LeetCode-Add Two Numbers

Description：
You are given two non-empty linked lists representing two non-negative integers. 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.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Example
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

Explanation:
342 + 465 = 807.

题意：给定两个由整数组成的链表，链表的每个节点都带有一个[0,9]的数字，每个链表都表示一个数字，头结点为数字的低位，尾结点为高位，要求将两个链表表示的数字相加，并以链表的形式返回；

解法：其实，就是对链表的各个相应位置的数字相加，注意有进位的可能，以及两个链表的长度可能不一致、链表可能为空的可能；

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode addResult = new ListNode(0);//头结点
        ListNode addResultNext = addResult;//尾结点
        int sum = 0;//两个数的和
        int cf = 0;//进位标志
        while(l1 != null || l2 != null){
            int x = l1 == null ? 0 : l1.val;
            int y = l2 == null ? 0 : l2.val;
            sum = x + y + cf;
            cf = sum / 10;
            addResultNext.next = new ListNode(sum % 10);
            addResultNext = addResultNext.next;
            if(l1 != null) l1 = l1.next;
            if(l2 != null) l2 = l2.next;
        }
        if(cf != 0){
            addResultNext.next = new ListNode(cf);
        }//最高位相加仍有进位
        return addResult.next;
    }
}