Let me give a few examples with some alternatives to avoid a ConcurrentModificationException.

Suppose we have the following collection of books

List<Book> books = new ArrayList<Book>();
books.add(new Book(new ISBN("0-201-63361-2")));
books.add(new Book(new ISBN("0-201-63361-3")));
books.add(new Book(new ISBN("0-201-63361-4")));

Collect and Remove

The first technique consists in collecting all the objects that we want to delete (e.g. using an enhanced for loop) and after we finish iterating, we remove all found objects.

ISBN isbn = new ISBN("0-201-63361-2");
List<Book> found = new ArrayList<Book>();
for(Book book : books){
    if(book.getIsbn().equals(isbn)){
        found.add(book);
    }
}
books.removeAll(found);

This is supposing that the operation you want to do is "delete".

If you want to "add" this approach would also work, but I would assume you would iterate over a different collection to determine what elements you want to add to a second collection and then issue an addAll method at the end.

Using ListIterator

If you are working with lists, another technique consists in using a ListIterator which has support for removal and addition of items during the iteration itself.

ListIterator<Book> iter = books.listIterator();
while(iter.hasNext()){
    if(iter.next().getIsbn().equals(isbn)){
        iter.remove();
    }
}

Again, I used the "remove" method in the example above which is what your question seemed to imply, but you may also use its add method to add new elements during iteration.

Using JDK >= 8

For those working with Java 8 or superior versions, there are a couple of other techniques you could use to take advantage of it.

You could use the new removeIf method in the Collection base class:

ISBN other = new ISBN("0-201-63361-2");
books.removeIf(b -> b.getIsbn().equals(other));

Or use the new stream API:

ISBN other = new ISBN("0-201-63361-2");
List<Book> filtered = books.stream()
                           .filter(b -> b.getIsbn().equals(other))
                           .collect(Collectors.toList());

In this last case, to filter elements out of a collection, you reassign the original reference to the filtered collection (i.e. books = filtered) or used the filtered collection to removeAll the found elements from the original collection (i.e. books.removeAll(filtered)).

Use Sublist or Subset

There are other alternatives as well. If the list is sorted, and you want to remove consecutive elements you can create a sublist and then clear it:

books.subList(0,5).clear();

Since the sublist is backed by the original list this would be an efficient way of removing this subcollection of elements.

Something similar could be achieved with sorted sets using NavigableSet.subSet method, or any of the slicing methods offered there.

Considerations:

What method you use might depend on what you are intending to do

• The collect and removeAl technique works with any Collection (Collection, List, Set, etc).
• The ListIterator technique obviously only works with lists, provided that their given ListIterator implementation offers support for add and remove operations.
• The Iterator approach would work with any type of collection, but it only supports remove operations.
• With the ListIterator/Iterator approach the obvious advantage is not having to copy anything since we remove as we iterate. So, this is very efficient.
• The JDK 8 streams example don't actually removed anything, but looked for the desired elements, and then we replaced the original collection reference with the new one, and let the old one be garbage collected. So, we iterate only once over the collection and that would be efficient.
• In the collect and removeAll approach the disadvantage is that we have to iterate twice. First we iterate in the foor-loop looking for an object that matches our removal criteria, and once we have found it, we ask to remove it from the original collection, which would imply a second iteration work to look for this item in order to remove it.
• I think it is worth mentioning that the remove method of the Iterator interface is marked as "optional" in Javadocs, which means that there could be Iterator implementations that throw UnsupportedOperationException if we invoke the remove method. As such, I'd say this approach is less safe than others if we cannot guarantee the iterator support for removal of elements.

There are several ways to do this. Let's look at the alternatives:

Iterating over a copy, removing from original

This is a simple solution for the underlying problem of your first code: A ConcurrentModificationException is thrown because you iterate through the list and removing from it at the same time.

Easy solution is to create a copy of the list and iterate through that.

for (Integer integer : new ArrayList<>(nums)) {
    if (integer < 3) {
        nums.remove(integer);
    }
}

Down-sides of this approach:

• Creates a copy of the original list, which requires memory and an operation which performance depends on the type of the list (ArrayList, LinkedList, etc.)
• Additionally, nums.remove(value) is a \ operation. Making this loop overall \

Java 8 Streams

List<Integer> filteredList = nums.stream().filter(i -> i >= 3).collect(Collectors.toList());

Down-sides:

• Does not actually modify the existing list, so if references to the list are spread around various variables, you still have some old elements that just shouldn't be in that list.
• Creates various stream-related objects which might not be the most effective option.

On the up-side, this is among the fastest for bigger lists.

If you're not using Java 8:

List<Object> originalList;
List<Object> newList = new YourFavoriteListType<>();
for (Object obj : originalList) {
    if (shouldKeep(obj)) {
        newList.add(obj);
    }
}

Java 8 method

nums.removeIf(i -> i < 3);

Java 8 introduced the default method removeIf on the Collection interface. This allows different implementations to have implementation-specific performance-optimized implementations of this method.

Iterator.remove()

Iterator<Integer> it = nums.iterator();
while (it.hasNext()) {
    Integer integer = it.next();
    if (integer < 3) {
        it.remove();
    }
}

The only down-side of this approach is that you need to switch your for-each to a while. However, this approach is the most efficient one, especially for LinkedList where it is \ (it's \ for ArrayList because it has to copy array data on each remove(index) call). This is the approach I would recommend in most cases.

Note: Instead of using a while-loop it can also be written as:

for (Iterator<Integer> it = list.iterator(); it.hasNext(); ) {
    Integer integer = it.next();
    ...

Conclusion

If you want to mutate the existing list, removeIf is the solution I would go with. If you like functional programming and prefer a new list instead of mutating the existing one, then go with the list.stream().filter(...).collect(Collectors.toList()) approach.

See also

"When to use LinkedList over ArrayList?" on Stack Overflow

You need to think about how list.remove(index) works. When remove(idx) gets called, the element at idx index gets deleted and all the next elements gets shifted to left. So, suppose you have a list containing 2, 3, 3, 4, 5. Now you want to remove all the 3s from this list. But if you use your current approach, it will remove only the first occurrence of 3. Because after removing 1st occurrence of 3 which is at position 1 your contents will get shifted to left and will be like this 2, 3, 4, 5. But now your for loop will increment the current index to 2 which contains 4 and not 3. That is why it is not advised to remove items while iterating, Because index of items gets changed after each removal.

Edit: Also if you are using a constant value in loop break condition like in above example i<10; you might get ArrayIndexOutOfBounds exception.