Checking if a LinkedList is a palindrome
As Ratchet Freak points out, your code uses an \$n\$ size stack to manage the reversal of the list, so it's \$O(n)\$ space complexity. The question is whether there's a better way too, though.

First up, your comparison here is very dubious:


if (s.pop() != iterator2.next()) {



That's doing a native object comparison, and you are comparing Integer, and not int. This is bound to fail, and I am surprised it works for you (I presume Java is reusing Objects for your int value boxing). You should be calling this instead:

if (!s.pop().equals(iterator2.next())) {


Then, your isPalindrome variable is useless. It's never set to false. Instead you return immediately if it's not a palindrome (which is a good thing).

Further, consider that your input is a LinkedList which is a double-linked implementation (as you pointed out) which means a reversed iteration is "easy".

public boolean isPalindromeList(LinkedList&lt;Integer&gt; list) {
    ListIterator&lt;Integer&gt; forward = list.listIterator(0);
    ListIterator&lt;Integer&gt; reverse = list.listIterator(list.size());
    while (forward.hasNext()) {
        if (!forward.next().equals(reverse.previous())) {
            return false;
        }
    }
    return true;
}


You can see this running in ideone: https://ideone.com/LmfQrB