Leetcode: Add Two Numbers II

You are given two linked lists representing two non-negative numbers. The most significant digit comes first and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Follow up:
What if you cannot modify the input lists? In other words, reversing the lists is not allowed.

Example:

Input: (7 -> 2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 8 -> 0 -> 7

The key of this problem is to think of using Stack

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        Stack<ListNode> st1 = new Stack<ListNode>();
        Stack<ListNode> st2 = new Stack<ListNode>();
        while (l1 != null) {
            st1.push(l1);
            l1 = l1.next;
        }
        while (l2 != null) {
            st2.push(l2);
            l2 = l2.next;
        }
        ListNode dummy = new ListNode(-1);
        int carry = 0;
        while (st1.size()!=0 || st2.size()!=0 || carry!=0) {
            int sum = 0;
            if (st1.size() != 0) sum += st1.pop().val;
            if (st2.size() != 0) sum += st2.pop().val;
            if (carry != 0) sum += carry;
            ListNode newNode = new ListNode(sum%10);
            newNode.next = dummy.next;
            dummy.next = newNode;
            carry = sum / 10;
        }
        return dummy.next;
    }
}