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

　题意，反向表达，相加之后还是以链表的形式表达。

class Solution {
public:
ListNode *addTwoNumbers(ListNode *l1, ListNode *l2)
{
int forw=0;//进位标志
//建立一个新的链表存储答案
ListNode root(0);
ListNode *output=&root;
while(l1||l2)
{
int v1 = (l1 ? l1->val : 0);
int v2 = (l2 ? l2->val : 0);
int sum=v1+v2+forw;
//检验sum是否为两位数
forw=sum/10;
sum=sum%10;
ListNode *coNode=new ListNode(sum);
output->next=coNode;
output=coNode;
if(l1)l1=l1->next;
if(l2)l2=l2->next;
}
if(forw>0)
{
ListNode *coNode=new ListNode(forw);
output->next=coNode;
output=coNode;
}
return root.next;
}
};

Add Two Numbers 解题报告