Add element during arrayList iteraction (add to final of the list)

Question

I want to add some elements to an arrayList during iteraction, but I want to add in the final of the list, I have tried with ListIterator but only adds after the actual element on interaction...

Example:

ArrayList<Integer> arr = new ArrayList<>(Arrays.asList(new Integer[]{1,2,3,4,5}));      
for (ListIterator<Integer> i = arr.listIterator(); i.hasNext();) {          
    i.add(i.next() + 10);               
}       
System.out.println(arr);

That prints: [1, 11, 2, 12, 3, 13, 4, 14, 5, 15]

What I have to do to get: [1, 2, 3, 4, 5, 11, 12, 13, 14, 15] ?

My problem cannot be solved creating another list and using addAll() after... My explanation of the problem was poor, let me explain better:

ArrayList<SomeClass> arr = new ArrayList<>();               
        int condition = 12; // example contidition number   
        for (ListIterator<SomeClass> i = arr.listIterator(); i.hasNext();) {                        
            if (i.next().conditionNumber == condition) {                
                // add to final of the list to process after all elements.                                                                 
            } else {                
                // process the element.     
                // change the contidition               
                condition = 4; // another example, this number will change many times               
            }                           
        }

I can't create a separated list because the elements add to final may enter in the condition again

Thanks

Alexis C. · Accepted Answer · 2015-05-11 13:04:53Z

4

Since you're using Java 8, you could use the Stream API. Concat two streams of the original list, with one that maps each element by adding 10 to it.

List<Integer> arr = Arrays.asList(1, 2, 3, 4, 5);
List<Integer> result = Stream.concat(arr.stream(), arr.stream().map(i -> i + 10))
                             .collect(Collectors.toList());

As you updated your question, I don't see why you can't create another list that you update continuously. This is what I'd do in your case.

Here's a solution that update the list you are iterating using an index-based for loop, but don't forget to cache the size before starting to iterate, so that you only process the elements that were in the list before the processing.

ArrayList<SomeClass> arr = new ArrayList<>();
int condition = 12; // example contidition number
final int prevSize = arr.size();
for (int i = 0; i < prevSize; i++) {
    SomeClass element = arr.get(i);
    if (element.conditionNumber == condition) {
        //probably update or create a new element here
        arr.add(someNewElement);
    } else {
        // process the element.
        // change the contidition
        condition = 4; // another example, this number will change many times
    }
}

I can't create a separated list because the elements add to final may enter in the condition again

It seems to me like a dangerous behavior, that may add elements indefinitely to the list, but you probably know the real problem you are facing. If you can try to avoid this that would be better, IMO.

edited May 11, 2015 at 13:04

answered May 11, 2015 at 12:46

Alexis C.

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3 Comments

Tagir Valeev Over a year ago

An alternative would be to use arr.addAll(arr.stream().map(i -> i+10).collect(Collectors.toList()). This does not recreate the original array (probably this matters in authors case).

Alexis C. Over a year ago

@TagirValeev Yes if you want to modify the list in place this is probably the way to go.

Johnny Willer Over a year ago

I have changed my question, my explanation was poor. Maybe you understand my problem now...

Saurabh Jhunjhunwala · Accepted Answer · 2015-05-11 12:38:17Z

0

Try:

This is in addition to what dasblinkenlight says: as add will add to its current position. you need to use

ArrayList<Integer> arr = new ArrayList<>(Arrays.asList(new Integer[]{1,2,3,4,5}));

ArrayList<Integer> arr1 =new ArrayList<Integer>();

for (ListIterator<Integer> i = arr.listIterator(); i.hasNext();) {          
    arr1.add(i.next() + 10);               
}       

arr.addAll(arr1);
System.out.println(arr);

answered May 11, 2015 at 12:38

Saurabh Jhunjhunwala

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Comments

Tagir Valeev · Accepted Answer · 2015-05-11 12:56:34Z

0

Why not using plain old for?

ArrayList<SomeClass> arr = new ArrayList<>();               
int condition = 12; // example contidition number
for (int i = 0; i < arr.size(); i++) {
    if (arr.get(i).conditionNumber == condition) {
        // add to final of the list to process after all elements.
        arr.add(someNewElement);
    } else {
        // process the element.
        // change the contidition
        condition = 4; // another example, this number will change many times
    }                           
}

Not so cool, but should work in your case.

answered May 11, 2015 at 12:56

Tagir Valeev

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2 Comments

Alexis C. Over a year ago

You'd need to cache the size before the loop, otherwise you may add elements indefinitely.

Alexis C. Over a year ago

Nevermind I re-read the OP's question, "I can't create a separated list because the elements add to final may enter in the condition again". Seems like a dangerous behavior.

Shirish Bari · Accepted Answer · 2015-05-11 13:46:16Z

You could use the collect method which can join two list with your customization, like the one I have done below

  List<Integer> arr = Arrays.asList(1,2,3,4,5);                                                                

  List<Integer> newList = arr.stream()                                                                         
          .collect(ArrayList<Integer>::new, (list, num) -> {                                                   
              list.add(num);                                                                                   
              list.add(num + 10);                                                                              
          }, ArrayList::addAll)       // this will give us the final list of element num and num+10            
          .stream()                                                                                            
          .sorted()                                                                                            
          .collect(Collectors.toList());                                                                       

  System.out.println(newList );

key here is using the below variation of collect method of stream api

collect(Supplier<R> supplier, BiConsumer<R, ? super T> accumulator,BiConsumer<R, R> combiner);
Please let me know if it can solve the problem.

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