在迭代期间向集合添加元素

How about building a Queue with the elements you want to iterate over; when you want to add elements, enqueue them at the end of the queue, and keep removing elements until the queue is empty. This is how a breadth-first search usually works.

In general, it's not safe, though for some collections it may be. The obvious alternative is to use some kind of for loop. But you didn't say what collection you're using, so that may or may not be possible.

There are two issues here:

The first issue is, adding to an Collection after an Iterator is returned. As mentioned, there is no defined behavior when the underlying Collection is modified, as noted in the documentation for Iterator.remove:

... The behavior of an iterator is unspecified if the underlying collection is modified while the iteration is in progress in any way other than by calling this method.

The second issue is, even if an Iterator could be obtained, and then return to the same element the Iterator was at, there is no guarantee about the order of the iteratation, as noted in the Collection.iterator method documentation:

... 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).

For example, let's say we have the list [1, 2, 3, 4].

Let's say 5 was added when the Iterator was at 3, and somehow, we get an Iterator that can resume the iteration from 4. However, there is no guarentee that 5 will come after 4. The iteration order may be [5, 1, 2, 3, 4] -- then the iterator will still miss the element 5.

As there is no guarantee to the behavior, one cannot assume that things will happen in a certain way.

One alternative could be to have a separate Collection to which the newly created elements can be added to, and then iterating over those elements:

Collection<String> list = Arrays.asList(new String[]{"Hello", "World!"});
Collection<String> additionalList = new ArrayList<String>();


for (String s : list) {
// Found a need to add a new element to iterate over,
// so add it to another list that will be iterated later:
additionalList.add(s);
}


for (String s : additionalList) {
// Iterate over the elements that needs to be iterated over:
System.out.println(s);
}

Edit

Elaborating on Avi's answer, it is possible to queue up the elements that we want to iterate over into a queue, and remove the elements while the queue has elements. This will allow the "iteration" over the new elements in addition to the original elements.

Let's look at how it would work.

Conceptually, if we have the following elements in the queue:

[1, 2, 3, 4]

And, when we remove 1, we decide to add 42, the queue will be as the following:

[2, 3, 4, 42]

As the queue is a FIFO (first-in, first-out) data structure, this ordering is typical. (As noted in the documentation for the Queue interface, this is not a necessity of a Queue. Take the case of PriorityQueue which orders the elements by their natural ordering, so that's not FIFO.)

The following is an example using a LinkedList (which is a Queue) in order to go through all the elements along with additional elements added during the dequeing. Similar to the example above, the element 42 is added when the element 2 is removed:

Queue<Integer> queue = new LinkedList<Integer>();
queue.add(1);
queue.add(2);
queue.add(3);
queue.add(4);


while (!queue.isEmpty()) {
Integer i = queue.remove();
if (i == 2)
queue.add(42);


System.out.println(i);
}

The result is the following:

1
2
3
4
42

As hoped, the element 42 which was added when we hit 2 appeared.

You may also want to look at some of the more specialised types, like ListIterator, NavigableSet and (if you're interested in maps) NavigableMap.

Using iterators...no, I don't think so. You'll have to hack together something like this:

    Collection< String > collection = new ArrayList< String >( Arrays.asList( "foo", "bar", "baz" ) );
int i = 0;
while ( i < collection.size() ) {


String curItem = collection.toArray( new String[ collection.size() ] )[ i ];
if ( curItem.equals( "foo" ) ) {
collection.add( "added-item-1" );
}
if ( curItem.equals( "added-item-1" ) ) {
collection.add( "added-item-2" );
}


i++;
}


System.out.println( collection );

Which yeilds:
[foo, bar, baz, added-item-1, added-item-2]

I prefer to process collections functionally rather than mutate them in place. That avoids this kind of problem altogether, as well as aliasing issues and other tricky sources of bugs.

So, I would implement it like:

List<Thing> expand(List<Thing> inputs) {
List<Thing> expanded = new ArrayList<Thing>();


for (Thing thing : inputs) {
expanded.add(thing);
if (needsSomeMoreThings(thing)) {
addMoreThingsTo(expanded);
}
}


return expanded;
}

IMHO the safer way would be to create a new collection, to iterate over your given collection, adding each element in the new collection, and adding extra elements as needed in the new collection as well, finally returning the new collection.

Besides the solution of using an additional list and calling addAll to insert the new items after the iteration (as e.g. the solution by user Nat), you can also use concurrent collections like the CopyOnWriteArrayList.

The "snapshot" style iterator method uses a reference to the state of the array at the point that the iterator was created. This array never changes during the lifetime of the iterator, so interference is impossible and the iterator is guaranteed not to throw ConcurrentModificationException.

With this special collection (usually used for concurrent access) it is possible to manipulate the underlying list while iterating over it. However, the iterator will not reflect the changes.

Is this better than the other solution? Probably not, I don't know the overhead introduced by the Copy-On-Write approach.

Actually it is rather easy. Just think for the optimal way. I beleive the optimal way is:

for (int i=0; i<list.size(); i++) {
Level obj = list.get(i);


//Here execute yr code that may add / or may not add new element(s)
//...


i=list.indexOf(obj);
}

The following example works perfectly in the most logical case - when you dont need to iterate the added new elements before the iteration element. About the added elements after the iteration element - there you might want not to iterate them either. In this case you should simply add/or extend yr object with a flag that will mark them not to iterate them.

public static void main(String[] args)
{
// This array list simulates source of your candidates for processing
ArrayList<String> source = new ArrayList<String>();
// This is the list where you actually keep all unprocessed candidates
LinkedList<String> list = new LinkedList<String>();


// Here we add few elements into our simulated source of candidates
// just to have something to work with
source.add("first element");
source.add("second element");
source.add("third element");
source.add("fourth element");
source.add("The Fifth Element"); // aka Milla Jovovich


// Add first candidate for processing into our main list
list.addLast(source.get(0));


// This is just here so we don't have to have helper index variable
// to go through source elements
source.remove(0);


// We will do this until there are no more candidates for processing
while(!list.isEmpty())
{
// This is how we get next element for processing from our list
// of candidates. Here our candidate is String, in your case it
// will be whatever you work with.
String element = list.pollFirst();
// This is where we process the element, just print it out in this case
System.out.println(element);


// This is simulation of process of adding new candidates for processing
// into our list during this iteration.
if(source.size() > 0) // When simulated source of candidates dries out, we stop
{
// Here you will somehow get your new candidate for processing
// In this case we just get it from our simulation source of candidates.
String newCandidate = source.get(0);
// This is the way to add new elements to your list of candidates for processing
list.addLast(newCandidate);
// In this example we add one candidate per while loop iteration and
// zero candidates when source list dries out. In real life you may happen
// to add more than one candidate here:
// list.addLast(newCandidate2);
// list.addLast(newCandidate3);
// etc.


// This is here so we don't have to use helper index variable for iteration
// through source.
source.remove(0);
}
}
}

Given a list List<Object> which you want to iterate over, the easy-peasy way is:

while (!list.isEmpty()){
Object obj = list.get(0);


// do whatever you need to
// possibly list.add(new Object obj1);


list.remove(0);
}

So, you iterate through a list, always taking the first element and then removing it. This way you can append new elements to the list while iterating.

Forget about iterators, they don't work for adding, only for removing. My answer applies to lists only, so don't punish me for not solving the problem for collections. Stick to the basics:

    List<ZeObj> myList = new ArrayList<ZeObj>();
// populate the list with whatever
........
int noItems = myList.size();
for (int i = 0; i < noItems; i++) {
ZeObj currItem = myList.get(i);
// when you want to add, simply add the new item at last and
// increment the stop condition
if (currItem.asksForMore()) {
myList.add(new ZeObj());
noItems++;
}
}

I tired ListIterator but it didn't help my case, where you have to use the list while adding to it. Here's what works for me:

Use LinkedList.

LinkedList<String> l = new LinkedList<String>();
l.addLast("A");


while(!l.isEmpty()){
String str = l.removeFirst();
if(/* Condition for adding new element*/)
l.addLast("<New Element>");
else
System.out.println(str);
}

This could give an exception or run into infinite loops. However, as you have mentioned

I'm pretty sure it won't in my case

checking corner cases in such code is your responsibility.

This is what I usually do, with collections like sets:

Set<T> adds = new HashSet<T>, dels = new HashSet<T>;
for ( T e: target )
if ( <has to be removed> ) dels.add ( e );
else if ( <has to be added> ) adds.add ( <new element> )


target.removeAll ( dels );
target.addAll ( adds );

This creates some extra-memory (the pointers for intermediate sets, but no duplicated elements happen) and extra-steps (iterating again over changes), however usually that's not a big deal and it might be better than working with an initial collection copy.

For examle we have two lists:

  public static void main(String[] args) {
ArrayList a = new ArrayList(Arrays.asList(new String[]{"a1", "a2", "a3","a4", "a5"}));
ArrayList b = new ArrayList(Arrays.asList(new String[]{"b1", "b2", "b3","b4", "b5"}));
merge(a, b);
a.stream().map( x -> x + " ").forEach(System.out::print);
}
public static void merge(List a, List b){
for (Iterator itb = b.iterator(); itb.hasNext(); ){
for (ListIterator it = a.listIterator() ; it.hasNext() ; ){
it.next();
it.add(itb.next());


}
}


}

a1 b1 a2 b2 a3 b3 a4 b4 a5 b5

Even though we cannot add items to the same list during iteration, we can use Java 8's flatMap, to add new elements to a stream. This can be done on a condition. After this the added item can be processed.

Here is a Java example which shows how to add to the ongoing stream an object depending on a condition which is then processed with a condition:

List<Integer> intList = new ArrayList<>();
intList.add(1);
intList.add(2);
intList.add(3);


intList = intList.stream().flatMap(i -> {
if (i == 2) return Stream.of(i, i * 10); // condition for adding the extra items
return Stream.of(i);
}).map(i -> i + 1)
.collect(Collectors.toList());


System.out.println(intList);

The output of the toy example is:

[2, 3, 21, 4]

Use ListIterator as follows:

List<String> l = new ArrayList<>();
l.add("Foo");
ListIterator<String> iter = l.listIterator(l.size());
while(iter.hasPrevious()){
String prev=iter.previous();
if(true /*You condition here*/){
iter.add("Bah");
iter.add("Etc");
}
}

The key is to iterate in reverse order - then the added elements appear on the next iteration.

I know its been quite old. But thought of its of any use to anyone else. Recently I came across this similar problem where I need a queue that is modifiable during iteration. I used listIterator to implement the same much in the same lines as of what Avi suggested -> Avi's Answer. See if this would suit for your need.

ModifyWhileIterateQueue.java

import java.util.ArrayList;
import java.util.List;
import java.util.ListIterator;


public class ModifyWhileIterateQueue<T> {
ListIterator<T> listIterator;
int frontIndex;
List<T> list;


public ModifyWhileIterateQueue() {
frontIndex = 0;
list =  new ArrayList<T>();
listIterator = list.listIterator();
}


public boolean hasUnservicedItems () {
return frontIndex < list.size();
}


public T deQueue() {
if (frontIndex >= list.size()) {
return null;
}
return list.get(frontIndex++);
}


public void enQueue(T t) {
listIterator.add(t);
}


public List<T> getUnservicedItems() {
return list.subList(frontIndex, list.size());
}


public List<T> getAllItems() {
return list;
}
}

ModifyWhileIterateQueueTest.java

    @Test
public final void testModifyWhileIterate() {
ModifyWhileIterateQueue<String> queue = new ModifyWhileIterateQueue<String>();
queue.enQueue("one");
queue.enQueue("two");
queue.enQueue("three");


for (int i=0; i< queue.getAllItems().size(); i++) {
if (i==1) {
queue.enQueue("four");
}
}


assertEquals(true, queue.hasUnservicedItems());
assertEquals ("[one, two, three, four]", ""+ queue.getUnservicedItems());
assertEquals ("[one, two, three, four]", ""+queue.getAllItems());
assertEquals("one", queue.deQueue());


}