链表求和

你有两个用链表代表的整数，其中每个节点包含一个数字。数字存储按照在原来整数中相反的顺序，使得第一个数字位于链表的开头。写出一个函数将两个整数相加，用链表形式返回和。

样例

给出两个链表 3->1->5->null 和 5->9->2->null，返回 8->0->8->null

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    /**
     * @param l1: the first list
     * @param l2: the second list
     * @return: the sum list of l1 and l2
     */
    public ListNode addLists(ListNode l1, ListNode l2) {
        // write your code here
        if(l1 == null)  return l2;
        if(l2 == null)  return l1;

        ListNode result = null;
        ListNode pNode = null;
        ListNode pNext = null;
        ListNode p = l1;
        ListNode q = l2;
        int up=0;
        while(p!=null&&q!= null){
            pNext = new ListNode(p.val+q.val +up);
            up = pNext.val/10;
            pNext.val = pNext.val%10;

            if(result == null){
                result = pNode = pNext;
            }
            else{
                pNode.next = pNext;
                pNode = pNext;
            }
            p = p.next;
            q = q.next;
        }

        while(p!= null){
            pNext = new ListNode(p.val + up);
            up = pNext.val/10;
            pNext.val = pNext.val%10;
            pNode.next = pNext;
            pNode = pNext;
            p = p.next;
        }
         while(q!= null){
            pNext = new ListNode(q.val + up);
            up = pNext.val/10;
            pNext.val = pNext.val%10;
            pNode.next = pNext;
            pNode = pNext;
            q = q.next;
        }
        if(up!= 0){
            pNext = new ListNode(up);
            pNode.next = pNext;
        }
        return result;
    }
}