Removing contiguous duplicates in ArrayList [Java]

Question

I am doing a simple Java program and I need to remove all of the contiguous duplicates in a String ArrayList.

My String ArrayList is something like this:

list = [a,b,c,c,d,a,b,c,d]

My goal is removing all (and only!) the contiguous duplicates so that the result would be: [a,b,c,d,a,b,c,d]. As you can see, one of the two contiguous "c" has been removed.

I tried something like this:

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

        if (list.get(i).compareTo(list.get(i+1))==0) {
            positionToRemove.add(i);
        }

    }

Where positionToRemove will at the end contain all the position of the contiguous elements which I will then remove using list.remove() (still not done)

Unfortunately I get

java.lang.IndexOutOfBoundsException

I am quite positive there is a very simple way to achieve this but I can't remember it at the moment!

user2422457 · Accepted Answer · 2014-03-13 11:41:54Z

1

There is no need to store the indexes of the elements you need to remove. Just remove it directly by:

int size = list.size();
for (int i = size - 1; i >= 1; i--) {
    if (list.get(i).compareTo(list.get(i - 1)) == 0) {
        list.remove(i);
    }
}

answered Mar 13, 2014 at 11:41

user2422457

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Comments

Rahul · Accepted Answer · 2014-03-13 11:36:07Z

0

In the last iteration of the for loop, list.get(i+1) goes beyond the bounds of the list and thus the IndexOutOfBoundsException. In any array/arraylist, the maximum accessible index is always size/length - 1.

To fix that you need to change your logic a bit.

for (int i = 1; i<list.size(); i++) {
    if (list.get(i-1).compareTo(list.get(i))==0) {
        positionToRemove.add(i);
    }
}

answered Mar 13, 2014 at 11:36

Rahul

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1 Comment

NoobNe0 Over a year ago

I have always problem with bounds exceptions, but this proved to be very useful for me in understanding how loops works. Many thanks!

Kevin Bowersox · Accepted Answer · 2014-03-13 11:36:38Z

0

When you access the List using index + 1 or i+1, you overstep the bounds of the List on the last iteration. You can fix this by setting the conditional for your for loop to i < list.size() -1.

for (int i = 0; i < list.size() -1; i++) {  
        if (list.get(i).compareTo(list.get(i+1))==0) {
            positionToRemove.add(i);
        }
 }

answered Mar 13, 2014 at 11:36

Kevin Bowersox

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Comments

ديناصور الأمة · Accepted Answer · 2014-03-13 11:43:09Z

0

It should not reach the length of your list. You must shutdown the traverses at size -1.

for (int i = 0; i<list.size() - 1; i++) {
    if (list.get(i).compareTo(list.get(i+1))==0) {
        positionToRemove.add(i);
    }
}

answered Mar 13, 2014 at 11:43

ديناصور الأمة

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Comments

Alexis C. · Accepted Answer · 2014-03-13 11:49:22Z

0

Where positionToRemove will at the end contain all the position of the contiguous elements which I will then remove using list.remove() (still not done)

Instead of storing each position, you could start from the end of the list and directly remove the current element if it's the same as its left neighbour. Using this, you don't need to create another list containing the indexes of the objects you want to remove.

List<String> list = new ArrayList<>(Arrays.asList("a","b","c","c","d","a","b","c","d"));
for(int i = list.size() - 1; i > 0; i--){
    if(list.get(i).compareTo(list.get(i-1)) == 0){
        list.remove(i);
    }
}
System.out.println(list);

Which outputs:

[a, b, c, d, a, b, c, d]

edited Mar 13, 2014 at 11:49

answered Mar 13, 2014 at 11:41

Alexis C.

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Comments

Radu Linu · Accepted Answer · 2020-03-12 14:41:25Z

0

You can do this with Java 7, using iterator.

Iterator<Integer> iterator = collection.values().iterator();
Integer previousValue = null;

while(iterator.hasNext()) {
    Integer currentValue = iterator.next();
    if(currentValue.equals(previousValue)){
        iterator.remove();
    }
    previousValue = currentValue;
}

answered Mar 12, 2020 at 14:41

Radu Linu

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