List<List<Integer>> ArrayList initialization does not work with new operator Java

Question

I have been struggling on this problem for two days. This problem comes from some problem I am doing. Basically, when I use

List<List<Integer>> temp = new ArrayList<>(result);

to create a new ArrayList copy of result, the result will change when I try to change temp within an advanced for loop. For example,

List<List<Integer>> result = new ArrayList<>();
result.add(new ArrayList<>());
List<List<Integer>> temp = new ArrayList<>(result);
int j = 0;
for (List<Integer> list: temp) {
    list.add(x[j]);
    j ++;
}

I did nothing to the result inside the loop but the result ends up with [[1]], which is the same as temp.

Why is this happening? Thanks a lot.

Update: thank everybody for answering my question. I learn the shallow copy was the reason. However, I still run into a similar problem. The result is updated while I try to change the temp in the following code:

List<List<Integer>> result = new ArrayList<>();
result.add(new ArrayList<>());
List<List<Integer>> temp = new ArrayList<>();
for (List<Integer> list: result) {
    list.add(10000);
    temp.add(new ArrayList(list));
}

I do not know why the result turns out to be [[10000]] as well as the temp. Is there anything wrong with the add method like temp.add(new ArrayList(list))?

new ArrayList<>(result) only does a shallow copy of result, i.e. it copies the "outer" list, but not its elements. — Andy Turner
– Andy Turner, Commented Jul 19, 2016 at 20:20

Tamas Rev · Accepted Answer · 2016-07-19 20:22:51Z

4

This is happening because the List<List<Integer>> temp = new ArrayList<>(result); statement copies only the top-level list. It will be the new list that contains references to the original items (aka sublists) from the original result.

You can fix this with a deep copy:

List<List<Integer>> temp = new ArrayList<>(); // empty list
for (List<Integer> sublist : result) {
    temp.add(new ArrayList<Integer>(result)); // copying a sublist and adding that
}

answered Jul 19, 2016 at 20:22

Tamas Rev

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1 Comment

Appalachian Math Over a year ago

Hi Sir, if you got time, could you please have a further look at my new similar problem? I updated the problem in the end of my post.

Andy Turner · Accepted Answer · 2016-07-19 20:22:25Z

1

temp = new ArrayList<>(result) only does a shallow copy of result, i.e. it copies the "outer" list, but not its elements.

temp.get(0) == result.get(0) - and I don't mean equals - they are exactly the same instance.

As such, anything you add to temp.get(0) will also appear in result.get(0).

answered Jul 19, 2016 at 20:22

Andy Turner

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1 Comment

Appalachian Math Over a year ago

Hi Sir, if you got time, could you please have a further look at my new similar problem? I updated the problem in the end of my post.

Draken · Accepted Answer · 2016-07-20 08:35:05Z

1

This is not surprising because you have only one list to iterate upon. If you add a second list you will see a second digit coming in.

int[] x = new int[]{1,2,3,4,5}; 

    List<List<Integer>> result = new ArrayList<>();
    result.add(new ArrayList());
result.add(new ArrayList());
    List<List<Integer>> temp = new ArrayList<>(result);
    for (Integer xx: x) {
        result.add(new ArrayList(xx));
    }
    System.out.println(result.toString());

If you try this code it will display you:

[[1],[2]]

edited Jul 20, 2016 at 8:35

Draken

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answered Jul 19, 2016 at 20:20

Alexander Petrov

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