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.

Input:(2 -> 4 -> 3) + (5 -> 6 -> 4)
Output:7 -> 0 -> 8

思路分析：一：兩個連結串列節點想加外產生新節點, 

     二：新連結串列需要逆序，只需用頭插法建立新連結串列即可

程式碼如下

#include <stdio.h>


*Definition for singly-linked list.
struct ListNode {
     int val;
     struct ListNode *next;
  };
 
struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2) {
	if(l1==null||l2==null)
		printf("error");
		return;
	ListNode sumNode;
	while(l1!=null&&l2!=null){
		int temp;
		ListNode tempNode;
		temp = *l1.val+*l2.val;
		tempNode.val = temp;
		tempNode.next = *sumNode;
		sumNode = *tempNode;
		l1++;
		l2++;
		
	}
	return sumNode;
}