leetcode hot 23

解题思路：我的思路对两个不等长的链表进行补0，短链表前面补0直到与长链表相等,然后对两个链表同步遍历，在list1上保存结果，用sum存储中间加法值，add保存进位，直到遍历结束，最后如果add=1，就在后面节点再加1，最后返回list1的头结点。

/*** 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 head = l1;ListNode p = l1;ListNode q = l2;while(p.next!=null || q.next!=null){if(p.next==null) p.next = new ListNode(0,null);if(q.next==null) q.next = new ListNode(0,null);p = p.next;q = q.next;}int add = 0;int sum = 0;while(l1!=null){sum=0;sum = l1.val+l2.val+add;if(sum>=10){l1.val = sum%10;add=1;}else{l1.val = sum;add=0;}l1=l1.next;l2=l2.next;}if(add==1) p.next = new ListNode(1,null);return head;}
}