Add Two Numbers--LeetCode

题目：

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

思路：遍历两个链表，直到这两个链表都为空，然后相加链表节点中的值。

void AddTwoNumber(List*& first,List*& second,List*& newlist)
{
	List* l1=first;
	List* l2=second;
	List* cur,*node;
	int carry=0;
	int tmp=0;
	newlist = NULL;
	if(first == NULL)
	{
		newlist = second;
		return ;
	}
	if(second == NULL)
	{
		newlist = first;
		return ;
	} 
	while(l1 !=NULL || l2!=NULL)
	{
		tmp = carry;
		if( l1 !=NULL)
		{
			tmp += l1->value;
			l1 = l1->next;
		}
		if(l2 != NULL)
		{
			tmp += l2->value;
			l2 = l2->next;
		}
		carry = tmp/10;
		tmp = tmp%10;
		if(newlist == NULL)
		{
			newlist = new List;
			newlist->next = NULL;
			newlist->value = tmp;
			cur = newlist;
		}
		else
		{
			node = new List;
			node->next = NULL;
			node->value= tmp;
			cur->next = node;
			cur = cur->next;
		}
	}
	if(carry ==1)
	{
		node = new List;
		node->next = NULL;
		node->value = carry;
		cur->next = node;
	}
}

int main()
{
	int array[]={5,1,2,7};
	int array1[]={8,4,3,6,9,9};
	List* first;
	
	Init_List(first,array,sizeof(array)/sizeof(int));
	List* second;
	Init_List(second,array1,sizeof(array1)/sizeof(int));
	List* newlist=NULL;
	AddTwoNumber(first,second,newlist);
	print_list(newlist); 
	return 0;
}