题目描述：
You are given two non-empty linked lists representing two non-negative integers. The most significant digit comes first 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.
Follow up:
What if you cannot modify the input lists? In other words, reversing the lists is not allowed.
Example:
Input: (7 -> 2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 8 -> 0 -> 7题目要求是不能翻转链表再求和，那么可以用栈作为辅助将链表里面的元素顺序翻转。
class Solution {
public:
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
ListNode* p=l1;
ListNode* q=l2;
stack<int> s1;
stack<int> s2;
while(p!=NULL)
{
s1.push(p->val);
p=p->next;
}
while(q!=NULL)
{
s2.push(q->val);
q=q->next;
}
ListNode* head=new ListNode(0);
int sum=0;
int carry=0;
while(s1.size()>0&&s2.size()>0)
{
sum=s1.top()+s2.top()+carry;
carry=sum/10;
ListNode* p=new ListNode(sum%10);
p->next=head->next;
head->next=p;
s1.pop();
s2.pop();
}
while(s1.size()>0)
{
sum=s1.top()+carry;
carry=sum/10;
ListNode* p=new ListNode(sum%10);
p->next=head->next;
head->next=p;
s1.pop();
}
while(s2.size()>0)
{
sum=s2.top()+carry;
carry=sum/10;
ListNode* p=new ListNode(sum%10);
p->next=head->next;
head->next=p;
s2.pop();
}
if(carry==1)
{
ListNode* p=new ListNode(1);
p->next=head->next;
head->next=p;
}
return head->next;
}
};