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

此题和Add Two Numbers不同之处在于，此题的链表头是位高的位置，和之前那道题刚好相反。我思考了很久并没有想到用什么方法来解决，后来看提示说用栈来做，下面是我自己想出来的解法：

/**

 * 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<Integer> s1 = new Stack<Integer>();

        Stack<Integer> s2 = new Stack<Integer>();

        Stack<Integer> s3 = new Stack<Integer>();

        while(l1!=null){

            s1.push(l1.val);

            l1 = l1.next;

        }

        while(l2!=null){

            s2.push(l2.val);

            l2 = l2.next;

        }

        int carry = 0;

        while(!s1.isEmpty()||!s2.isEmpty()||carry!=0){

            if(!s1.isEmpty()){

                carry+=s1.pop();

            }

            if(!s2.isEmpty()){

                carry+=s2.pop();

            }

            s3.push(carry%10);

            carry/=10;

        }

        ListNode dummy = new ListNode(0);

        ListNode node = dummy;

        while(!s3.isEmpty()){

            node.val = s3.pop();

            if(!s3.isEmpty()){

                ListNode next = new ListNode(0);

                node.next = next;

                node = node.next;

            }

        }

        return dummy;

    }

}

但是后来看了答案的解法发现根本不需要用第三个栈，解法如下：