题解一：

        迭代：首先判断整数0，然后分别遍历两段链表，将对应位数的值相加并存入新链表，再遍历新链表，将节点值val>=10的减10，并且其下一节点值val+=1。需要注意最后一位节点进位是将下一位节点值设置为1。

class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        if (l1.val == 0 && l1.next == null) return l2;//判断0
        else if (l2.val == 0 && l2.next == null) return l1;

        ListNode result = new ListNode();
        
        ListNode temp = result;//相加
        while (l1 != null && l2 != null) {
            temp.next = new ListNode();
            temp.next.val = (l1.val + l2.val);
            l1 = l1.next;
            l2 = l2.next;
            temp = temp.next;
        }
        while (l1 != null) {
            temp.next = new ListNode();
            temp.next.val = l1.val;
            l1 = l1.next;
            temp = temp.next;
        }
        while (l2 != null) {
            temp.next = new ListNode();
            temp.next.val = l2.val;
            l2 = l2.next;
            temp = temp.next;
        }


        temp = result.next;//进位
        while (temp != null) {
            if (temp.val >= 10) {
                temp.val -= 10;
                if (temp.next != null) {
                    temp.next.val++;
                } else {
                    temp.next = new ListNode();
                    temp.next.val = 1;
                    temp.next.next = null;
                    break;
                }
            }
            temp = temp.next;
        }

        return result.next;
    }
}