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How to remove duplicate characters from a string in Java

3 Answers

0 votes
import java.util.*; 

public class MyClass {
    public static String remove_duplicate_characters(String s) {
       StringBuilder sb = new StringBuilder();
        for (int i = 0; i < s.length(); i++) {
            char ch = s.charAt(i);
            if (s.indexOf(ch, i + 1) == -1) {
                sb.append(ch);
            }
        }
        
        return sb.toString();
    }
    
    public static void main(String args[]) {
        String s = "jaaaaaaaavva";
        
        s = remove_duplicate_characters(s);
        
        System.out.println(s);
    }
}



/*
run:

jva

*/

 



answered Oct 27, 2019 by avibootz
edited Oct 27, 2019 by avibootz
0 votes
import java.util.*; 

public class MyClass {
    public static String remove_duplicate_characters(String s) {
        char[] char_arr = s.toCharArray();
        Set<Character> char_set = new LinkedHashSet<Character>();
        for (char ch : char_arr) {
            char_set.add(ch);
        }

        StringBuilder sb = new StringBuilder();
        for (Character ch : char_set) {
            sb.append(ch);
        }
        
        return sb.toString();
    }
    
    public static void main(String args[]) {
        String s = "jaaaaaaaavva";
        
        s = remove_duplicate_characters(s);
        
        System.out.println(s);
    }
}



/*
run:

jav

*/

 



answered Oct 27, 2019 by avibootz
edited Oct 27, 2019 by avibootz
0 votes
public class MyClass {
    public static String remove_duplicate_characters(String s) {
        String no_dup = "";
         
        for (int i = 0; i < s.length(); i++) {
            if(!no_dup.contains(String.valueOf(s.charAt(i)))) {
                no_dup += String.valueOf(s.charAt(i));
            }
        }
        return no_dup;
    }

    public static void main(String args[]) {
        String s = "jaaaaaaaavva";
         
        s = remove_duplicate_characters(s);
         
        System.out.println(s);
    }
}



/*
run:

jav

*/

 



answered Oct 30, 2019 by avibootz

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