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for my compsci assignment we're supposed to take a string given to us with integers and letters in it, then create a method that takes that string and converts it into an integer array with the integers in it. For some reason my method is not adding ints to the array, I'm not sure why.

For the LETTERS given in the string, we're supposed to discard them, so we have an array with ONLY int values; ex. input: abs3131afas312 the array would have {3131,312}

This is the link to the assignment.

Here's my method:

public static int[] intParse(String a){
    int[] array1 = new int[a.length()];
    int b = 0;
    for(int i = 0; i < a.length(); ++i)
    {
        int g = a.charAt(i);

        if(g == 1 || g == 2 || g == 3 || g == 4 || g == 5 || g == 6 || g == 7 || g == 8 || g == 9 || g == 0)
        {

            String c;
            for(int j = i; j < a.length(); ++j)
            {
                int k = a.charAt(j);
                if(k != 1 && k != 2 && k != 3 && k != 4 && k != 5 && k != 6 && k != 7 && k != 8 && k != 9 && k != 0)
                {
                    c = a.substring(j,k-1);
                    array1[b] += Integer.parseInt(c);
                    b++;
                    j = (a.length());
                    i = a.charAt(j);

                }
                else
                {
                    c = a.substring(j,a.length());
                    array1[b] = Integer.parseInt(c);
                    j = a.length();
                }
            }

        }


    }

        return array1;
}
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what do you intend to happen to the characters that are letters? –  Justin Gingy McDonald Dec 11 '12 at 1:12
 
You should check if g == '1', or Character.isDigit(...). –  MouseEvent Dec 11 '12 at 1:43
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4 Answers

up vote 2 down vote

Rather than comparing your characters to integers and using Integer.parseInt, you should be using the following very useful utility methods:

Character.isDigit(int codepoint)
Character.getNumericValue(int codepoint)

Also, your logic seems a little sketchy. When k is a digit code point, you are trying to parse the entire rest of the string. That doesn't seem consistent with what you're trying to do with the outer loop.

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My professor told us to use the Integer.parseInt, I tried using your code but I'm confused as how to implement it. But it would work if I created a string to hold the one character, then convert that to a numeric value to check, correct? –  user1804833 Dec 11 '12 at 1:32
 
@user1804833 - Yes, that will work. However, don't make a string out of one character. You sort of had the right idea, I think, in your original code. Once you find a digit, scan ahead to the first non-digit or the end of the input, whichever comes first, and use parseInt on the substring and save the result. Then advance the scanning index to past the input you just consumed. I think that's what you were trying to do with your current code. However, you need to be much more careful about the bookkeeping, because what you have now just isn't going to do that job. But use isDigit()! –  Ted Hopp Dec 11 '12 at 1:52
 
Right now I have: for(int i = 0; i < a.length(); ++i) { String zx; int g; zx = a.substring(i,i+1); g = Integer.parseInt(zx); but it's still not working for some reason, what am I doing wrong? –  user1804833 Dec 11 '12 at 2:05
 
@user1804833 - Well, that doesn't look much like your original code. What's not working about it? Are you testing whether the character at i is a digit? If you step through with the debugger, you should be able to see where it's going wrong. –  Ted Hopp Dec 11 '12 at 3:52
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up vote 0 down vote

First of all, a is a string, and contains ASCII characters not integers. Character '1' is not equal to the integer 1. It is equal to the ASCII value of '1' which happens to be 49.

So first thing you should do is change that long if condition to:

char c = a.charAt(i);
if (c >= '0' && c <='9')
{
   ...
}

What you should do then is keep a string (a new string each time you encounter a non-numeric character and then a numeric character, and keep appending c to it until the character you find is non-numeric.

Then you can simply do Integer.parseInt(yourString) to get the number in an integer.

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up vote 0 down vote

a fix for you: 

char g = a.charAt(i);
if(g == '1' || g == '2' || g == '3' || g == '4' || g == '5' || g == '6' || g == '7' || g == '8' || g == '9' || g == '0') {

or much nicer:

char g = a.charAt(i);
if(g >= '0' && g <= '9') {

Your other if needs to be fixed, too

char k = a.charAt(j);
if(k < '0' || k > '9') {
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up vote 0 down vote

You might prefer this approach:

      Scanner sc = new Scanner("abs3131afas312");
      String match;
      while ((match = sc.findInLine("(\\d+)"))!=null) {
              // instead of printing it, put it in your array
              // or a list (and then convert to array)
              System.out.println(Integer.parseInt(match));
      }         
      sc.close();

Output:

3131
312
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