Java ArrayList -- What's going on here

Question

New to Java and came across this example online. The function removeZeros is supposed to remove all Integers in the ArrayList that are zero, but for some reason, it doesn't remove them if they're consecutive. I'm really confused by this, as I cannot see why it does not strictly remove all occurrences of 0 in the array. The reason is probably really obvious but I just cannot see it...

public static void main(String args[]) 
{
    List<Integer> a = new ArrayList<Integer>();
    a.add(0);
    a.add(0);
    a.add(4);

    System.out.println(a);  // prints [0 , 0 , 4]
    removeZeros(a);
    System.out.println(a);  // prints [0 , 4] ??   why not just [4]?
}

// function to remove all zeros from an integer list
public static void removeZeros(List<Integer> nums) 
{
  for (int i = 0; i < nums.size(); i++)
    if (nums.get(i) == 0)
        nums.remove(i);
}

Any insight is appreciated.

When you iterate over a List you should do so in reverse order — jrmullen, Oct 21 '17 at 18:39
"came across this example online" I suggest you look for better examples to learn from. — Andy Turner, Oct 21 '17 at 18:40
@AndyTurner it was a question on a computer science exam that I found online. Sorry, I guess "example" was a misleading word :0 — travisjayday, Oct 21 '17 at 18:43
@alfasin The duplicate shows how to solve the surrounding problem but OP wants an **explanation** why his **explicit code is not working**. **Not** how to solve the problem in general... — Zabuzard, Oct 21 '17 at 18:56
@Zabuza you guys are right, should have marked it as dup of: https://stackoverflow.com/questions/6958478/how-to-modify-a-collection-while-iterating-using-for-each-loop-without-concurren my bad. — Nir Alfasi, Oct 21 '17 at 19:14

Zabuzard · Accepted Answer · 2017-10-21T18:53:46.680

Explanation

Its simply because your index-logic is not correct.

Let's take a look at the following example:

[1, 0, 0, 4, 5, 0]

We have zeros at index 1, 2 and 5. Your code starts iterating, index 0 is fine. Now i = 1 and we'll remove it from the list using list.remove(i). The list now looks like:

[1, 0, 4, 5, 0]

Now note that the indices have changed! The remaining zeros are now at indices 1 and 4 instead of 2 and 5. But your loop advances i and the next element you check will be i = 2 thus you miss the zero at position 1.

So the problem is that the removal of an element influences the index logic of your loop.

Solution

An easy fix is to do the procedure reversely:

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

Because it doesn't influence the lower indices if the size decrements.

Or alternatively do it forward but don't advance i if you remove something:

int i = 0;
while (i < nums.size()) {
    if (nums.get(i) == 0) {
        nums.remove(i);
    } else {
        i++;
    }
}

score 1 · Answer 2 · answered Oct 21 '17 at 18:39

1

It's because once you've removed the first zero, the array now looks like:

[0, 4]

We then move to index 1, which is set to 4, so it's not removed.

As mentioned by user @Andy Turner, you should iterate the array in reverse order

answered Oct 21 '17 at 18:39

user184994

17,791
1
46
52

1

Or use `list.removeIf(x -> x == 0);`. – Andy Turner Oct 21 '17 at 18:41

BaSsGaz · Answer 3 · 2017-10-22T20:08:57.477

0

First iteration, index is 0 and nums has :

-> 0
0
4

first 0 was removed!

Second iteration, index is 1 and nums has :

0
-> 4

and since 4!=0, nothing will be removed, so you're left with [0, 4]

edited Oct 22 '17 at 20:08

answered Oct 21 '17 at 18:40

BaSsGaz

666
1
18
31

Java ArrayList -- What's going on here

3 Answers3

Explanation

Solution