LeetCode – Valid Palindrome (Java)

Given a string, determine if it is a palindrome, considering only alphanumeric characters and ignoring cases.

For example, "Red rum, sir, is murder" is a palindrome, while "Programcreek is awesome" is not.

Note:
Have you consider that the string might be empty? This is a good question to ask during an interview.

For the purpose of this problem, we define empty string as valid palindrome.

Thoughts

From start and end loop though the string, i.e., char array. If it is not alpha or number, increase or decrease pointers. Compare the alpha and numeric characters. The solution below is pretty straightforward.

Java Solution 1 - Naive

public class Solution {
 
    public  boolean isPalindrome(String s) {
 
        if(s == null) return false;
        if(s.length() < 2) return true;
 
        char[] charArray = s.toCharArray();
        int len = s.length();
 
        int i=0;
        int j=len-1;
 
        while(i<j){
            char left, right;
 
            while(i<len-1 && !isAlpha(left) && !isNum(left)){
                i++;
                left =  charArray[i];
            }
 
            while(j>0 && !isAlpha(right) && !isNum(right)){
                j--;
                right = charArray[j];
            }
 
            if(i >= j)
            	break;
 
            left =  charArray[i];
            right = charArray[j];
 
            if(!isSame(left, right)){
                return false;
            }
 
            i++;
            j--;
        }
        return true;
    }
 
    public  boolean isAlpha(char a){
        if((a >= 'a' && a <= 'z') || (a >= 'A' && a <= 'Z')){
            return true;
        }else{
            return false;
        }
    }
 
    public  boolean isNum(char a){
        if(a >= '0' && a <= '9'){
            return true;
        }else{
            return false;
        }
    }
 
    public  boolean isSame(char a, char b){
        if(isNum(a) && isNum(b)){
            return a == b;
        }else if(Character.toLowerCase(a) == Character.toLowerCase(b)){
            return true;
        }else{
            return false;
        }
    }
}

Java Solution 2 - Using Stack

This solution removes the special characters first. (Thanks to Tia)

public boolean isPalindrome(String s) {
	s = s.replaceAll("[^a-zA-Z0-9]", "").toLowerCase();
 
	int len = s.length();
	if (len < 2)
		return true;
 
	Stack<Character> stack = new Stack<Character>();
 
	int index = 0;
	while (index < len / 2) {
		stack.push(s.charAt(index));
		index++;
	}
 
	if (len % 2 == 1)
		index++;
 
	while (index < len) {
		if (stack.empty())
			return false;
 
		char temp = stack.pop();
		if (s.charAt(index) != temp)
			return false;
		else
			index++;
	}
 
	return true;
}

Java Solution 3 - Using Two Pointers

In the discussion below, April and Frank use two pointers to solve this problem. This solution looks really simple.

public class ValidPalindrome {
	public static boolean isValidPalindrome(String s){
		if(s==null||s.length()==0) return false;
 
		s = s.replaceAll("[^a-zA-Z0-9]", "").toLowerCase();
		System.out.println(s);
 
		for(int i = 0; i < s.length() ; i++){
			if(s.charAt(i) != s.charAt(s.length() - 1 - i)){
				return false;
			}
		}
 
		return true;
	}
 
	public static void main(String[] args) {
		String str = "A man, a plan, a canal: Panama";
 
		System.out.println(isValidPalindrome(str));
	}
}
Category >> Algorithms  
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  • Varun Shetty

    public class Solution {

    public boolean isPalindrome(String s) {

    s = s.replaceAll(“[^a-zA-Z0-9]”, “”).toLowerCase();

    String rev = new StringBuilder(s).reverse().toString();

    if(s.equals(rev))

    return true;

    else return false;

    }
    }

    How is this is as a solution? Any flaws?

  • Diego

    Hey this is my solution, is not a big deal hehe just a simple for and StringBuilder

    public boolean isPalindrome(String s) {

    s = s.toLowerCase();

    StringBuilder builder = new StringBuilder();

    for(int index = 0; index < s.length(); index++) {

    Character temp = s.charAt(index);

    if(Character.isLetterOrDigit(temp)) {

    builder.append(temp.toString());

    }

    }

    String tempString = builder.toString();

    String otherString = builder.reverse().toString();

    if(tempString.equals(otherString)) return true;

    return false;

    }

  • Sunil Vurity

    how about ?

    if (s.Length % 2 != 0)

    {

    return false;

    }

  • Arnaud Hebert

    Your solution 1 will throw:
    Error:(.. , ..) java: variable left might not have been initialized
    Error:(.. , ..) java: variable right might not have been initialized

    Your solution 3:
    The for loop is not best choice here. You are comparing two times the same character.

    When i = 0,
    you compare charAt(0) with charAt(s.length() – 1)
    When i = s.length() – 1

    you compare charAt(s.length() – 1) and charAt(0)

    While loop is recommended

  • Chris

    Cool!

  • Tiago Pinho

    We could use too ():

    public static boolean isPalindrome(String s) {
    String pureString = s.replaceAll(“[^A-Za-z0-9]”, “”);
    if(pureString.length() == 0)
    return false;
    return(pureString.toLowerCase().equals(new StringBuilder(pureString.toLowerCase()).reverse().toString()));
    }

  • chenk

    public static boolean isPolindrome(String s) {
    s = s.replaceAll(“[^a-zA-Z]”, “”);
    return (s.equalsIgnoreCase(new StringBuilder(s).reverse().toString()));
    }

  • JOY

    I have a question that if the string is empty the statement s.length()==0 would be true, it will return false. Is this wrong?

  • eranyanay

    Solution 3 is the most intuitive one when I think of this problem, but you can easily make its performance split in half by doing the iteration from i=0 to s.length() / 2, because at the moment each character is compared twice, i.e for “abcd” you compare “a” == “d” but at the end also “d” == “a”

  • ryanlr

    Right. Thanks for pointing out this problem. I have changed the code.

  • ryanlr

    Great, Thanks! It is added!

  • April

    Here is an easier solution using two pointers

    public static boolean isPalindrome1(String s) {

    s = s.replaceAll(“[^0-9a-zA-Z]”, “”).toLowerCase();

    int start = 0;

    int end = s.length() – 1;

    while (start < end){

    if (s.charAt(start) != s.charAt(end)){

    return false;

    }

    start++;

    end–;

    }

    return true;

    }

  • Jarvis

    this is a great solution, thanks for sharing!

  • Frank

    public class ValidPalindrome {

    public static boolean isValidPalindrome(String s){

    if(s==null||s.length()==0) return false;

    s = s.replaceAll(“[^a-zA-Z0-9]”, “”).toLowerCase();

    System.out.println(s);

    for(int i = 0; i < s.length() ; i++){

    if(s.charAt(i) != s.charAt(s.length() – 1 – i)){

    return false;

    }

    }

    return true;

    }

    public static void main(String[] args) {

    String str = "A man, a plan, a canal: Panama";

    System.out.println(isValidPalindrome(str));

    }

    }

  • Diana Du

    Hi, I found that there are two redundant lines in your first solution. These two lines

    “left = charArray[i];

    right = charArray[j];”

    seem to be redundant in the largest while loop, because you have already changed them in the above two while loops.

  • ryanlr

    Great Solution! Thanks.

  • tia

    Here is the idea that remove non-letter first

    public class Solution {

    public boolean isPalindrome(String s) {

    s = s.replaceAll(“[^a-zA-Z0-9]”, “”);

    s = s.toLowerCase();

    int len = s.length();

    if(len<2)

    return true;

    Stack sta = new Stack();

    int index = 0;

    while(index < len/2){

    sta.push(s.charAt(index));

    index ++;

    }

    if (len%2 ==1)

    index++;

    while(index < len){

    if(sta.empty())

    return false;

    char temp = sta.pop();

    if(s.charAt(index) != temp)

    return false;

    else

    index ++;

    }

    return true;

    }

    }