You are given two non-empty linked lists representing two non-negative integers. 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

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { val = x; } * } */class Solution {    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {        Stack
     
       stack1 = new Stack<>();        Stack
      
        stack2 = new Stack<>();        while (l1 != null) {            stack1.push(l1.val);            l1 = l1.next;        }        while (l2 != null) {            stack2.push(l2.val);            l2 = l2.next;        }        //node的下一个结点为head        ListNode head = new ListNode(0); //head记录结果        while (!stack1.isEmpty() || !stack2.isEmpty()) {            if (!stack1.isEmpty()) head.val += stack1.pop();            if (!stack2.isEmpty()) head.val += stack2.pop();            ListNode node = new ListNode(head.val / 10);  //node记录进位            head.val %= 10; //head存储结果            node.next = head; //head往前移动，指向node            head = node;        }        //前导0的情况,head为node的引用，可能为0        return head.val == 0 ? head.next : head;    }}