Telephone Key Pad Combination

Question

For homework I am trying to print all possible combinations of letters that can be represented using the numeric keypad on a telephone. This problem is the same as the one posted here: How can I print out all possible letter combinations a given phone number can represent?

I keep getting a String index out of range: 0 problem but I'm not 100% sure why. Here is the code:

public class TelephoneKeyPad2 {
public static String keyPad[][] = {
        {"0"}, {"1"}, {"A", "B", "C"}, {"D", "E", "F"}, {"G", "H", "I"},
        {"J", "K", "L"}, {"M", "N", "O"}, {"P", "Q", "R", "S"},
        {"T", "U", "V"}, {"W", "X", "Y", "Z"}
};

public static void tpK(String num, StringBuilder combination) {
    if (num.length() == 0) {
        System.out.println(combination);
    }

    for (String letter: keyPad[Character.getNumericValue(num.charAt(0))]) {
        tpK(num.substring(1), combination.append(letter));
    }
}

public static void main(String[] args) {
    tpK("345", new StringBuilder());
} }

I have a feeling it's because of the if statement...there is no way to break the recursion after num.length == 0 is confirmed to be true, so the string index out of range. I am not sure how to fix this problem and any help would be appreciated.

EDIT: So it seems like it may be just a logic problem then afterwards.

DGJ

DGJK

DGJKL

DGJKLHJ

DGJKLHJK

are example outputs so now I'm really confused on how to solve this problem as well.

"There is no way to break the recursion"--use a `return` statement or an else statement in order to allow the second half of the `tpK` method run *only* when `num.length()` isn't zero. — FThompson, Feb 03 '15 at 05:22
Thanks for the responses...however it also seems like my logic is off as well. After adding the else statement in (or return statement), it returns: DGJ DGJK DGJKL DGJKLHJ DGJKLHJK DGJKLHJKL DGJKLHJKLIJ DGJKLHJKLIJK DGJKLHJKLIJKL DGJKLHJKLIJKLEGJ DGJKLHJKLIJKLEGJK DGJKLHJKLIJKLEGJKL DGJKLHJKLIJKLEGJKLHJ DGJKLHJKLIJKLEGJKLHJK DGJKLHJKLIJKLEGJKLHJKL — Senescence, Feb 03 '15 at 05:31

score 0 · Answer 1 · answered Feb 03 '15 at 05:28

Add an else statement so that the for loop only gets executed when there is more than 0 characters:

public static void tpK(String num, StringBuilder combination) {
    if (num.length() == 0) {
        System.out.println(combination);
    } else {

        for (String letter: keyPad[Character.getNumericValue(num.charAt(0))]) {
            tpK(num.substring(1), combination.append(letter));
        }
    }
}

score 0 · Answer 2 · answered Feb 07 '15 at 20:16

I solved the other problem I had. Turns it out the StringBuilder seems to create a static object(?). When the program goes through recursion, it only instantiates one object. I changed the StringBuilder type to String and used the concat method instead, solving the problem.

score -1 · Answer 3 · answered Feb 03 '15 at 05:21

-1

If num.length() is 0, you still go on to ask for num.charAt(0).

answered Feb 03 '15 at 05:21

John3136

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While this is a true statement, it seems that the OP is asking about how to break the recursion in order to prevent exactly what you're pointing out. – FThompson Feb 03 '15 at 05:23
`I keep getting a String index out of range: 0 problem but I'm not 100% sure why` - this is why. Perhaps OP should be asking 2 different question ;-) – John3136 Feb 03 '15 at 05:24

Telephone Key Pad Combination

3 Answers3