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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;
}
}
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原文地址：http://www.cnblogs.com/wangnanabuaa/p/5134847.html
