如何摆脱此String []中的空格？

Question

Originally, I was just trying to get the characters in a string. I used split to isolate my characters to count them. I have my characters, but I can't get rid of the spaces it's printing in the array.

I tried everything that I saw in other stack overflow posts.

public class Test {

 public static void main(String[] args) {
    String str = "+d+=3=+s+";
    String[] alpha = str.split("[^a-z]");
    for(int i = 0; i < alpha.length; i++) {
      alpha[i] = alpha[i].replaceAll("\\s+", "");
         System.out.println(alpha[i]);
    }

       // System.out.println(alpha.length);
    }

}

I'm just trying to count characters without spaces and using more loops.

Answer 1

Do this way

public class Test {

    public static void main(String[] args) {
        String str = "+d+=3=+s+";
        System.out.println(str);

        str = str.replaceAll("[^a-zA-Z]", "");
        System.out.println(str);
        System.out.println("count "+str.length());
    }

}

Output will be

ds
count 2

output doesn't contains whitespace so you can get rid of loops

Answer 2

If you're referring to the blank lines in your output, those are not spaces , they are empty strings in the alpha array.

Since you regex only matching a single character, the split operation will create empty entries, ie a plain split will produce the following array:

{ "", "d", "", "", "", "", "s", "" }

Since split by default removes trailing empty strings, what you actually get is:

{ "", "d", "", "", "", "", "s" }

As you can see, there are no spaces anywhere.

You can remove most of the empty strings by changing "[^az]" to "[^az]+" , so multiple consecutive non-letters will be a single separator, which will give you:

{ "", "d", "s" }

To remove the leading empty string, remove the leading non-letters from the original string first:

String[] alpha = str.replaceFirst("[^a-z]+", "").split("[^a-z]+");

That will give you:

{ "d", "s" }

Which of course will eliminate all the blank lines in your output:

d
s

And you can remove that replaceAll("\\\\s+", "") call, since it's not doing anything.