OK, it’s time for test results of proposed approaches. Here what approaches I have tested (name of each approach is also class name in my sources): NaiveRemoveManyPerformer – ArrayList with iterator and remove – first and naive implementation given in my question. BetterNaiveRemoveManyPerformer – ArrayList with backward iteration and removal from end to front. LinkedRemoveManyPerformer … Read more
I would skip the data structures exercise and just use an SQL database. Why write another custom data structure that you have to analyze and debug, just use a database. They are really good at answering queries like this.
The key in List<T>.slice function is the .toList() at the end, which will create a new List with the elements, using a shallow element copy. For summary: .slice() will create a new List with the shallow copy of the subset of elements; if the original list contains objects, changes to values within the objects will … Read more
There is no specified maximum size of these structures. The actual practical size limit is probably somewhere in the region of Integer.MAX_VALUE (i.e. 2147483647, roughly 2 billion elements), as that’s the maximum size of an array in Java. A HashSet uses a HashMap internally, so it has the same maximum size as that A HashMap … Read more
Here are the characteristic differences of some important Map implementations: LinkedHashMap: “with predictable iteration order […] which is normally the order in which keys were inserted into the map (insertion-order).” HashMap: “makes no guarantees as to the order of the map” TreeMap: “is sorted according to the natural ordering of its keys, or by a … Read more
In brief: A list is an ordered list of objects, where the same object may well appear more than once. For example: [1, 7, 1, 3, 1, 1, 1, 5]. It makes sense to talk about the “third element” in a list. You can add an element anywhere in the list, change an element anywhere … Read more
You can use Java 8 Optional: String[] arr = null; List<String> list = Arrays.asList(Optional.ofNullable(arr).orElse(new String[0]));
List(Of String) will handle that, mostly – though you need to either use AddRange to add a collection of items, or Add to add one at a time: lstOfString.Add(String1) lstOfString.Add(String2) lstOfString.Add(String3) lstOfString.Add(String4) If you’re adding known values, as you show, a good option is to use something like: Dim inputs() As String = { “some … Read more
You use in. if element in thetuple: #whatever you want to do.