如何检查排除非字母数字字符的回文？

Question

Here's the code that I attempted

public String isPalindrome(String s) {


    String trimmed = s.replaceAll("[^A-Za-z0-9]", "");


    String reversed = "";

    int len = trimmed.length();


    for (int i = len - 1; i >= 0; i--) {

        char[] allChars = trimmed.toCharArray();

        reversed += allChars[i];

    }


    if (trimmed.equalsIgnoreCase(reversed)) {
        return "true";
    } else {
        return "false";
    }

}

Sample Input 1 A man, a plan, a canal: Panama

Sample Output 1 true

Explanation 1 The given string is palindrome when considering only alphanumeric characters.

Sample Input 2 race a car

Sample Output 2 false

Explanation 2 The given string is not a palindrome when considering alphanumeric characters.

Answer 1

You can return boolean instead of String :

public static boolean isPalindrome(String s) {
    String trimmed = s.replaceAll("[^A-Za-z0-9]", "").toLowerCase();

    int from = 0, to = trimmed.length() - 1;
    while (from < to) {
        if (trimmed.charAt(from) != trimmed.charAt(to)) {
            return false;
        }
        from++;
        to--;
    }
    return true;
}

Answer 2

You can use StringBuilder to reverse a String:

public static void main(String[] args) {
    String input = "a#b!b^a";
    String clean = input.replaceAll("[^A-Za-z0-9]", "");
    String reverse = new StringBuilder(clean).reverse().toString();
    boolean isPalindrome = reverse.equals(clean);
    System.out.println(isPalindrome);
}

Answer 3

Your variable len comes from the length of the String s . But you use the value on the array coming from trimmed .

So if you want to remove the IndexOutOfBoundsException you should change your len declaration to:

int len = trimmed.length();

Answer 4

You can do like this in linear time as the loops are driven by the presence of non-alphabetic/digit characters. Also, no trimming or reversal of the string is required.

String[] test =  {"A man,  a plan, a canal: Panama",
         "race a car","foobar", "ABC2CEc2cba"};

for (String s : test) {
     System.out.printf("%5b -> %s%n", isPalindrome(s), s);
}

prints

 true ->    A man,  a plan, a canal: Panama 
false -> race a car
false -> foobar
 true -> ABC2CEc2cba 

The outer while loop drives then entire process until the indices cross or are equal. The inner loops simply skip over non-alphabetic/digit characters.

public static boolean isPalindrome(String s) {
    int k = s.length() - 1;
    int i = 0;
    char c1 = '#';
    char c2 = '#';
    while (i <= k) {
        while (!Character.isLetterOrDigit(c1 = s.charAt(i++)));
        while (!Character.isLetterOrDigit(c2 = s.charAt(k--)));
        if (Character.toLowerCase(c1) != Character.toLowerCase(c2)) {
            return false;
        }
    }
    return true;
}