Question

处理有关Java的面试问题，直到....回文问题为止，一切都很好。

我决定使用递归方法，并且正确编写了代码（或者我以为自己做了！）。但是，我收到了无效的方法消息，并且我不明白为什么。

我完全感到困惑吗？

其次，我发现此解决方案颇费周折，所以我想知道是否还有另一种较短的递归方法容易记住（我知道您可以使用字符串缓冲区，但我读到它不很健壮才是真的。在面试压力下，这会不会很重要？

这是代码。我期待您的答复：

public class HowToCheckForAPalindromeString {
    //Logic to solve this problem:  If the characters at the beginning and the end are the same = Palindrome          
    //Method : Using Recursion: in which a method calls itself to solve some problem.

    public static void main(String[] args) {


        String pal = "Rotator"; 
        boolean result = isPalindrome(pal);
        System.out.println(pal + "is Palindrome ="+result);


        //Body of the Palindrome Method

        public static boolean isPalindrome(String pal){

            //step 1. Check for exception handling for edge cases
            if(pal==null){
                return false; //check if the beginning and the of the palindrome are equal
            }
            if(pal.length()<=1){
                return true;  //if a person enters an empty string that's a palindrome (very important!)
            }

            //step 2. Finding the Palindrome 
            String firstLet = pal.substring(0, 1); //this will check the first characters of the string
            String lastLet = pal.substring(pal.length()-1, pal.length());// check the starting and the ending index 

            if(!firstLet.equals(lastLet)){//if the the first and last letters are NOT the same = not palindome
                return false;

            }else{
                return isPalindrome(pal.length()-1, pal.length()); //if they are the same= Palimdrome- Yay!
            }



        }//end of method




    }

}

Answer 1

在递归调用中，您尝试传递2个整数，该整数应为字符串。假设您要传递剪切第一个和最后一个字母的子字符串，这是固定代码：

public class HowToCheckForAPalindromeString {
//Logic to solve this problem:  If the characters at the beginning and the end are  the same = Palindrome          
//Method : Using Recursion: in which a method calls itself to solve some problem.

  public static void main(String[] args) {
      String pal = "Rotator"; 
      boolean result = isPalindrome(pal);
      System.out.println(pal + " is Palindrome = " + result);
  }

  //Body of the Palindrome Method
  public static boolean isPalindrome(String pal){

        //step 1. Check for exception handling for edge cases
        if(pal==null){
            return false; //check if the beginning and the of the palindrome are equal
        }
        if(pal.length()<=1){
            return true;  //if a person enters an empty string that's a palindrome (very important!)
        }

        //step 2. Finding the Palindrome 
        char firstLet = pal.charAt(0);
        char lastLet = pal.charAt(pal.length()-1);//simpler way to get first and last chars

        if(!firstLet.equals(lastLet)){//if the the first and last letters are NOT the same = not palindome
            return false;

        }else{
            return isPalindrome(pal.substring(1,pal.length()-1)); //this will cut first and last letter
        }
    }//end of method
 }//end of class

如您所见，您需要从索引1（删除索引0处的字母）到索引长度1（在substring方法中是唯一的，因此它将删除最后一个字母）的子字符串。

Answer 2

嗨，我能够自己修改代码。我注意到我错过了重要但很少的信息。我将发布代码给其他可能处于同一位置的人。

使用递归方法对回文进行关键处理：

该类中有2个方法。 1用于输出，另一用于找到回文

请小心，因为您可能会遗漏代码信息或将其放置在错误的位置。很难发现并修复。

String pal = "rotator"; 
boolean result = isPalindrome(pal);
System.out.println(pal + " is Palindrome ="+result);
}
//End of method1


//Body of the Palindrome Method

//方法2：使用递归

公共静态布尔isPalindrome（字符串pal）{

    //step 1. Check for exception handling and edge cases
    if(pal==null){
        return false; //check if the beginning and the end of the palindrome are same
    }
    if(pal.length()<=1){
        return true;  //if a person enters an empty string that's a palindrome (very important!)
    }

    //step 2. Finding the Palindrome 
    String firstLet= pal.substring(0,1);
    String lastLet = pal.substring(pal.length()-1, pal.length());

    if(!firstLet.equals(lastLet)){//if the the first and last letters are NOT the same = not palindome
        return false;

    }else{
        return isPalindrome(pal.substring(1, pal.length()-1)) ; //first and last letter of string are same
    }//end of if loop



}//end of  recursive method

Java面试问：我的回文症获得无效的方法

2 个答案: