Java access ArrayList without for loop

Question

I'm trying to make a hangman type game but I'm having difficulty with writing the method that returns and updates the word which the user is trying to guess. So for example if the word to guess was 'hello' I would want it to be printed as * but if someone guesses 'e', it should update to e**. I have kind of got this to work but it loops through two ArrayLists which is not exactly what I want:

public String getVisible() {
    String rtn = null;
    for(Character b: brokenphrases){

        for(Character c:characters){
            if(b == c){
                rtn = rtn + b;

            }
        }

        if(b.toString().equals(" ")){
            rtn = rtn +" ";

        }
        else{

            rtn = rtn + "*";
        }

    }

    return rtn;
}

The idea is that the ArrayList of characters contains the input that the user puts in, and the ArrayList of brokenphrases has the word that needs to be guessed, broken down into single characters so they can compare the two. What happens now is when I run the program I get my word to guess, for example hello printed as * but as I guess letters the the text becomes doubled up instead of being replaced. So if I was to type in h ,e ,l ,o , the system prints out h*e*l*l*o* instead out getting rid of the asterix. I figure that it's to do with using two for loops but I'm not sure on how to mend it. Any advice would be appreciated. Thanks.

robertoia · Accepted Answer · 2014-04-08 11:45:20Z

1

You can simplify your code to get something like this:

public String getVisible() {
    String rtn = null;

    for(Character b: brokenphrases){
        if(characters.contains(b)){
            rtn = rtn + b;
        } else {
            rtn = rtn + "*";
        }
    }

    return rtn;
}

You iterate over all characters of your target word (brokenphrases), for example hello. If the character is in characters, you add the character to the output. If not, you add *.

If you guessed h and l you would get h*ll*.

No need to add blankspace.

edited Apr 8, 2014 at 11:45

answered Apr 8, 2014 at 11:37

robertoia

2,36125 silver badges29 bronze badges

Sign up to request clarification or add additional context in comments.

5 Comments

Lukasz Medza Over a year ago

This doesn't seem to work too well. Every time I add a letter it doubles the whole string in size. So for hello it starts as ***** then becomes ********** then *************** and so on.

robertoia Over a year ago

@LukaszMedza Sorry, i think i found the problem. No need for two loops now. Just check if the letter is in characters.

Lukasz Medza Over a year ago

System.out.println("Word to guess:" + getVisible());

Lukasz Medza Over a year ago

How would I check if the letter is in characters without using a for loop?

robertoia Over a year ago

@LukaszMedza characters.contains(b) checks if b is in characters.

blalasaadri · Accepted Answer · 2014-04-08 11:46:19Z

0

You don't really need the characters to be sorted, so you can use a Set (e.g. a TreeSet) for them instead of an ArrayList. Then you can change your code to something like this:

public String getVisible() {
    String rtn = null;
    for(Character b: brokenphrases){

        if(characters.contains(b)){
            rtn = rtn + b;
        }

        if(b.toString().equals(" ")){
            rtn = rtn +" ";

        }
        else {

            rtn = rtn + "*";
        }

    }

    return rtn;
}

answered Apr 8, 2014 at 11:46

blalasaadri

6,2166 gold badges40 silver badges63 bronze badges

Collectives™ on Stack Overflow

Java access ArrayList without for loop

2 Answers 2

5 Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

5 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related