out.println("Enter How Many number of Input you want : "); int numOfInt = in.nextInt(); original = in.nextLine(); do { if (numOfInt == 0) { System.out.println("Your Input Conplete"); } else { System.out.println("Enter a string to check palindrome"); original = in.nextLine(); StringBuffer buffer = new StringBuffer(original); reverse = buffer.reverse().toString(); if (original.equalsIgnoreCase(reverse)) { System.out.println("The entered string is Palindrome:"+reverse); } else { System.out.println("The entered string is not Palindrome:"+reverse); } } numOfInt--; } while (numOfInt >= 0); } } 

IMO, the recursive way is the simplest and clearest.

public static boolean isPal(String s) { if(s.length() == 0 || s.length() == 1) return true; if(s.charAt(0) == s.charAt(s.length()-1)) return isPal(s.substring(1, s.length()-1)); return false; } 

here, checking for the largest palindrome in a string, always starting from 1st char.

public static String largestPalindromeInString(String in) { int right = in.length() - 1; int left = 0; char[] word = in.toCharArray(); while (right > left && word[right] != word[left]) { right--; } int lenght = right + 1; while (right > left && word[right] == word[left]) { left++; right--; } if (0 >= right - left) { return new String(Arrays.copyOf(word, lenght )); } else { return largestPalindromeInString( new String(Arrays.copyOf(word, in.length() - 1))); } } 

Code Snippet:

import java.util.Scanner; class main { public static void main(String []args) { Scanner sc = new Scanner(System.in); String str = sc.next(); String reverse = new StringBuffer(str).reverse().toString(); if(str.equals(reverse)) System.out.println("Pallindrome"); else System.out.println("Not Pallindrome"); } } 

For-loop contains sub.length() / 2 - 1 . It has to be subtracted with 1 as the element in the middle of the string does not have to checked.

For example, if we have to check an string with 7 chars (1234567), then 7/2 => 3 and then we subtrack 1, and so the positions in the string will become (0123456). The chars checked with be the 0, 1, 2 element with the 6, 5, 4 respectively. We do not care about the element at the position 3 as it is in the exact middle of the string.

 private boolean isPalindromic(String sub) { for (int i = 0; i <= sub.length() / 2 - 1; i++) { if (sub.charAt(i) != sub.charAt(sub.length() - 1 - i)) { return false; } } return true; }