Converting and extracting integers from a string

Question

I'm trying to do something like this:

Input: a string like (the numbers are random):

"<12, 3>, <4, 5>, <23, 5>, <33, 56>"

Output: an array of integers like this one

[12, 3, 4, 5, 23, 5, 33, 56]

I tried using some grammars and some splits but I'm always ending with string or chars.

yes , the string is only of that format , only numbers can change — nescionomen93, Mar 22 '15 at 23:36
Why don't you just remove the "<" and ">" chars and split by "," afterwards? — Stephen Reindl, Mar 22 '15 at 23:42
possible duplicate of [Convert String to int array in java](http://stackoverflow.com/questions/7646392/convert-string-to-int-array-in-java) — Joel, Mar 22 '15 at 23:47
if i would just remove "<>" and split by "," shouldn i found only {12,3,4,5,23,5,33,56}? — nescionomen93, Mar 22 '15 at 23:48

score 0 · Accepted Answer · edited Jan 29 '16 at 05:19

This should do what you want to do:

package split;

public class Main {

    public static void main(String[] args){

        String mString = "<12,3>,<4,5>,<23,5>,<33,56>"; 
        mString = mString.replace("<", ""); 
        mString = mString.replace(">", ""); 
        String[] stringArray = mString.split(","); 

        int[] intArray = new int[stringArray.length];
        int index = 0; 
        for(String mS : stringArray){
            intArray[index] = Integer.parseInt(mS); 
            index++;
        }

        for(int i : intArray){
            System.out.println(i); 
        }
    }
}

To change a String to an int you use Integer.parseIn("your String");.

score 0 · Answer 2 · answered Mar 23 '15 at 01:03

Sometimes the trick with regular expressions is not to try to match the complete format of the thing you are trying to parse, but instead just to find the things you are looking for.

In your case, since you want to flatten the list of numbers, all you need to look for is a sequence of numbers. So that is easy to do.

List<Integer> findNumbers(String s) {
    List<Integer> result = new ArrayList<Integer>();
    Pattern pattern = Pattern.compile("\\d+");
    Matcher matcher = pattern.matcher(s);
    while (matcher.find()) {
        result.add(Integer.parseInt(matcher.group()));
    }
    return result;
}

Converting and extracting integers from a string

2 Answers2