Problem：
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.
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Resolution
public ListNode addTwoNumbers(ListNode l1,ListNode l2){
int carry =0;//每一位的数值
ListNode p,result = new ListNode(0);
p=result;
while(l1!=null||l2!=null||carry!=0){
if(l1!=null){
carry+=l1.val;
l1=l1.next;
}
if (l2!=null){
carry+=l2.val;
l2=l2.next;
}
p.next=new ListNode(carry%10);
carry/10;
p=p.next;
}
return result.next;
}
注：
①需要保证l1或l2非空。在carry！=0时也要继续进行计算的原因是当l1与l2均进行完加法，到达链表末尾，不会出现溢出问题。
②处理方式并非相对l1与l2的val加和在赋值给carry，而是carry+=l1.val，在carry+=l2.val；
原文：http://www.cnblogs.com/yumiaomiao/p/7029202.html