How to convert Set containing values into string

Question

public static void main(String[] args) {
        // TODO Auto-generated method stub
        String str = "abcdaa";
        dups(str);
    }
    public static void dups(String str){
        HashSet hs = new HashSet();
        char[] ch = str.toCharArray();
        for(int i=0; i < ch.length;i++){
            hs.add(ch[i]);
        }
        System.out.println(hs);
    }

The above code return Output: [a,b,c,d]

But I want to print the Set values into a string so I can return a string value return value looks like this:Expected Output: abcd

Take a look at `java.util.StringJoiner` – Steve Kuo May 12 '17 at 17:33 — Steve Kuo, May 12 '17 at 17:33

score 2 · Accepted Answer · edited May 23 '17 at 12:18

 public static void main(String[] args) {
        // TODO Auto-generated method stub
        String str = "abcdaa";
        dups(str);
    }

    public static void dups(String str) {
        HashSet<Character> hs = new HashSet<Character>();
        char[] ch = str.toCharArray();
        StringBuilder sb = new StringBuilder();
        for (int i = 0; i < ch.length; i++) {
            if(hs.add(ch[i])){
                sb.append(ch[i]);
            }
        }
        System.out.println(sb);
    }

EDIT

public static void dups(String str) {
            HashSet<Character> hs = new HashSet<Character>();
            StringBuilder sb = new StringBuilder();
            for (Character character : str.toCharArray()) {
                if(hs.add(character)){
                    sb.append(character);
                }
            }
            System.out.println(sb);
        }

I can't think a better way to do this... It's better use StringBuilder instead of String, check this answer https://stackoverflow.com/a/1532483/6949032

Why not use modern language features? `System.out.println(hs.stream().collect(Collectors.joining()));` — DavidW, May 12 '17 at 17:07
thanks @ErickMaia.. it was working.. But I am looking to optimize that instead of using StringBuilder. — user1918566, May 12 '17 at 17:14

score 2 · Answer 2 · answered May 15 '17 at 04:06

2

Not sure why no one had mentioned this yet, but it's pretty simple in Java 8:

System.out.println(String.join("", hs));

Note that if you want to preserve the original order, you'll need to use LinkedHashSet instead.

answered May 15 '17 at 04:06

shmosel

49,289
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Vasyl Lyashkevych · Answer 3 · 2017-05-12T17:30:41.957

0

public static void main(String[] args) {
    // TODO Auto-generated method stub
    String str = "abcdaa";
    String result = dups(str);
    System.out.println(result);
}
public static String dups(String str){
    HashSet hs = new HashSet();
    String strss = "";
    char[] ch = str.toCharArray();
    for(int i=0; i < ch.length;i++){
        if(hs.add(ch[i])){
            strss +=ch[i];
        }
    }
    return strss;
}

edited May 12 '17 at 17:30

answered May 12 '17 at 16:57

Vasyl Lyashkevych

1,920
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the return is not `abcd`=( – Erick Maia May 12 '17 at 17:03
Yes .. Return is not abcd – user1918566 May 12 '17 at 17:27
1

I have checked, it returns "abcd" – Vasyl Lyashkevych May 12 '17 at 17:30
1

`strss +=` is a bad idea as a new String is allocated every time. This is exactly what StringBuilder is for – Steve Kuo May 12 '17 at 17:38
a new String, I think, it's a good idea, because you will have always new instance, sometimes it can untie misunderstanding – Vasyl Lyashkevych May 12 '17 at 17:43
Strings are immutable so there's no value in a new instance – Steve Kuo May 12 '17 at 18:48
Yes, you're right, in this example, "strss" will have no value. I was considering the case where we have a new instance – Vasyl Lyashkevych May 12 '17 at 19:32

Donald Raab · Answer 4 · 2017-05-15T04:04:27.590

If you can use a third-party library, the CharAdapter class in Eclipse Collections can solve the problem.

String str = "abcdaa";
CharAdapter distinct = CharAdapter.adapt(str).distinct();
System.out.println(distinct);

You can see what the code does here. Using a java.util.HashSet will box the char values as Character objects. The distinct method in CharAdapter uses a CharHashSet which does not need to box the char values.

Note: I am a committer for Eclipse Collections.

How to convert Set containing values into string

4 Answers4