从Java中的Arraylist中的列表中删除重复项

Question

I checked many examples but i could not applied for my variables. I have a ArratyList Of lists of Strings.

ArrayList<List<String>> bulkUploadList = new ArrayList<List<String>>();

and it's look like this:

[id, title, tags, descriptions]

[4291483113.0000000000000, Camden, camdentown;london, NoValue]
[4292220054.0000000000000, IMG_2720, NoValue, NoValue]
[4292223824.0000000000000, IMG_2917, london;camdentown, NoValue]
[4292224728.0000000000000, IMG_2945, London;CamdenTown, NoValue]

I want to remove those rows which have the same titles and the same tags. I do not know how work with HashSet since I have a ArrayList of List of Strings.

Answer 1

Not best solution, but you can start with this:

    ArrayList<List<String>> bulkUploadList = new ArrayList<List<String>>();
    ArrayList<List<String>> result = new ArrayList<List<String>>();

    HashSet<String> hashSet = new HashSet<>();

    for(List<String> item : bulkUploadList) {
        String title = item.get(1);
        String tags = item.get(2);
        String uniqueString = (title + "#" + tags).trim().toUpperCase();

        if(!hashSet.contains(uniqueString)) {
            result.add(item);
            hashSet.add(uniqueString);
        } else {
            System.out.println("Filtered element " + uniqueString);
        }
    }

Answer 2

As suggested in one of the comments, you should create a class for the data, make that class implement equals(), and then use HashSet to remove dups. Like this.

class Foo {
String id;
String title;
String tags;
String description;

public boolean equals(Foo this, Foo other) {
   return this.id.equals(other.id) 
      && this.title.equals(other.title)
      && etc.
}

then you can remove dups with

 Set<Foo> set = new LinkedHashSet<Foo>(list);

as Sets do not allow duplication, and the equals() method is used to check.

You should use a linkedHashSet here because you want to preserve the order (according to a comment you made on elsewhere).

You should also implement a hashcode() method consistent with equals().