迭代时从ArrayList中删除奇数元素

Question

I have an ArrayList of Strings and I'm trying to remove the odd elements of the ArrayList, ie list.remove(1), list.remove(3), list.remove(5) etc.

This is the code I'm attempting to use which throws up an IllegalStateException error:

int i = 0;
    for (Iterator<String> it = words.iterator(); it.hasNext(); )
    {
        if (i % 2 != 0 && it.hasNext())
        {
            it.remove();
        }
        i++;
    }

Is there a better (working) way to do this?

Answer 1

int i = 0;
    for (Iterator<String> it = words.iterator(); it.hasNext(); )
    {
        it.next(); // Add this line in your code
        if (i % 2 != 0)
        {
            it.remove();
        }
        i++;
    }

Answer 2

You can try something like this to remove every second element starting from words[1] . No need to check whether the index is odd, when we remove an element we can just increment i and that will be the next odd number.

int i = 1;

while (i < words.size()) {
    words.remove(i++);
}

Answer 3

You need to clone this Array or copy that odd element into another array. During iterate time it was used same object so if you remove its index and state was changes.

int i = 0;
    List<String> list = new ArrayList<String>();
    List<String> words = new ArrayList<String>();
    for (String word:words)
    {
        if (i % 2 != 0)
        {

            //it.remove();
            list.add(word);
        }

        i++;
    }
    words.removeAll(list);

now just remove this all by passing this list to your words list object

words.removeAll(list);

Answer 4

I wouldn't use an iterator, because you don't need to examine the elements. Just use List.remove(index) :

for (int i = words.size() - 1; i >=0; i--) {
    if (i % 2 == 1) {
        words.remove(i);
    {
}

Note that you must count down, not up with this approach because removing an element shuffles e everything after to the left

If your list is immutable (explains the exception) make a copy first:

words = new ArrayList(words);

Answer 5

Just use this.

for (int i = 1; i < list.size(); i++) {
    list.remove(i);
}

Answer 6

The below also works fine

public static void main(String[] args) {
        ArrayList<Integer> list = new ArrayList<Integer>(Arrays.asList(1, 2, 3,
                4, 5, 6, 7, 8, 9));

        for (int i = 0; i < list.size(); i++) {
            System.out.println(i);
            list.remove(i);
        }
        System.out.println(list);
    }

yielding

[2, 4, 6, 8]

Answer 7

You need to use next() and then call remove() ---

    int counter = 0;
    for (final Iterator<String> iterator = list.iterator(); iterator.hasNext();) {
        iterator.next();
        if (++counter % 2 != 0) {
            iterator.remove();
        }
    }

Answer 8

public static List<Integer> removeImpairElements(List<Integer> integers){
int j = 0; 
for(int i = 0 ; i < integers.size(); i++){
  if( i % 2 == 0){
    integers.set(j, integers.get(i));
    j++;
  }
}
int half = integers.size()%2==0 ? integers.size()/2 : integers.size()/2 + 1;  
integers.subList(half , integers.size()).clear();
return integers;

}