如何递归获取输入字符串并返回每个字符重复的字符串？

Question

Basically I have an assignment that requires me to write a method called stutter . The stutter method is supposed to take an input String s and return the string with each character repeated. For example, if the input string was "help" then the result of running this method should be "hheellpp". I have tried a bunch of different things and can't get it to work. Here is my code:

import java.util.*;

public class Stutter {  
    static String stutterString = "";

    public static String stutter ( String s ) {
        char ch = s.charAt (0);
        String tempString = String.valueOf ( ch );
        if ( s.length() == 0 ) {
            return stutterString;
        } else {
            stutterString += tempString + tempString;
            return stutter ( s.substring (1) );
        }
    }

    public static void main ( String [] args ) {
        Scanner inputScanner = new Scanner ( System.in );
        System.out.println ( "What word would you like to stutter?" );
        String userInput = inputScanner.next();     
        inputScanner.close();       
        System.out.println ( stutter ( userInput ) );
    }
}

I get an error that I'm not sure what to do with. This is the error:

Exception in thread "main" java.lang.StringIndexOutOfBoundsException: String index out of range: 0
    at java.lang.String.charAt(Unknown Source)
    at Stutter.stutter(Stutter.java:12)
    at Stutter.stutter(Stutter.java:23)
    at Stutter.main(Stutter.java:41)

Any help would be appreciated. This isn't a huge program. As you can see, I've posted the entire Stutter class that I'm using. It's just bugging me because I'm sure there is a simple fix to this, but I'm not seeing it.

Answer 1

You need to change this line

char ch = s.charAt (0);

to

char ch = s.length() > 0 ? s.charAt(0) : ' ';

And your code will work as expected.

A better and clearer solution would be:

if (s.length() == 0) {
    return stutterString;
} else {
    char ch = s.charAt(0);
    String tempString = String.valueOf(ch);
    stutterString += tempString + tempString;
    return stutter(s.substring (1));
}

What word would you like to stutter?
>> abcdefg
>> aabbccddeeffgg

Explanation:

What will happen when you try to s.charAt(0) when s is an empty String? You're not verifying that s is not empty, adding the simple check s.length() > 0 is what you're missing.

Tip: Always use the debugger, it's there to help you, you'll better understand the flow of your program when you use it. Also when writing a recursion, using a pencil and a paper to draw the calls will help you to understand it.

Answer 2

I did it like this:

My base case is when the length of the string is less than 1, in which case it will return "".

Otherwise it will print of the first character of the string twice, and then call the stutter method again.

I pass in the original string as a parameter, except I have removed the first character from it.In this way the 2nd character of the original string will be printed out twice next and the string gets shorter.

import java.util.*;

public class Stutter {  

    public static String stutter ( String s ) {
        if(s.length() < 1) return "";
        else{
            return "" + s.charAt(0) + s.charAt(0) + stutter(s.substring(1,s.length()));
        }
    }

    public static void main ( String [] args ) {
        Scanner s = new Scanner ( System.in );
        System.out.println ( "What word would you like to stutter?" );
        String userInput = s.nextLine();
        System.out.println(stutter(userInput));
    }
}

Answer 3

提示：如果s.length()为零，则s.charAt(0)将抛出异常...因为您正在尝试获取超出零长度字符串末尾的字符。

Answer 4

Check the length of your s variable before s.charAt (0) . For example, move

char ch = s.charAt (0);
String tempString = String.valueOf ( ch );

to else block

Answer 5

How about this:

public class Stutter {
    private static String head(String str) {
        return str.substring(0,1);
    }

    private static String tail(String str) {
        return str.substring(1);
    }

    private static String stutter(String str) {
        if (str.length() > 0)
            return head(str)+head(str)+stutter(tail(str));
        else
            return "";
    }

    public static void main(String args[]) throws Exception  {
        if (args.length > 0) {
            System.out.println(stutter(args[0]));
        }
    }
}

Answer 6

This can be achieved very simply using String.replaceAll() :

public class Stutter {
    public static void main (String args[]) {
        String TEST_STRING = "abcdefg";
        System.out.println(stutter(TEST_STRING));
    }

    private static String stutter(String s) {
        return s.replaceAll("(.)", "$1$1");
    }
}

Answer 7

Editted.

import java.util.*;

public class Stutter {

    static String stutterString = "";

    public static String stutter ( String s ) {
        if(s.length() > 0)
        {
            return stutterString.concat(s.substring(0,1)).concat(s.substring(0,1)).concat(stutter(s.substring(1)));
        }
        else
        {
            return stutterString;
        }
    }

    public static void main ( String [] args ) {


        Scanner inputScanner = new Scanner ( System.in );

        System.out.println ( "What word would you like to stutter?" );
        String userInput = inputScanner.next();

        inputScanner.close();

        System.out.println ( stutter ( userInput ) );

    }

}