HashSet In the.NET Framework 3.5, the C# HashSet data structure was introduced. The HashSet MSDN page has a complete list of the implemented members.
Asked by 001
The HashSet class in C# takes the first technique, which means that the order of the elements is not preserved. It’s a lot quicker than a standard List. When working with fundamental types, HashSet is noticeably faster, according to some simple benchmarks (int, double, bool, etc.). When working with class objects, it is much faster. That is to say, HashSet is quick.
The only disadvantage of HashSet is that it does not support indices. You can either use an enumerator to retrieve elements or use the built-in method to transform the HashSet to a List and iterate through it. Take a look at this.
Answered by kamaci
A HashSet has an internal structure (hash) that allows things to be rapidly searched and identified. The disadvantage is that iterating through a HashSet (or retrieving an item by index) takes a long time.
So, why would someone want to know if an entry in a set already exists?
A HashSet can be beneficial in obtaining different values from a list that may contain duplicates. Once an item is added to the HashSet, determining whether or not it exists is simple (Contains operator).
IntersectWith, IsSubsetOf, IsSupersetOf, Overlaps, SymmetricExceptWith, and UnionWith are some of the HashSet’s other features.
These set operations will be recognized if you are familiar with the object constraint language. You’ll also see that it’s one step closer to an executable UML implementation.
Answered by k rey
Simply put, and without exposing any kitchen secrets, a set is a collection of items that contain no duplicates and are not arranged in any specific sequence. As a result, a HashSetT> is identical to a generic ListT>, except that it is optimized for quick lookups (through hashtables, as the name implies) at the expense of order.
Answered by Stacked
If you just need to avoid duplicates in your application, HashSet is the way to go because its Lookup, Insert, and Remove complexity are all O(1) – constant. This means that it doesn’t matter how many elements HashSet has; it will take the same amount of time to verify if an element exists or not, and because you’re inserting items at O(1) as well, it’s ideal for this.
Answered by Matas Vaitkevicius
Post is based on https://stackoverflow.com/questions/4558754/define-what-is-a-hashset