leetCode | Add Two Numbers

You are given two linked lists representing two non-negative numbers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
题目的意思是两个非负数的链表相加，只不过数字是倒过来存的，并且每个结点仅仅包含一个单个数字，最后返回一个相加之和的链表，实际上也就是大数字相加减，但是要注意到进位，以及当两个链表的长度不同的时候还需继续对长的链表进行遍历，处理进位，直到最后。
注意：两个链表相加可能出现最高位的进位，那么如果最高位有进位，还需要增加一个最高位的进位结点放在最后来存储进位
代码如下：

`/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { *         val = x; *         next = null; *     } * } */public class Solution {    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {        ListNode resultList = null;        ListNode p = null;        ListNode pNode = null;        int up = 0;        if (l1 == null)            return l2;        if (l2 == null)            return l1;        while (l1 != null && l2 != null) {            int sum = l1.val + l2.val + up;            pNode = new ListNode(sum % 10);            up = sum / 10;            if (resultList == null) {                resultList = p = pNode;            } else {                p.next = pNode;                p = p.next;            }            l1 = l1.next;            l2 = l2.next;        }        while (l1 != null) {            int sum = l1.val + up;            pNode = new ListNode(sum % 10);            up = sum / 10;            p.next = pNode;            p = p.next;            l1 = l1.next;        }        while (l2 != null) {            int sum = l2.val + up;            pNode = new ListNode(sum % 10);            up = sum / 10;            p.next = pNode;            p = p.next;            l2 = l2.next;        }        if (up > 0) {            pNode = new ListNode(up);            p.next = pNode;        }        return resultList;    }}`