题目：
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
这题边界情况要处理好：空值；两个链表长度可能不等；最后两个为空还可能留有一个进位木有处理。
这次做题，自己做的不太好：变量名取的渣，取个next，搞的自己逻辑混乱，所以取变量名要取好；还有如果一个为null可以不往下循环，直接指针指向不为空的内个链表就可以了。
代码：
/**
* 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) {
if(l1 == null || l2 == null)
return null;
int temp1 = 0;
int temp2 = 0;
int count = 0;
int sum = 0;
ListNode first = null;
ListNode next = null;
while(l1 != null || l2 != null) {
temp1 = l1 != null ? l1.val : 0;
temp2 = l2 != null ? l2.val : 0;
sum = temp1 + temp2 + count;
if(sum > 9) {
sum -= 10;
count = 1;
} else {
count = 0;
}
ListNode list = new ListNode(sum);
if(next == null) {
first = list;
next = first;
} else {
next.next = list;
next = list;
}
l1 = l1 == null ? null :l1.next;
l2 = l2 == null ? null :l2.next;
}
if(count == 1) {
next.next = new ListNode(1);
}
return first;
}
}