Simplest: dump the whole collection into a Set (using the Set(Collection) constructor or Set.addAll), then see if the Set has the same size as the ArrayList. List<Integer> list = …; Set<Integer> set = new HashSet<Integer>(list); if(set.size() < list.size()){ /* There are duplicates */ } Update: If I’m understanding your question correctly, you have a 2d … Read more
Yes: use ArrayList. In Java, “normal” arrays are fixed-size. You have to give them a size and can’t expand them or contract them. To change the size, you have to make a new array and copy the data you want – which is inefficient and a pain for you. Fortunately, there are all kinds of … Read more
In Java8 you can use streams: public static Carnet findByCodeIsIn(Collection<Carnet> listCarnet, String codeIsIn) { return listCarnet.stream().filter(carnet -> codeIsIn.equals(carnet.getCodeIsin())).findFirst().orElse(null); } Additionally, in case you have many different objects (not only Carnet) or you want to find it by different properties (not only by cideIsin), you could build an utility class, to ecapsulate this logic in it: … Read more
Use Collection#retainAll(). listA.retainAll(listB); // listA now contains only the elements which are also contained in listB. If you want to avoid that changes are being affected in listA, then you need to create a new one. List<Integer> common = new ArrayList<Integer>(listA); common.retainAll(listB); // common now contains only the elements which are contained in listA and … Read more
According to the reference, arrays are created in the following way: For Java’s primitive types there are distinct types IntArray, DoubleArray etc. which store unboxed values. They are created with the corresponding constructors and factory functions: val arrayOfZeros = IntArray(size) //equivalent in Java: new int[size] val numbersFromOne = IntArray(size) { it + 1 } val … Read more
Here’s a tutorial about ordering objects: The Java Tutorials – Collections – Object Ordering Although I will give some examples, I would recommend to read it anyway. There are various way to sort an ArrayList. If you want to define a natural (default) ordering, then you need to let Contact implement Comparable. Assuming that you … Read more
Custom Comparator should help Collections.sort(list, new Comparator<String>() { @Override public int compare(String s1, String s2) { return s1.compareToIgnoreCase(s2); } }); Or if you are using Java 8: list.sort(String::compareToIgnoreCase);
Implement custom adapter for your class: public class MyClassAdapter extends ArrayAdapter<MyClass> { private static class ViewHolder { private TextView itemView; } public MyClassAdapter(Context context, int textViewResourceId, ArrayList<MyClass> items) { super(context, textViewResourceId, items); } public View getView(int position, View convertView, ViewGroup parent) { if (convertView == null) { convertView = LayoutInflater.from(this.getContext()) .inflate(R.layout.listview_association, parent, false); viewHolder = … Read more
If I read the JSONArray constructors correctly, you can build them from any Collection (arrayList is a subclass of Collection) like so: ArrayList<String> list = new ArrayList<String>(); list.add(“foo”); list.add(“baar”); JSONArray jsArray = new JSONArray(list); References: jsonarray constructor: http://developer.android.com/reference/org/json/JSONArray.html#JSONArray%28java.util.Collection%29 collection: http://developer.android.com/reference/java/util/Collection.html
The method arrayList.size() returns the number of items in the list – so if the index is greater than or equal to the size(), it doesn’t exist. if(index >= myList.size() || index < 0){ //index does not exists }else{ // index exists }