445 Add Two Numbers II

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

我们可以利用栈来保存所有的元素，然后利用栈的后进先出的特点就可以从后往前取数字了，我们首先遍历两个链表，将所有数字分别压入两个栈s1和s2中，我们建立一个值为0的res节点，然后开始循环，如果栈不为空，则将栈顶数字加入sum中，然后将res节点值赋为sum%10，然后新建一个进位节点head，赋值为sum/10，如果没有进位，那么就是0，然后我们head后面连上res，将res指向head，这样循环退出后，我们只要看res的值是否为0，为0返回res->next，不为0则返回res即可

/**
 * 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<ListNode> s1 = new Stack<>();
        Stack<ListNode> s2 = new Stack<>();
        
        while (l1 != null) {
            s1.push(l1);
            l1 = l1.next;
        }
        while (l2 != null) {
            s2.push(l2);
            l2 = l2.next;
        }
        
        int sum = 0;
        ListNode res = new ListNode(0);
        while (!s1.isEmpty() || !s2.isEmpty()) {
            if (!s1.isEmpty()) {
                sum += s1.pop().val;
            }
            if (!s2.isEmpty()) {
                sum += s2.pop().val;
            }
            res.val = sum % 10;
            sum /= 10;
            ListNode head = new ListNode(sum);
            head.next = res;
            res = head;
        }
        return res.val == 0 ? res.next : res;
    }
}