Java - When to use LinkedList over ArrayList in Java

arraylistcollectionsjavalinked-list

I've always been one to simply use:

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

I use the interface as the type name for portability, so that when I ask questions such as these I can rework my code.

When should LinkedList be used over ArrayList and vice-versa?

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Summary ArrayList with ArrayDeque are preferable in many more use-cases than LinkedList. If you're not sure — just start with ArrayList.

TLDR, in ArrayList accessing an element takes constant time [O(1)] and adding an element takes O(n) time [worst case]. In LinkedList adding an element takes O(n) time and accessing also takes O(n) time but LinkedList uses more memory than ArrayList.

LinkedList and ArrayList are two different implementations of the List interface. LinkedList implements it with a doubly-linked list. ArrayList implements it with a dynamically re-sizing array.

As with standard linked list and array operations, the various methods will have different algorithmic runtimes.

For LinkedList<E>

get(int index) is O(n) (with n/4 steps on average), but O(1) when index = 0 or index = list.size() - 1 (in this case, you can also use getFirst() and getLast()). One of the main benefits of LinkedList<E>
add(int index, E element) is O(n) (with n/4 steps on average), but O(1) when index = 0 or index = list.size() - 1 (in this case, you can also use addFirst() and addLast()/add()). One of the main benefits of LinkedList<E>
remove(int index) is O(n) (with n/4 steps on average), but O(1) when index = 0 or index = list.size() - 1 (in this case, you can also use removeFirst() and removeLast()). One of the main benefits of LinkedList<E>
Iterator.remove() is O(1). One of the main benefits of LinkedList<E>
ListIterator.add(E element) is O(1). One of the main benefits of LinkedList<E>

^{Note: Many of the operations need n/4 steps on average, constant number of steps in the best case (e.g. index = 0), and n/2 steps in worst case (middle of list)}

For ArrayList<E>

get(int index) is O(1). Main benefit of ArrayList<E>
add(E element) is O(1) amortized, but O(n) worst-case since the array must be resized and copied
add(int index, E element) is O(n) (with n/2 steps on average)
remove(int index) is O(n) (with n/2 steps on average)
Iterator.remove() is O(n) (with n/2 steps on average)
ListIterator.add(E element) is O(n) (with n/2 steps on average)

^{Note: Many of the operations need n/2 steps on average, constant number of steps in the best case (end of list), n steps in the worst case (start of list)}

LinkedList<E> allows for constant-time insertions or removals using iterators, but only sequential access of elements. In other words, you can walk the list forwards or backwards, but finding a position in the list takes time proportional to the size of the list. Javadoc says "operations that index into the list will traverse the list from the beginning or the end, whichever is closer", so those methods are O(n) (n/4 steps) on average, though O(1) for index = 0.

ArrayList<E>, on the other hand, allow fast random read access, so you can grab any element in constant time. But adding or removing from anywhere but the end requires shifting all the latter elements over, either to make an opening or fill the gap. Also, if you add more elements than the capacity of the underlying array, a new array (1.5 times the size) is allocated, and the old array is copied to the new one, so adding to an ArrayList is O(n) in the worst case but constant on average.

So depending on the operations you intend to do, you should choose the implementations accordingly. Iterating over either kind of List is practically equally cheap. (Iterating over an ArrayList is technically faster, but unless you're doing something really performance-sensitive, you shouldn't worry about this -- they're both constants.)

The main benefits of using a LinkedList arise when you re-use existing iterators to insert and remove elements. These operations can then be done in O(1) by changing the list locally only. In an array list, the remainder of the array needs to be moved (i.e. copied). On the other side, seeking in a LinkedList means following the links in O(n) (n/2 steps) for worst case, whereas in an ArrayList the desired position can be computed mathematically and accessed in O(1).

Another benefit of using a LinkedList arises when you add or remove from the head of the list, since those operations are O(1), while they are O(n) for ArrayList. Note that ArrayDeque may be a good alternative to LinkedList for adding and removing from the head, but it is not a List.

Also, if you have large lists, keep in mind that memory usage is also different. Each element of a LinkedList has more overhead since pointers to the next and previous elements are also stored. ArrayLists don't have this overhead. However, ArrayLists take up as much memory as is allocated for the capacity, regardless of whether elements have actually been added.

The default initial capacity of an ArrayList is pretty small (10 from Java 1.4 - 1.8). But since the underlying implementation is an array, the array must be resized if you add a lot of elements. To avoid the high cost of resizing when you know you're going to add a lot of elements, construct the ArrayList with a higher initial capacity.

If the data structures perspective is used to understand the two structures, a LinkedList is basically a sequential data structure which contains a head Node. The Node is a wrapper for two components : a value of type T [accepted through generics] and another reference to the Node linked to it. So, we can assert it is a recursive data structure (a Node contains another Node which has another Node and so on...). Addition of elements takes linear time in LinkedList as stated above.

An ArrayList, is a growable array. It is just like a regular array. Under the hood, when an element is added at index i, it creates another array with a size which is 1 greater than previous size (So in general, when n elements are to be added to an ArrayList, a new array of size previous size plus n is created). The elements are then copied from previous array to new one and the elements that are to be added are also placed at the specified indices.

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Java - What are the differences between a HashMap and a Hashtable in Java

There are several differences between HashMap and Hashtable in Java:

Hashtable is synchronized, whereas HashMap is not. This makes HashMap better for non-threaded applications, as unsynchronized Objects typically perform better than synchronized ones.
Hashtable does not allow null keys or values. HashMap allows one null key and any number of null values.
One of HashMap's subclasses is LinkedHashMap, so in the event that you'd want predictable iteration order (which is insertion order by default), you could easily swap out the HashMap for a LinkedHashMap. This wouldn't be as easy if you were using Hashtable.

Since synchronization is not an issue for you, I'd recommend HashMap. If synchronization becomes an issue, you may also look at ConcurrentHashMap.

Java - Is Java “pass-by-reference” or “pass-by-value”

Java is always pass-by-value. Unfortunately, when we deal with objects we are really dealing with object-handles called references which are passed-by-value as well. This terminology and semantics easily confuse many beginners.

It goes like this:

public static void main(String[] args) {
    Dog aDog = new Dog("Max");
    Dog oldDog = aDog;
    // we pass the object to foo
    foo(aDog);
    // aDog variable is still pointing to the "Max" dog when foo(...) returns
    aDog.getName().equals("Max"); // true
    aDog.getName().equals("Fifi"); // false
    aDog == oldDog; // true
}
public static void foo(Dog d) {
    d.getName().equals("Max"); // true
    // change d inside of foo() to point to a new Dog instance "Fifi"
    d = new Dog("Fifi");
    d.getName().equals("Fifi"); // true
}

In the example above aDog.getName() will still return "Max". The value aDog within main is not changed in the function foo with the Dog "Fifi" as the object reference is passed by value. If it were passed by reference, then the aDog.getName() in main would return "Fifi" after the call to foo.

Likewise:

public static void main(String[] args) {
    Dog aDog = new Dog("Max");
    Dog oldDog = aDog;
    foo(aDog);
    // when foo(...) returns, the name of the dog has been changed to "Fifi"
    aDog.getName().equals("Fifi"); // true
    // but it is still the same dog:
    aDog == oldDog; // true
}
public static void foo(Dog d) {
    d.getName().equals("Max"); // true
    // this changes the name of d to be "Fifi"
    d.setName("Fifi");
}

In the above example, Fifi is the dog's name after call to foo(aDog) because the object's name was set inside of foo(...). Any operations that foo performs on d are such that, for all practical purposes, they are performed on aDog, but it is not possible to change the value of the variable aDog itself.

For more information on pass by reference and pass by value, consult the following SO answer: https://stackoverflow.com/a/430958/6005228. This explains more thoroughly the semantics and history behind the two and also explains why Java and many other modern languages appear to do both in certain cases.

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