3

Consider the following code:

final Set<String> allPaths = new HashSet<String>();
for (final String path: paths) {
        allPaths.add(path);
}
final MyData d = new MyData(new ArrayList<String>(allPaths));

MyData is some class I should not touch. It should get an ArrayList as an argument. Before this day, we used that way because we didn't care about the order, that is why we used Set (so there will not be duplicated). But now, I would like to keep the order of the elements, so after some research, I found out that I can use the data structure LinkedHashSet in order to do so. So I did:

final LinkedHashSet<String> allPaths = new LinkedHashSet<String>();
for (final String path: paths) {
        allPaths .add(path);
}
final MyData d = new MyData(new ArrayList<String>(allPaths));

Problem is, I'm not sure how to convert LinkedHashSet to ArrayList. Also, I thought of using ArrayList instead of LinkedHashSet so I won't have to convert it, but I'll have to iterate over the array (O(n)).

What good, clean and effiect way should I use?

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9
  • Possible duplicate of Convert Set to List without creating new List Commented Dec 30, 2018 at 17:23
  • Change MyData to use a Collection. Why was it hardcoded to use an ArrayList in the first place? Commented Dec 30, 2018 at 17:24
  • 1
    What is wrong with the way you are doing it? There should not be a difference in the creation of the ArrayList between using a LinkedHashSet and using any other Set, except that it will use the proper iteration order. Commented Dec 30, 2018 at 17:25
  • what's the problem with what you're doing right now? I mean, creating a list fro ma set is basically what you did and this is fine. what does it have to do with O(n)? I don't get this point. Commented Dec 30, 2018 at 17:27
  • 1
    ...and I wonder what paths is in your code? Is that a List or Set? Commented Dec 30, 2018 at 17:30

4 Answers 4

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10

Just use the public boolean addAll(Collection<? extends E> c) method, on the arrayList, it accepts any Collection.

You have your LinkedHashSet:

final LinkedHashSet<String> allPaths = new LinkedHashSet<String>();
for (final String path: paths) {
        allPaths .add(path);
}

and then do (you can use this even if mylist is not empty):

List<String> myList = new ArrayList<>();
mylist.addAll(allPaths);

or for even a simpler approach:

List<String> myList = new ArrayList<>(allPaths);
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6 Comments

does new ArrayList<String>(<???>) know to handle with a collection (like LinkedHashSet)? If so, will my way still work?
@TTaJTa4 yes, that is the whole point of the Collection interface. It is an abstract idea, that has many implementations, but it is basically the same thing, a collection of things. Java uses Collection, instead of hard coded implementations (like ArrayList for example), to be able to swap from one implementation to another - just like in your case, when you want to use LinkedHashSet instead of HashSet, without the need to re-write the method/class you are passing the argument into.
addAll is merely an extra over what the OP already did. There is no need to use it, it can be used the way it was originally.
@RealSkeptic you call use the addAll method even if the list is not empty, in case he needs it, as I mentioned in my answer.
I'm wondering what's different in your answer from what he already does in his question?!
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2

Why dont you just convert your paths to an LinkedHashSet like that (assuming that paths is a Collection?

final MyData d = new MyData(new ArrayList<>(new LinkedHashSet<>(paths)));

In case paths is an array, you can use Arrays.asList(paths) inside the conversion above

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0 
@TTaJTa4 you can use the code below as an example. Both ways are fine.



import java.util.ArrayList;
import java.util.LinkedHashSet;
import java.util.Set;

public class ConvertLinkedHashSetToArrayList
{

  public static void main(String[] args)
  {
    Set<String> testStrings = new LinkedHashSet<>();
    testStrings.add("String 1");
    testStrings.add("String 2");
    testStrings.add("String 3");
    testStrings.add("String 4");
    testStrings.add("String 5");

    System.out.println("** Printing LinkedHashSet: " + testStrings);
    ArrayList<String> linkedHashSetToArrayList1 = new ArrayList<>(testStrings);
    System.out.println("** Printing linkedHashSetToArrayList1:" + 
    linkedHashSetToArrayList1);

    ArrayList<String> linkedHashSetToArrayList2 = new ArrayList<>();
    linkedHashSetToArrayList2.addAll(testStrings);
    System.out.println("** Printing linkedHashSetToArrayList2:" + 
    linkedHashSetToArrayList2);
  }
}

Results are like:

** Printing LinkedHashSet: [String 1, String 2, String 3, String 4, String 5]
** Printing linkedHashSetToArrayList1:[String 1, String 2, String 3, String 4, String 5]
** Printing linkedHashSetToArrayList2:[String 1, String 2, String 3, String 4, String 5]

Follow the GitHub link for the full java project: GitHub example

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Comments

0

If paths is a collection you can get an array list no duplicates:

ArrayList<String> p = paths.stream().distinct().collect(Collectors.toCollection(ArrayList::new));

If paths is an array:

ArrayList<String> p = Stream.of(paths).distinct().collect(Collectors.toCollection(ArrayList::new));

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