解法一：申请三个指针，分别指向l1、l2、new_l。先设置第一个节点；再l1、l2循环设置后续节点；若某一条链未结束，继续循环设置；若最后enter1（下一位是否进1）仍为真，则多了一位还要新加一位1。

/*** Definition for singly-linked list.* public class ListNode {*     int val;*     ListNode next;*     ListNode() {}*     ListNode(int val) { this.val = val; }*     ListNode(int val, ListNode next) { this.val = val; this.next = next; }* }*/
class Solution {public ListNode addTwoNumbers(ListNode l1, ListNode l2) {ListNode sum_list = new ListNode();ListNode p1 = l1.next, p2 = l2.next, ps = sum_list;int sum = l1.val+l2.val;boolean enter1 = sum>=10 ? true : false;// 先设置第一个节点sum_list.val = sum%10;sum_list.next = null;// 设置后续节点while(p1!=null && p2!=null){sum = enter1 ? p1.val+p2.val+1 : p1.val+p2.val;enter1 = false;if(sum >= 10){enter1 = true;}ListNode temp = new ListNode();temp.val = sum%10;temp.next = ps.next;ps.next = temp;ps = ps.next;p1 = p1.next;p2 = p2.next;}// 某一条链未结束p1 = p1!=null ? p1 : p2;while(p1!=null){sum = enter1 ? p1.val+1 : p1.val;enter1 = false;if(sum >= 10){enter1 = true;}ListNode temp = new ListNode();temp.val = sum%10;temp.next = ps.next;ps.next = temp;ps = ps.next;p1 = p1.next;}// 多了一位，还要进1if(enter1){ListNode temp = new ListNode();temp.val = 1;temp.next = ps.next;ps.next = temp;}return sum_list;}
}