Develop Reference

Sorted Lists in Java [duplicate]

In Java there are the SortedSet and SortedMap interfaces. Both belong to the Java Collections framework and provide a sorted way to access the elements.
However, in my understanding there is no SortedList in Java. You can use java.util.Collections.sort() to sort a list.
Any idea why it is designed like that?

List iterators guarantee first and foremost that you get the list's elements in the internal order of the list (aka. insertion order). More specifically it is in the order you've inserted the elements or on how you've manipulated the list. Sorting can be seen as a manipulation of the data structure, and there are several ways to sort the list.
I'll order the ways in the order of usefulness as I personally see it:
1. Consider using Set or Bag collections instead
NOTE: I put this option at the top because this is what you normally want to do anyway.
A sorted set automatically sorts the collection at insertion, meaning that it does the sorting while you add elements into the collection. It also means you don't need to manually sort it.
Furthermore if you are sure that you don't need to worry about (or have) duplicate elements then you can use the TreeSet<T> instead. It implements SortedSet and NavigableSet interfaces and works as you'd probably expect from a list:
TreeSet<String> set = new TreeSet<String>();
set.add("lol");
set.add("cat");
// automatically sorts natural order when adding
for (String s : set) {
System.out.println(s);
}
// Prints out "cat" and "lol"
If you don't want the natural ordering you can use the constructor parameter that takes a Comparator<T>.
Alternatively, you can use Multisets (also known as Bags), that is a Set that allows duplicate elements, instead and there are third-party implementations of them. Most notably from the Guava libraries there is a TreeMultiset, that works a lot like the TreeSet.
2. Sort your list with Collections.sort()
As mentioned above, sorting of Lists is a manipulation of the data structure. So for situations where you need "one source of truth" that will be sorted in a variety of ways then sorting it manually is the way to go.
You can sort your list with the java.util.Collections.sort() method. Here is a code sample on how:
List<String> strings = new ArrayList<String>()
strings.add("lol");
strings.add("cat");
Collections.sort(strings);
for (String s : strings) {
System.out.println(s);
}
// Prints out "cat" and "lol"
Using comparators
One clear benefit is that you may use Comparator in the sort method. Java also provides some implementations for the Comparator such as the Collator which is useful for locale sensitive sorting strings. Here is one example:
Collator usCollator = Collator.getInstance(Locale.US);
usCollator.setStrength(Collator.PRIMARY); // ignores casing
Collections.sort(strings, usCollator);
Sorting in concurrent environments
Do note though that using the sort method is not friendly in concurrent environments, since the collection instance will be manipulated, and you should consider using immutable collections instead. This is something Guava provides in the Ordering class and is a simple one-liner:
List<string> sorted = Ordering.natural().sortedCopy(strings);
3. Wrap your list with java.util.PriorityQueue
Though there is no sorted list in Java there is however a sorted queue which would probably work just as well for you. It is the java.util.PriorityQueue class.
Nico Haase linked in the comments to a related question that also answers this.
In a sorted collection you most likely don't want to manipulate the internal data structure which is why PriorityQueue doesn't implement the List interface (because that would give you direct access to its elements).
Caveat on the PriorityQueue iterator
The PriorityQueue class implements the Iterable<E> and Collection<E> interfaces so it can be iterated as usual. However, the iterator is not guaranteed to return elements in the sorted order. Instead (as Alderath points out in the comments) you need to poll() the queue until empty.
Note that you can convert a list to a priority queue via the constructor that takes any collection:
List<String> strings = new ArrayList<String>()
strings.add("lol");
strings.add("cat");
PriorityQueue<String> sortedStrings = new PriorityQueue(strings);
while(!sortedStrings.isEmpty()) {
System.out.println(sortedStrings.poll());
}
// Prints out "cat" and "lol"
4. Write your own SortedList class
NOTE: You shouldn't have to do this.
You can write your own List class that sorts each time you add a new element. This can get rather computation heavy depending on your implementation and is pointless, unless you want to do it as an exercise, because of two main reasons:
It breaks the contract that List<E> interface has because the add methods should ensure that the element will reside in the index that the user specifies.
Why reinvent the wheel? You should be using the TreeSet or Multisets instead as pointed out in the first point above.
However, if you want to do it as an exercise here is a code sample to get you started, it uses the AbstractList abstract class:
public class SortedList<E> extends AbstractList<E> {
private ArrayList<E> internalList = new ArrayList<E>();
// Note that add(E e) in AbstractList is calling this one
#Override
public void add(int position, E e) {
internalList.add(e);
Collections.sort(internalList, null);
}
#Override
public E get(int i) {
return internalList.get(i);
}
#Override
public int size() {
return internalList.size();
}
}
Note that if you haven't overridden the methods you need, then the default implementations from AbstractList will throw UnsupportedOperationExceptions.

Because the concept of a List is incompatible with the concept of an automatically sorted collection. The point of a List is that after calling list.add(7, elem), a call to list.get(7) will return elem. With an auto-sorted list, the element could end up in an arbitrary position.

Since all lists are already "sorted" by the order the items were added (FIFO ordering), you can "resort" them with another order, including the natural ordering of elements, using java.util.Collections.sort().
EDIT:
Lists as data structures are based in what is interesting is the ordering in which the items where inserted.
Sets do not have that information.
If you want to order by adding time, use List. If you want to order by other criteria, use SortedSet.

Set and Map are non-linear data structure. List is linear data structure.
The tree data structure SortedSet and SortedMap interfaces implements TreeSet and TreeMap respectively using used Red-Black tree implementation algorithm. So it ensure that there are no duplicated items (or keys in case of Map).
List already maintains an ordered collection and index-based data structure, trees are no index-based data structures.
Tree by definition cannot contain duplicates.
In List we can have duplicates, so there is no TreeList(i.e. no SortedList).
List maintains elements in insertion order. So if we want to sort the list we have to use java.util.Collections.sort(). It sorts the specified list into ascending order, according to the natural ordering of its elements.

JavaFX SortedList
Though it took a while, Java 8 does have a sorted List.
http://docs.oracle.com/javase/8/javafx/api/javafx/collections/transformation/SortedList.html
As you can see in the javadocs, it is part of the JavaFX collections, intended to provide a sorted view on an ObservableList.
Update: Note that with Java 11, the JavaFX toolkit has moved outside the JDK and is now a separate library. JavaFX 11 is available as a downloadable SDK or from MavenCentral. See https://openjfx.io

For any newcomers, as of April 2015, Android now has a SortedList class in the support library, designed specifically to work with RecyclerView. Here's the blog post about it.

Another point is the time complexity of insert operations.
For a list insert, one expects a complexity of O(1).
But this could not be guaranteed with a sorted list.
And the most important point is that lists assume nothing about their elements.
For example, you can make lists of things that do not implement equals or compare.

Think of it like this: the List interface has methods like add(int index, E element), set(int index, E element). The contract is that once you added an element at position X you will find it there unless you add or remove elements before it.
If any list implementation would store elements in some order other than based on the index, the above list methods would make no sense.

In case you are looking for a way to sort elements, but also be able to access them by index in an efficient way, you can do the following:
Use a random access list for storage (e.g. ArrayList)
Make sure it is always sorted
Then to add or remove an element you can use Collections.binarySearch to get the insertion / removal index. Since your list implements random access, you can efficiently modify the list with the determined index.
Example:
/**
* #deprecated
* Only for demonstration purposes. Implementation is incomplete and does not
* handle invalid arguments.
*/
#Deprecated
public class SortingList<E extends Comparable<E>> {
private ArrayList<E> delegate;
public SortingList() {
delegate = new ArrayList<>();
}
public void add(E e) {
int insertionIndex = Collections.binarySearch(delegate, e);
// < 0 if element is not in the list, see Collections.binarySearch
if (insertionIndex < 0) {
insertionIndex = -(insertionIndex + 1);
}
else {
// Insertion index is index of existing element, to add new element
// behind it increase index
insertionIndex++;
}
delegate.add(insertionIndex, e);
}
public void remove(E e) {
int index = Collections.binarySearch(delegate, e);
delegate.remove(index);
}
public E get(int index) {
return delegate.get(index);
}
}
(See a more complete implementation in this answer)

First line in the List API says it is an ordered collection (also known as a sequence). If you sort the list you can't maintain the order, so there is no TreeList in Java.
As API says Java List got inspired from Sequence and see the sequence properties http://en.wikipedia.org/wiki/Sequence_(mathematics)
It doesn't mean that you can't sort the list, but Java strict to his definition and doesn't provide sorted versions of lists by default.

I think all the above do not answer this question due to following reasons,
Since same functionality can be achieved by using other collections such as TreeSet, Collections, PriorityQueue..etc (but this is an alternative which will also impose their constraints i.e. Set will remove duplicate elements. Simply saying even if it does not impose any constraint, it does not answer the question why SortedList was not created by java community)
Since List elements do not implements compare/equals methods (This holds true for Set & Map also where in general items do not implement Comparable interface but when we need these items to be in sorted order & want to use TreeSet/TreeMap,items should implement Comparable interface)
Since List uses indexing & due to sorting it won't work (This can be easily handled introducing intermediate interface/abstract class)
but none has told the exact reason behind it & as I believe these kind of questions can be best answered by java community itself as it will have only one & specific answer but let me try my best to answer this as following,
As we know sorting is an expensive operation and there is a basic difference between List & Set/Map that List can have duplicates but Set/Map can not.
This is the core reason why we have got a default implementation for Set/Map in form of TreeSet/TreeMap. Internally this is a Red Black Tree with every operation (insert/delete/search) having the complexity of O(log N) where due to duplicates List could not fit in this data storage structure.
Now the question arises we could also choose a default sorting method for List also like MergeSort which is used by Collections.sort(list) method with the complexity of O(N log N). Community did not do this deliberately since we do have multiple choices for sorting algorithms for non distinct elements like QuickSort, ShellSort, RadixSort...etc. In future there can be more. Also sometimes same sorting algorithm performs differently depending on the data to be sorted. Therefore they wanted to keep this option open and left this on us to choose. This was not the case with Set/Map since O(log N) is the best sorting complexity.

https://github.com/geniot/indexed-tree-map
Consider using indexed-tree-map . It's an enhanced JDK's TreeSet that provides access to element by index and finding the index of an element without iteration or hidden underlying lists that back up the tree. The algorithm is based on updating weights of changed nodes every time there is a change.

We have Collections.sort(arr) method which can help to sort ArrayList arr. to get sorted in desc manner we can use Collections.sort(arr, Collections.reverseOrder())

SortedList that maintains order like SortedSet but also permits duplicate elements [duplicate]

In Java there are the SortedSet and SortedMap interfaces. Both belong to the Java Collections framework and provide a sorted way to access the elements.
However, in my understanding there is no SortedList in Java. You can use java.util.Collections.sort() to sort a list.
Any idea why it is designed like that?

List iterators guarantee first and foremost that you get the list's elements in the internal order of the list (aka. insertion order). More specifically it is in the order you've inserted the elements or on how you've manipulated the list. Sorting can be seen as a manipulation of the data structure, and there are several ways to sort the list.
I'll order the ways in the order of usefulness as I personally see it:
1. Consider using Set or Bag collections instead
NOTE: I put this option at the top because this is what you normally want to do anyway.
A sorted set automatically sorts the collection at insertion, meaning that it does the sorting while you add elements into the collection. It also means you don't need to manually sort it.
Furthermore if you are sure that you don't need to worry about (or have) duplicate elements then you can use the TreeSet<T> instead. It implements SortedSet and NavigableSet interfaces and works as you'd probably expect from a list:
TreeSet<String> set = new TreeSet<String>();
set.add("lol");
set.add("cat");
// automatically sorts natural order when adding
for (String s : set) {
System.out.println(s);
}
// Prints out "cat" and "lol"
If you don't want the natural ordering you can use the constructor parameter that takes a Comparator<T>.
Alternatively, you can use Multisets (also known as Bags), that is a Set that allows duplicate elements, instead and there are third-party implementations of them. Most notably from the Guava libraries there is a TreeMultiset, that works a lot like the TreeSet.
2. Sort your list with Collections.sort()
As mentioned above, sorting of Lists is a manipulation of the data structure. So for situations where you need "one source of truth" that will be sorted in a variety of ways then sorting it manually is the way to go.
You can sort your list with the java.util.Collections.sort() method. Here is a code sample on how:
List<String> strings = new ArrayList<String>()
strings.add("lol");
strings.add("cat");
Collections.sort(strings);
for (String s : strings) {
System.out.println(s);
}
// Prints out "cat" and "lol"
Using comparators
One clear benefit is that you may use Comparator in the sort method. Java also provides some implementations for the Comparator such as the Collator which is useful for locale sensitive sorting strings. Here is one example:
Collator usCollator = Collator.getInstance(Locale.US);
usCollator.setStrength(Collator.PRIMARY); // ignores casing
Collections.sort(strings, usCollator);
Sorting in concurrent environments
Do note though that using the sort method is not friendly in concurrent environments, since the collection instance will be manipulated, and you should consider using immutable collections instead. This is something Guava provides in the Ordering class and is a simple one-liner:
List<string> sorted = Ordering.natural().sortedCopy(strings);
3. Wrap your list with java.util.PriorityQueue
Though there is no sorted list in Java there is however a sorted queue which would probably work just as well for you. It is the java.util.PriorityQueue class.
Nico Haase linked in the comments to a related question that also answers this.
In a sorted collection you most likely don't want to manipulate the internal data structure which is why PriorityQueue doesn't implement the List interface (because that would give you direct access to its elements).
Caveat on the PriorityQueue iterator
The PriorityQueue class implements the Iterable<E> and Collection<E> interfaces so it can be iterated as usual. However, the iterator is not guaranteed to return elements in the sorted order. Instead (as Alderath points out in the comments) you need to poll() the queue until empty.
Note that you can convert a list to a priority queue via the constructor that takes any collection:
List<String> strings = new ArrayList<String>()
strings.add("lol");
strings.add("cat");
PriorityQueue<String> sortedStrings = new PriorityQueue(strings);
while(!sortedStrings.isEmpty()) {
System.out.println(sortedStrings.poll());
}
// Prints out "cat" and "lol"
4. Write your own SortedList class
NOTE: You shouldn't have to do this.
You can write your own List class that sorts each time you add a new element. This can get rather computation heavy depending on your implementation and is pointless, unless you want to do it as an exercise, because of two main reasons:
It breaks the contract that List<E> interface has because the add methods should ensure that the element will reside in the index that the user specifies.
Why reinvent the wheel? You should be using the TreeSet or Multisets instead as pointed out in the first point above.
However, if you want to do it as an exercise here is a code sample to get you started, it uses the AbstractList abstract class:
public class SortedList<E> extends AbstractList<E> {
private ArrayList<E> internalList = new ArrayList<E>();
// Note that add(E e) in AbstractList is calling this one
#Override
public void add(int position, E e) {
internalList.add(e);
Collections.sort(internalList, null);
}
#Override
public E get(int i) {
return internalList.get(i);
}
#Override
public int size() {
return internalList.size();
}
}
Note that if you haven't overridden the methods you need, then the default implementations from AbstractList will throw UnsupportedOperationExceptions.

Because the concept of a List is incompatible with the concept of an automatically sorted collection. The point of a List is that after calling list.add(7, elem), a call to list.get(7) will return elem. With an auto-sorted list, the element could end up in an arbitrary position.

Since all lists are already "sorted" by the order the items were added (FIFO ordering), you can "resort" them with another order, including the natural ordering of elements, using java.util.Collections.sort().
EDIT:
Lists as data structures are based in what is interesting is the ordering in which the items where inserted.
Sets do not have that information.
If you want to order by adding time, use List. If you want to order by other criteria, use SortedSet.

Set and Map are non-linear data structure. List is linear data structure.
The tree data structure SortedSet and SortedMap interfaces implements TreeSet and TreeMap respectively using used Red-Black tree implementation algorithm. So it ensure that there are no duplicated items (or keys in case of Map).
List already maintains an ordered collection and index-based data structure, trees are no index-based data structures.
Tree by definition cannot contain duplicates.
In List we can have duplicates, so there is no TreeList(i.e. no SortedList).
List maintains elements in insertion order. So if we want to sort the list we have to use java.util.Collections.sort(). It sorts the specified list into ascending order, according to the natural ordering of its elements.

JavaFX SortedList
Though it took a while, Java 8 does have a sorted List.
http://docs.oracle.com/javase/8/javafx/api/javafx/collections/transformation/SortedList.html
As you can see in the javadocs, it is part of the JavaFX collections, intended to provide a sorted view on an ObservableList.
Update: Note that with Java 11, the JavaFX toolkit has moved outside the JDK and is now a separate library. JavaFX 11 is available as a downloadable SDK or from MavenCentral. See https://openjfx.io

For any newcomers, as of April 2015, Android now has a SortedList class in the support library, designed specifically to work with RecyclerView. Here's the blog post about it.

Another point is the time complexity of insert operations.
For a list insert, one expects a complexity of O(1).
But this could not be guaranteed with a sorted list.
And the most important point is that lists assume nothing about their elements.
For example, you can make lists of things that do not implement equals or compare.

Think of it like this: the List interface has methods like add(int index, E element), set(int index, E element). The contract is that once you added an element at position X you will find it there unless you add or remove elements before it.
If any list implementation would store elements in some order other than based on the index, the above list methods would make no sense.

In case you are looking for a way to sort elements, but also be able to access them by index in an efficient way, you can do the following:
Use a random access list for storage (e.g. ArrayList)
Make sure it is always sorted
Then to add or remove an element you can use Collections.binarySearch to get the insertion / removal index. Since your list implements random access, you can efficiently modify the list with the determined index.
Example:
/**
* #deprecated
* Only for demonstration purposes. Implementation is incomplete and does not
* handle invalid arguments.
*/
#Deprecated
public class SortingList<E extends Comparable<E>> {
private ArrayList<E> delegate;
public SortingList() {
delegate = new ArrayList<>();
}
public void add(E e) {
int insertionIndex = Collections.binarySearch(delegate, e);
// < 0 if element is not in the list, see Collections.binarySearch
if (insertionIndex < 0) {
insertionIndex = -(insertionIndex + 1);
}
else {
// Insertion index is index of existing element, to add new element
// behind it increase index
insertionIndex++;
}
delegate.add(insertionIndex, e);
}
public void remove(E e) {
int index = Collections.binarySearch(delegate, e);
delegate.remove(index);
}
public E get(int index) {
return delegate.get(index);
}
}
(See a more complete implementation in this answer)

First line in the List API says it is an ordered collection (also known as a sequence). If you sort the list you can't maintain the order, so there is no TreeList in Java.
As API says Java List got inspired from Sequence and see the sequence properties http://en.wikipedia.org/wiki/Sequence_(mathematics)
It doesn't mean that you can't sort the list, but Java strict to his definition and doesn't provide sorted versions of lists by default.

I think all the above do not answer this question due to following reasons,
Since same functionality can be achieved by using other collections such as TreeSet, Collections, PriorityQueue..etc (but this is an alternative which will also impose their constraints i.e. Set will remove duplicate elements. Simply saying even if it does not impose any constraint, it does not answer the question why SortedList was not created by java community)
Since List elements do not implements compare/equals methods (This holds true for Set & Map also where in general items do not implement Comparable interface but when we need these items to be in sorted order & want to use TreeSet/TreeMap,items should implement Comparable interface)
Since List uses indexing & due to sorting it won't work (This can be easily handled introducing intermediate interface/abstract class)
but none has told the exact reason behind it & as I believe these kind of questions can be best answered by java community itself as it will have only one & specific answer but let me try my best to answer this as following,
As we know sorting is an expensive operation and there is a basic difference between List & Set/Map that List can have duplicates but Set/Map can not.
This is the core reason why we have got a default implementation for Set/Map in form of TreeSet/TreeMap. Internally this is a Red Black Tree with every operation (insert/delete/search) having the complexity of O(log N) where due to duplicates List could not fit in this data storage structure.
Now the question arises we could also choose a default sorting method for List also like MergeSort which is used by Collections.sort(list) method with the complexity of O(N log N). Community did not do this deliberately since we do have multiple choices for sorting algorithms for non distinct elements like QuickSort, ShellSort, RadixSort...etc. In future there can be more. Also sometimes same sorting algorithm performs differently depending on the data to be sorted. Therefore they wanted to keep this option open and left this on us to choose. This was not the case with Set/Map since O(log N) is the best sorting complexity.

https://github.com/geniot/indexed-tree-map
Consider using indexed-tree-map . It's an enhanced JDK's TreeSet that provides access to element by index and finding the index of an element without iteration or hidden underlying lists that back up the tree. The algorithm is based on updating weights of changed nodes every time there is a change.

We have Collections.sort(arr) method which can help to sort ArrayList arr. to get sorted in desc manner we can use Collections.sort(arr, Collections.reverseOrder())

If I iterate over a collection that is actually a list in java, will the collection have order?

List<String> stringList;
//fill with strings somehow
Collection<String> stringCollection = (Collection<String>) stringList;
for(String str : stringCollection){
//will this loop be guaranteed to iterate in the order found in stringList
}
I think it is guaranteed that this for-each loop will iterate in the correct order, since the syntactic sugar actually uses an iterator and the iterator() method is overridden in List to have an order. Since the run time type of stringCollection is a List, then it will use the overridden method which starts at the beginning of the list. Is this correct?

Yes, the enhanced for loop will use the iterator of the provided collection. So if b is really a list (runtime type), then the order will be guaranteed.
Note that with the new stream API (Java SE 8) this is a little bit different.
While b.stream() would still guarantee the order, b.parallelStream() wouldn't.
Also see: https://docs.oracle.com/javase/tutorial/collections/streams/parallelism.html#ordering

http://docs.oracle.com/javase/7/docs/api/java/util/Collection.html#iterator()
Returns an iterator over the elements in this collection. There are no
guarantees concerning the order in which the elements are returned
(unless this collection is an instance of some class that provides a
guarantee).

Yes.
Collection.iterator is implemented by the JDK implementations of Collection, like ArrayList. This is inherent to how object oriented programming works; if you call a method of an object where you only know one of it's interfaces, it will still call the method of the fully implemented class.

Neither the Collection nor List interface provide an implementation for the iterate() method, so this implementation must come from the run-time type of the object you're iterating over. So yes, the collection will iterate in a predictable order if you use an ordered list.

Need of Iterator class in Java?

The question might be pretty vague I know. But the reason I ask this is because the class must have been made with some thought in mind.
This question came into my mind while browsing through a few questions here on SO.
Consider the following code:
class A
{
private int myVar;
A(int varAsArg)
{
myVar = varAsArg;
}
public static void main(String args[])
{
List<A> myList = new LinkedList<A>();
myList.add(new A(1));
myList.add(new A(2));
myList.add(new A(3));
//I can iterate manually like this:
for(A obj : myList)
System.out.println(obj.myVar);
//Or I can use an Iterator as well:
for(Iterator<A> i = myList.iterator(); i.hasNext();)
{
A obj = i.next();
System.out.println(obj.myVar);
}
}
}
So as you can see from the above code, I have a substitute for iterating using a for loop, whereas, I could do the same using the Iterator class' hasNext() and next() method. Similarly there can be an example for the remove() method. And the experienced users had commented on the other answers to use the Iterator class instead of using the for loop to iterate through the List. Why?
What confuses me even more is that the Iterator class has only three methods. And the functionality of those can be achieved with writing a little different code as well.
Some people might argue that the functionality of many classes can be achieved by writing one's own code instead of using the class made for the purpose. Yes,true. But as I said, Iterator class has only three methods. So why go through the hassle of creating an extra class when the same job can be done with a simple block of code which is not way too complicated to understand either.
EDIT:
While I'm at it, since many of the answers say that I can't achieve the remove functionality without using Iterator,I would just like to know if the following is wrong, or will it have some undesirable result.
for(A obj : myList)
{
if(obj.myVar == 1)
myList.remove(obj);
}
Doesn't the above code snippet do the same thing as remove() ?

Iterator came long before the for statement that you show in the evolution of Java. So that's why it's there. Also if you want to remove something, using Iterator.remove() is the only way you can do it (you can't use the for statement for that).

First of all, the for-each construct actually uses the Iterator interface under the covers. It does not, however, expose the underlying Iterator instance to user code, so you can't call methods on it.
This means that there are some things that require explicit use of the Iterator interface, and cannot be achieved by using a for-each loop.
Removing the current element is one such use case.
For other ideas, see the ListIterator interface. It is a bidirectional iterator that supports inserting elements and changing the element under the cursor. None of this can be done with a for-each loop.
for(A obj : myList)
{
if(obj.myVar == 1)
myList.remove(obj);
}
Doesn't the above code snippet do the same thing as remove() ?
No, it does not. All standard containers that I know of will throw ConcurrentModificationException when you try to do this. Even if it were allowed to work, it is ambiguous (what if obj appears in the list twice?) and inefficient (for linked lists, it would require linear instead of constant time).

The foreach construct (for (X x: list)) actually uses Iterator as its implementation internally. You can feed it any Iterable as a source of elements.
And, as others already remarked: Iterator is longer in Java than foreach, and it provides remove().
Also: how else would you implement your own provider class (myList in your example)? You make it Iterable and implement a method that creates an Iterator.

For one thing, Iterator was created way before the foreach loop (shown in your code sample above) was introduced into Java. (The former came in Java2, the latter only in Java5).
Since Java5, indeed the foreach loop is the preferred idiom for the most common scenario (when you are iterating through a single Iterable at a time, in the default order, and do not need to remove or index elements). Note though that the foreach uses an iterator in the background for standard collection classes; in other words it is just syntactic sugar.

Iterator, listIterator both are used to allow different permission to user, like list iterator have 9 methods but iterator have only 3 methods, but have remove functionality which you can't achieve with for loop. Enumeration is another thing which is also used to give only read permissions.

Iterator is an implementation of the classical GoF design pattern. In that way you can achieve clear behaviour separation from the 'technical code' which iterates (the Iterator) and your business code.
Imagine you have to change the 'next' behaviour (say, by getting not the next element but the next EVEN element). If you rely only on for loops you will have to change manually every single for loop, in a way like this
for (int i; i < list.size(); i = i+2)
while if you use an Iterator you can simply override/rewrite the "next()" and "hasNext()" methods and the change will be visible everywhere in your application.

I think answer to your question is abstraction. Iterator is written because to abstract iterating over different set of collections.
Every collection has different methods to iterate over their elements. ArrayList has indexed access. Queues has poll and peek methods. Stack has pop and peek.
Usually you only need to iterate over elements so Iterator comes into play. You do not care about which type of Collection you need to iterate. You only call iterator() method and user Iterator object itself to do this.
If you ask why not put same methods on Collection interface and get rid of extra object creation. You need to know your current position in collection so you can not implement next method in Collection because you can not use it on different locations because every time you call next() method it will increment index (simplifying every collection has different implementation) so you will skip some objects if you use same collection at different places. Also if collection support concurrency than you can not write a multi-thread safe next() method in collection.
It is usually not safe to remove an object from collection iterating by other means than iterator. Iterator.remove() method is safest way to do it. For ArrayList example:
for(int i=0;i

Java data structures (simple question)

Say I want to work with a linked list in java. I thought that the best way to create one is by:
List list = new LinkedList();
But I noticed that this way I can only use methods on the list that are generic. I assume that the implementation is different among the different data structures.
So if I want to use the specific methods for linked list, I have to create the list by:
LinkedList list = new LinkedList();
What's the main reason for that?
Tnanks.

List is an interface that abstracts the underlying list implementation. It is also implemented by e.g. ArrayList.
However, if you specifically want a LinkedList, there is nothing wrong with writing LinkedList list. In fact, if you just pass it around as a list, people may (not knowing the implementation) unknowingly write algorithms like:
for(int i = 0; i < list.size(); i++)
{
// list.get(i) or list.set(i, obj)
}
which are linear on a random access list (e.g. ArrayList) but quadratic on a LinkedList (it would be preferable to use a iterator or list iterator). Java provides the RandomAccess marker interface so you can distinguish.
Of course, you can call these methods on a reference of type LinkedList too, but people should be more likely to consider the cost.
As a note, in .NET LinkedList does not implement IList for this reason.

The first idiom allows you to change the runtime type that list points to without modifying any client code that uses it.
What methods in LinkedList do you think you need that aren't in List? You can always cast for those.
But the whole idea behind interfaces is to shield clients from how the interface is implemented.
If you really need a LinkedList, so be it. But I prefer the first idiom, because most of the time I really just need List methods.

Every LinkedList is a List, too. That also means that you can do everything with a LinkedList that you can do with a List and that you can store a LinkedList as List. However, when you store it as List, you can only call methods of the LinkedList that a List also has.
By the way: This is not Generics. Generics are like this:
LinkedList<String> list = new LinkedList<String>();

List list = getSomeList();
Here you're saying that it's a list. You have no idea whether or not it's a LinkedList or an ArrayList or whatever. It is an abstract thing (I assume you mean "abstract" by the word "generic", since generics are a different thing entirely). Thus you can't treat it like it's an LinkedList -- you have to treat it like it's a List (which it is).
The fact that "you know" that it's a LinkedList is all well and good, and you can safely cast if you need to do it. But it might help to tell the compiler that it's a LinkedList, by declaring it as so, if it's always going to act as a LinkedList.