Nsryjdtyk

Question

A palindrome is a word, phrase, number or other sequence of units that can be read the same way in either direction.

To check whether a word is a palindrome I get the char array of the word and compare the chars. I tested it and it seems to work. However I want to know if it is right or if there is something to improve.

Here is my code:

public class Aufg1 {

    public static void main(String args) {

        String wort = "reliefpfpfeiller";

        char warray = wort.toCharArray(); 

        System.out.println(istPalindrom(warray));       

    }



    public static boolean istPalindrom(char wort){

        boolean palindrom = false;

        if(wort.length%2 == 0){

            for(int i = 0; i < wort.length/2-1; i++){

                if(wort[i] != wort[wort.length-i-1]){

                    return false;

                }else{

                    palindrom = true;

                }

            }

        }else{

            for(int i = 0; i < (wort.length-1)/2-1; i++){

                if(wort[i] != wort[wort.length-i-1]){

                    return false;

                }else{

                    palindrom = true;

                }

            }

        }

        return palindrom;

    }

}

Not sure if this is intentional but the string in your example - reliefpfpfeiller - isn't a palindrome — Nov 10 '10 at 1:47

Michael Myers♦ 155k37256280 · Accepted Answer · 2015-05-18 23:05:01Z

164

Why not just:

public static boolean istPalindrom(char word){

    int i1 = 0;

    int i2 = word.length - 1;

    while (i2 > i1) {

        if (word[i1] != word[i2]) {

            return false;

        }

        ++i1;

        --i2;

    }

    return true;

}

Example:

Input is "andna".

i1 will be 0 and i2 will be 4.

First loop iteration we will compare word[0] and word[4]. They're equal, so we increment i1 (it's now 1) and decrement i2 (it's now 3).

So we then compare the n's. They're equal, so we increment i1 (it's now 2) and decrement i2 (it's 2).

Now i1 and i2 are equal (they're both 2), so the condition for the while loop is no longer true so the loop terminates and we return true.

edited May 18 '15 at 23:05

Michael Myers♦

155k37256280

answered Nov 9 '10 at 21:32

dcp

42.7k16119146

1

instead of pre increment (++i1 and --i2), we can also use post increment (i1++, i2--)result is same i think !
– user0946076422
Aug 15 '15 at 3:44

@user0946076422 Yes. I too felt that way. Would be great if OP has a different explanation.
– Vijay Tholpadi
Mar 10 '17 at 3:11

2

@Vijay Tholpadi - It's just really a coding preference more than anything else. Post increment would accomplish same result in this particular example, but I always use pre increment unless there's a specific reason not to.
– dcp
Mar 15 '17 at 12:44

add a comment |

score 111 · Accepted Answer · 2016-07-13 20:10:25Z

111

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!

edited Jul 13 '16 at 20:10

answered Nov 9 '10 at 21:54

Greg

28.5k138297

10

Compare the complexity of your algorithm with respect to others.
– Fernando Pelliccioni
Feb 3 '14 at 14:24

2

@FernandoPelliccioni, I think it's the same complexity as the other solutions, no?
– aioobe
Aug 23 '15 at 8:16

1

@Fernando, as far as I can tell all answers have a linear complexity. Because of this there's no way to give a definitive answer to which solution is most efficient. You could run benchmarks but they would be specific for a particular JVM and JRE. Best of luck with your blog post. Looking forward to reading it.
– aioobe
Aug 26 '15 at 21:24

15

Here my analysis: componentsprogramming.com/palindromes
– Fernando Pelliccioni
Jun 6 '16 at 21:09

1

@FernandoPelliccioni nice analysis
– Saravana
Jul 12 '16 at 4:11

|
show 9 more comments

David Robles 6,59453242 · Accepted Answer · 2013-08-13 06:01:13Z

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;    

}

score 14 · Accepted Answer · 2016-10-30 13:46:09Z

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;

    else return s.charAt(0) != s.charAt(length - 1) ? false :

            isPalindrome(s.substring(1, length - 1));

}

Nice code, recursion makes it really easy and less lines on code. — Jan 31 '17 at 6:35
the shorter version can be simplified: return s.charAt(0) == s.charAt(l - 1) && isPalindrome(s.substring(1, l - 1)); — Apr 19 '18 at 23:05

Francisco Gutiérrez 9981022 · Accepted Answer · 2014-03-04 08:44:05Z

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

score 4 · Accepted Answer · 2010-11-09 21:41:20Z

public class Aufg1 {

    public static void main(String args) {

         String wort = "reliefpfpfeiller";

         char warray = wort.toCharArray(); 

         System.out.println(istPalindrom(warray));       

    }



    public static boolean istPalindrom(char wort){

        if(wort.length%2 == 0){

            for(int i = 0; i < wort.length/2-1; i++){

                if(wort[i] != wort[wort.length-i-1]){

                    return false;

                }

            }

        }else{

            for(int i = 0; i < (wort.length-1)/2-1; i++){

                if(wort[i] != wort[wort.length-i-1]){

                    return false;

                }

            }

        }

        return true;

    }

}

simplified a bit. but I like dcp's answer!
– Casey
Nov 9 '10 at 21:36 — Nov 9 '10 at 21:36

Mona Jalal 7,87926108208 · Accepted Answer · 2016-05-26 03:52:45Z

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;

    }

wumpz 4,60921522 · Accepted Answer · 2017-05-02 06:49:57Z

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

Why not allMatch with allMatch(i -> str.charAt(i) == str.charAt(str.length() - i - 1)) ? — Apr 5 '18 at 9:47

Abhilash Muthuraj 84472039 · Accepted Answer · 2017-06-02 17:27:40Z

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;

    }

user2039532 215 · Accepted Answer · 2013-02-08 12:50:20Z

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");

}

}

score 2 · Accepted Answer · 2017-05-24 15:52:30Z

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.

Absolutely nothing - I must not have had my coffee. Will fix my response - thanks!. — May 24 '17 at 15:51

Felix 442317 · Accepted Answer · 2018-01-18 16:33:34Z

Amazing how many different solutions to such a simple problem exist! Here's another one.

private static boolean palindrome(String s){

    String revS = "";

    String checkS = s.toLowerCase();

    String checkSArr = checkS.split("");



    for(String e : checkSArr){

        revS = e + revS;

    }



    return (checkS.equals(revS)) ? true : false;

}

wizzardz 3,40743157 · Accepted Answer · 2013-03-12 07:29:42Z

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");

            }

    }

gogobebe2 82214 · Accepted Answer · 2014-12-16 02:08:43Z

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;



            }



        }

    }

}

If the string is made of no letters or just one letter, it is a
palindrome.

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.

Amandeep Dhanjal 435 · Accepted Answer · 2015-05-28 23:58:43Z

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;

}

Madura Harshana 81662036 · Accepted Answer · 2015-06-24 11:18:20Z

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;

}

}

This approach is excellent. Time Complexity O(n), space Complexity O(1) — Jan 1 '18 at 4:06

Fernando Pelliccioni 595614 · Accepted Answer · 2016-11-14 14:28:41Z

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

capt.swag 5,98212425 · Accepted Answer · 2017-04-20 06:56:31Z

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;

}

aayushi 15627 · Accepted Answer · 2017-04-20 07:24:34Z

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");

        }

    }

}

as you must be knowing that, stack is of LIFO type that means you are basically pushing data through the beginning of the stack and retrieving data from the end of stack using pop(). Hope this helps! — May 16 '17 at 19:42

Chandara Chea 514 · Accepted Answer · 2017-05-26 12:33:37Z

 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;

   }



}

score 1 · Accepted Answer · 2018-12-02 04:15:35Z

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;

}

Mona Jalal 7,87926108208 · Accepted Answer · 2016-05-26 03:53:23Z

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;

                }



            }

        }

    }



}

Angela Sanchez 1351112 · Accepted Answer · 2016-10-24 07:43:39Z

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;

    }

Sotti 10.6k23839 · Accepted Answer · 2016-10-24 15:51:32Z

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.

huseyin 8701012 · Accepted Answer · 2016-12-08 07:44:08Z

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"));

}

score 0 · Accepted Answer · 2016-12-23 17:48:23Z

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);

}

}

john Smith 71022243 · Accepted Answer · 2017-01-08 21:45:28Z

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;

}

answered Jan 8 '17 at 21:45

john Smith

71022243

2

This already has been used in an answer: Check string for palindrome (just a note)
– Tom
Jan 8 '17 at 22:00

sorry, I missed it.
– john Smith
Jan 9 '17 at 13:48

add a comment |

juanmf 1,66411824 · Accepted Answer · 2017-02-03 16:21:36Z

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)));

    }

}

rashedcs 1,0791224 · Accepted Answer · 2017-03-28 18:23:37Z

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");

     }

}

Keshav Gera 2,4331728 · Accepted Answer · 2017-06-12 09:15:21Z

enter image description here

import java.util.Collections;

import java.util.HashSet;

import java.util.Scanner;

import java.util.Set;



public class GetAllPalindromes 

{

    static Scanner in;



    public static void main(String args) 

    {

        in = new Scanner(System.in);

        System.out.println("Enter a string n");

        String abc = in.nextLine();

        Set a = printAllPalindromes(abc);

        System.out.println("set is   " + a);

    }



    public static Set<CharSequence> printAllPalindromes(String input) 

    {

        if (input.length() <= 2) {

            return Collections.emptySet();

        }



        Set<CharSequence> out = new HashSet<CharSequence>();

        int length = input.length();



        for (int i = 1; i < length - 1; i++) 

        {

            for (int j = i - 1, k = i + 1; j >= 0 && k < length; j--, k++) 

            {

                if (input.charAt(j) == input.charAt(k)) {

                    out.add(input.subSequence(j, k + 1));

                } else {

                    break;

                }

            }

        }

        return out;

    }

}



**Get All Palindrome in s given string**

Output
D:Java>java GetAllPalindromes
Enter a string

Hello user nitin is my best friend wow !

Answer is set is [nitin, nitin , wow , wow, iti]

D:Java>

Michael Myers♦ 155k37256280 · Accepted Answer · 2015-05-18 23:05:01Z

164

Why not just:

public static boolean istPalindrom(char word){

    int i1 = 0;

    int i2 = word.length - 1;

    while (i2 > i1) {

        if (word[i1] != word[i2]) {

            return false;

        }

        ++i1;

        --i2;

    }

    return true;

}

Example:

Input is "andna".

i1 will be 0 and i2 will be 4.

First loop iteration we will compare word[0] and word[4]. They're equal, so we increment i1 (it's now 1) and decrement i2 (it's now 3).

So we then compare the n's. They're equal, so we increment i1 (it's now 2) and decrement i2 (it's 2).

Now i1 and i2 are equal (they're both 2), so the condition for the while loop is no longer true so the loop terminates and we return true.

edited May 18 '15 at 23:05

Michael Myers♦

155k37256280

answered Nov 9 '10 at 21:32

dcp

42.7k16119146

1

instead of pre increment (++i1 and --i2), we can also use post increment (i1++, i2--)result is same i think !
– user0946076422
Aug 15 '15 at 3:44

@user0946076422 Yes. I too felt that way. Would be great if OP has a different explanation.
– Vijay Tholpadi
Mar 10 '17 at 3:11

2

@Vijay Tholpadi - It's just really a coding preference more than anything else. Post increment would accomplish same result in this particular example, but I always use pre increment unless there's a specific reason not to.
– dcp
Mar 15 '17 at 12:44

add a comment |

score 111 · Accepted Answer · 2016-07-13 20:10:25Z

111

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!

edited Jul 13 '16 at 20:10

answered Nov 9 '10 at 21:54

Greg

28.5k138297

10

Compare the complexity of your algorithm with respect to others.
– Fernando Pelliccioni
Feb 3 '14 at 14:24

2

@FernandoPelliccioni, I think it's the same complexity as the other solutions, no?
– aioobe
Aug 23 '15 at 8:16

1

@Fernando, as far as I can tell all answers have a linear complexity. Because of this there's no way to give a definitive answer to which solution is most efficient. You could run benchmarks but they would be specific for a particular JVM and JRE. Best of luck with your blog post. Looking forward to reading it.
– aioobe
Aug 26 '15 at 21:24

15

Here my analysis: componentsprogramming.com/palindromes
– Fernando Pelliccioni
Jun 6 '16 at 21:09

1

@FernandoPelliccioni nice analysis
– Saravana
Jul 12 '16 at 4:11

|
show 9 more comments

David Robles 6,59453242 · Accepted Answer · 2013-08-13 06:01:13Z

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;    

}

score 14 · Accepted Answer · 2016-10-30 13:46:09Z

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;

    else return s.charAt(0) != s.charAt(length - 1) ? false :

            isPalindrome(s.substring(1, length - 1));

}

Nice code, recursion makes it really easy and less lines on code. — Jan 31 '17 at 6:35
the shorter version can be simplified: return s.charAt(0) == s.charAt(l - 1) && isPalindrome(s.substring(1, l - 1)); — Apr 19 '18 at 23:05

Francisco Gutiérrez 9981022 · Accepted Answer · 2014-03-04 08:44:05Z

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

score 4 · Accepted Answer · 2010-11-09 21:41:20Z

public class Aufg1 {

    public static void main(String args) {

         String wort = "reliefpfpfeiller";

         char warray = wort.toCharArray(); 

         System.out.println(istPalindrom(warray));       

    }



    public static boolean istPalindrom(char wort){

        if(wort.length%2 == 0){

            for(int i = 0; i < wort.length/2-1; i++){

                if(wort[i] != wort[wort.length-i-1]){

                    return false;

                }

            }

        }else{

            for(int i = 0; i < (wort.length-1)/2-1; i++){

                if(wort[i] != wort[wort.length-i-1]){

                    return false;

                }

            }

        }

        return true;

    }

}

simplified a bit. but I like dcp's answer!
– Casey
Nov 9 '10 at 21:36 — Nov 9 '10 at 21:36

Mona Jalal 7,87926108208 · Accepted Answer · 2016-05-26 03:52:45Z

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;

    }

wumpz 4,60921522 · Accepted Answer · 2017-05-02 06:49:57Z

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

Why not allMatch with allMatch(i -> str.charAt(i) == str.charAt(str.length() - i - 1)) ? — Apr 5 '18 at 9:47

Abhilash Muthuraj 84472039 · Accepted Answer · 2017-06-02 17:27:40Z

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;

    }

user2039532 215 · Accepted Answer · 2013-02-08 12:50:20Z

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");

}

}

score 2 · Accepted Answer · 2017-05-24 15:52:30Z

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.

Absolutely nothing - I must not have had my coffee. Will fix my response - thanks!. — May 24 '17 at 15:51

Felix 442317 · Accepted Answer · 2018-01-18 16:33:34Z

Amazing how many different solutions to such a simple problem exist! Here's another one.

private static boolean palindrome(String s){

    String revS = "";

    String checkS = s.toLowerCase();

    String checkSArr = checkS.split("");



    for(String e : checkSArr){

        revS = e + revS;

    }



    return (checkS.equals(revS)) ? true : false;

}

wizzardz 3,40743157 · Accepted Answer · 2013-03-12 07:29:42Z

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");

            }

    }

gogobebe2 82214 · Accepted Answer · 2014-12-16 02:08:43Z

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;



            }



        }

    }

}

If the string is made of no letters or just one letter, it is a
palindrome.

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.

Amandeep Dhanjal 435 · Accepted Answer · 2015-05-28 23:58:43Z

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;

}

Madura Harshana 81662036 · Accepted Answer · 2015-06-24 11:18:20Z

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;

}

}

This approach is excellent. Time Complexity O(n), space Complexity O(1) — Jan 1 '18 at 4:06

Fernando Pelliccioni 595614 · Accepted Answer · 2016-11-14 14:28:41Z

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

capt.swag 5,98212425 · Accepted Answer · 2017-04-20 06:56:31Z

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;

}

aayushi 15627 · Accepted Answer · 2017-04-20 07:24:34Z

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");

        }

    }

}

as you must be knowing that, stack is of LIFO type that means you are basically pushing data through the beginning of the stack and retrieving data from the end of stack using pop(). Hope this helps! — May 16 '17 at 19:42

Chandara Chea 514 · Accepted Answer · 2017-05-26 12:33:37Z

 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;

   }



}

score 1 · Accepted Answer · 2018-12-02 04:15:35Z

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;

}

Mona Jalal 7,87926108208 · Accepted Answer · 2016-05-26 03:53:23Z

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;

                }



            }

        }

    }



}

Angela Sanchez 1351112 · Accepted Answer · 2016-10-24 07:43:39Z

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;

    }

Sotti 10.6k23839 · Accepted Answer · 2016-10-24 15:51:32Z

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.

huseyin 8701012 · Accepted Answer · 2016-12-08 07:44:08Z

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"));

}

score 0 · Accepted Answer · 2016-12-23 17:48:23Z

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);

}

}

john Smith 71022243 · Accepted Answer · 2017-01-08 21:45:28Z

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;

}

answered Jan 8 '17 at 21:45

john Smith

71022243

2

This already has been used in an answer: Check string for palindrome (just a note)
– Tom
Jan 8 '17 at 22:00

sorry, I missed it.
– john Smith
Jan 9 '17 at 13:48

add a comment |

juanmf 1,66411824 · Accepted Answer · 2017-02-03 16:21:36Z

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)));

    }

}

rashedcs 1,0791224 · Accepted Answer · 2017-03-28 18:23:37Z

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");

     }

}

Keshav Gera 2,4331728 · Accepted Answer · 2017-06-12 09:15:21Z

enter image description here

import java.util.Collections;

import java.util.HashSet;

import java.util.Scanner;

import java.util.Set;



public class GetAllPalindromes 

{

    static Scanner in;



    public static void main(String args) 

    {

        in = new Scanner(System.in);

        System.out.println("Enter a string n");

        String abc = in.nextLine();

        Set a = printAllPalindromes(abc);

        System.out.println("set is   " + a);

    }



    public static Set<CharSequence> printAllPalindromes(String input) 

    {

        if (input.length() <= 2) {

            return Collections.emptySet();

        }



        Set<CharSequence> out = new HashSet<CharSequence>();

        int length = input.length();



        for (int i = 1; i < length - 1; i++) 

        {

            for (int j = i - 1, k = i + 1; j >= 0 && k < length; j--, k++) 

            {

                if (input.charAt(j) == input.charAt(k)) {

                    out.add(input.subSequence(j, k + 1));

                } else {

                    break;

                }

            }

        }

        return out;

    }

}



**Get All Palindrome in s given string**

Output
D:Java>java GetAllPalindromes
Enter a string

Hello user nitin is my best friend wow !

Answer is set is [nitin, nitin , wow , wow, iti]

D:Java>

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Nsryjdtyk

Check string for palindrome

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protected by Community♦ Apr 7 '13 at 17:01

34 Answers
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34 Answers
34

Code Snippet:

Code Snippet:

Code Snippet:

Code Snippet:

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