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0else 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;}
}