CODEWITHSUNDEEP
javastringchar
user412308
user412308

Reputation: 59

Java Method to find the occurencies of a certain character

my code should prompt the user to enter a string and a character, and tell where the character is located for instance "Welcome" and "e" returns "2, 7"

How can my code be fixed? Code is here. Thanks in advance (this is not homework, but some hint could be useful anyway if you don't want to post a solution).

import java.util.Scanner;
public class Test {
    public static void main(String[] args) {
        System.out.println("Please enter a string and a character");
        Scanner input = new Scanner(System.in);
        String s = input.nextLine();
        char ch = input.next().charAt(0);

        System.out.println(count(ch));

    }

    public static int count (String s, char a) {
        int count = 0;
        for (int i = 0; i < s.length(); i++) {
            if (s.charAt(i) == a) {
                count++;
            }
        }
        return count;

    }
}

Upvotes: 0

Views: 70

Answers (3)

Michał Szewczyk
Michał Szewczyk

Reputation: 8168

Since Java 8, you can do this by using streams:

public static String count(String s, char a) {
    return IntStream.range(0, s.length())
            .filter(i -> a == s.charAt(i)).mapToObj(i -> i + "")
            .collect(Collectors.joining(", "));
}

This code will print indexes of your Character, seprated by comma.
For more about streams, you can read here, in Java 8 documentation.

Upvotes: 1

Ashraful Islam
Ashraful Islam

Reputation: 12830

Just Change the count method : 

   public static ArrayList<Integer> count(String s, char a) {
        ArrayList<Integer> positions = new ArrayList<>();
        for (int i = 0; i < s.length(); i++) {
            if (s.charAt(i) == a) {
                positions.add(i+1);
            }
        }
        return positions;
   }

Upvotes: 0

ROMANIA_engineer
ROMANIA_engineer

Reputation: 56616

Some mistakes:

  1. Your code doesn't compile. Call:

    System.out.println(count(s, ch));

    instead of

    System.out.println(count(ch));

  2. You count the number of appearances. Instead, you should keep the indexes. You can use a String or you can add them to a list / array and convert it later to what you want.

    public static String count(String s, char a) {
    String result = "";
    for (int i = 0; i < s.length(); i++) {
        if (s.charAt(i) == a) {
            result += (i+1) + ", ";
        }
    }
    return result.substring(0, result.length() - 2);
}

    I used i+1 instead of i because the indexes start at 0 in Java.

    I also returned the string result.substring(0, result.length() - 2) without its last 2 characters, because I added , after every character.

Upvotes: 1

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