思路 遍历，当一个链表为空时，置0于另一个链表及进位符想加
当最高位产生进位时，根据carry位的状态 ，增加一个节点。
/**
* 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) {
int carry = 0;
ListNode sum1 = new ListNode(0,l1);
ListNode cur = sum1;
if(l1==null){
return l2;
}
if(l2==null){
return l1;
}
while(l1!=null||l2!=null){
int sum = 0;
if(l1==null){
sum = l2.val + carry;
}
if(l2==null){
sum = l1.val +carry;
}
if(l1!=null&&l2!=null){
sum = l1.val + l2.val +carry;
}
carry = sum/10;
sum = sum % 10;
ListNode a = new ListNode(sum);
sum1.next = a;
sum1 = sum1.next;
if(l1!=null){
l1 = l1.next;
}
if(l2!=null){
l2 = l2.next;
}
}
if(carry==1){
ListNode a = new ListNode(1);
sum1.next = a;
}
return cur.next;
}
}