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

大数相加，只保留各位，进位加到下一节点。

需注意，判断l1,l2是否为空。若两链表长度不同，需将剩余节点同样作进位处理，若最后剩有进位，需增加新节点。

1 /** 2  * Definition for singly-linked list. 3  * struct ListNode { 4  *     int val; 5  *     ListNode *next; 6  *     ListNode(int x) : val(x), next(NULL) {} 7  * }; 8  */ 9 class Solution {10 public:11     ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {12         if(l1==NULL) return l2;13         if(l2==NULL) return l1;14         ListNode* res=NULL,*pNode=NULL,*pnext=NULL;15         ListNode* p=l1,*q=l2;16         int up=0;17         while(p!=NULL&&q!=NULL)18         {19             pnext=new ListNode(p->val+q->val+up);20             up=pnext->val/10;21             pnext->val=pnext->val%10;22             if(res==NULL)23               res=pNode=pnext;24             else25               {26                   pNode->next=pnext;27                   pNode=pnext;28               }29             p=p->next;30             q=q->next;31         }32         while(p!=NULL)33         {34             pnext=new ListNode(p->val+up);35             up=pnext->val/10;36             pnext->val=pnext->val%10;37             pNode->next=pnext;38             pNode=pnext;39             p=p->next;40         }41         while(q!=NULL)42         {43             pnext=new ListNode(q->val+up);44             up=pnext->val/10;45             pnext->val=pnext->val%10;46             pNode->next=pnext;47             pNode=pnext;48             q=q->next;49         }50         if(up!=0)51         {52             pnext=new ListNode(up);53              pNode->next=pnext;54             pNode=pnext;55         }56         return res;57     }58 };