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;
    }
}