Java:How do I make the "char" parameter into a String parameter in this specific java method?

Question

public static int getNthOccurrence(int n, char find, String str)
    {
        int counter=0;
        for(int i=0;i<str.length();i++)
        {
            if(str.charAt(i)==find)
            {
                counter++;
                if(counter==n)
                    return i;
            }
        }
        return -1;
    }

I have already seen this thread Java: method to get position of a match in a String?

The code runs that:

int n=getNthOccurrence(3,'n',"you arent gonna find me");

output: 13

What exactly do you mean? What are you looking for it to do? — Andy Turner, Dec 17 '17 at 19:53
Are you asking how to replace `find` with a single character `String` or with a general `String`? — Eran, Dec 17 '17 at 19:53
Swapping the word "char" with "String"? What isn't working? What are you having trouble with? — Davy M, Dec 17 '17 at 19:54
@DavyM I have tried that, but how would I change the line (str.charAt(i)==find) so that it works with Strings? — Cool Dude89, Dec 17 '17 at 19:56

Andy Turner · Answer 1 · 2017-12-17T20:56:37.717

Once you've changed the parameter type to String:

public static int getNthOccurrence(int n, String find, String str)

Use String.indexOf(String, int):

i = -find.length();
for (int count = 0; count < n; ++count) {
  i = str.indexOf(find, i + find.length());
  if (i < 0) break;
}
return (i >= 0) ? i : -1;

(Note that String.indexOf(char, int) is a preferable way to do it with char, as in the case of the original code).

score 0 · Answer 2 · answered Dec 17 '17 at 20:07

0

You can make a char into a String by simply doing this

String theString = yourChar + "";

Your code could look something like this !spoiler below!

public static int getNthOccurrence(int n, char find, String str)
    {

        String f = find + "";
        int counter = 0;

        for (String c : str.split("")){

            if (c.equals(f)) n--;
            if (n < 1 ) return counter;
            counter++;
        }

        return -1;
    }
}

answered Dec 17 '17 at 20:07

adihrustic

1
2

I'm not convinced that this is quite what OP meant, because this offers no advantage over the current way. Moreover, this is just a poor approach: don't split `str`, just iterate over the indices and use `str.regionMatches(i, f, 0, f.length())`. – Andy Turner Dec 17 '17 at 20:14
What makes the approach poor? – adihrustic Dec 17 '17 at 20:42
Splitting the string to iterate it character-by-character. – Andy Turner Dec 17 '17 at 20:45
Yes, why is that poor? – adihrustic Dec 17 '17 at 20:48
Because you can iterate it character-by-character using `str.charAt`. Converting everything to a string requires you to create a string per character, as well as an array to hold the strings; but it's unnecessary because the only operation you do (equals, on a string that is always 1 char long) can be done equivalently with chars. – Andy Turner Dec 17 '17 at 20:53

Java:How do I make the "char" parameter into a String parameter in this specific java method?

2 Answers2