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.
Example:
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.
思路：使用一个ListNode进行保存结果，进行一直后移链表操作
class Solution { public: ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) { ListNode *res=new ListNode(0); ListNode *tmp=res; bool start=true; int num=0; while(l1||l2){ if(l1) { num+=l1->val; l1=l1->next; } if(l2){ num+=l2->val; l2=l2->next; } if(start) tmp->val=num%10; else { tmp->next=new ListNode(num%10); tmp=tmp->next; } start=false; num/=10; } if(num!=0) tmp->next=new ListNode(num); return res; } };
原文：https://www.cnblogs.com/littlepage/p/12117530.html