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
题目含义：做列表的加法（由低位开始）

    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        if (l1 == null) return l2;
        if (l2 == null) return l1;
        Stack<Integer> stack1 = new Stack<>();
        Stack<Integer> stack2 = new Stack<>();
        while (l1 != null) {
            stack1.push(l1.val);
            l1 = l1.next;
        }
        while (l2 != null) {
            stack2.push(l2.val);
            l2 = l2.next;
        }

        int sum = 0;
        ListNode list = new ListNode(0);
        while (!stack1.isEmpty() || !stack2.isEmpty()) {
            if (!stack1.isEmpty()) sum += stack1.pop();
            if (!stack2.isEmpty()) sum += stack2.pop();
            list.val = sum % 10;
            ListNode head = new ListNode(sum / 10);
            sum /= 10;
            head.next = list;
            list = head;
        }
        return list.val == 0 ? list.next : list;        
    }