[LeetCode]Add Two Numbers

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

注意点:改了又改，好多测试用例没有考虑到。代码很乱~

      1.{5}，{5}->{0,1}

      2.{1},{9,9}->{0,0,1}

      3.{9,8},{1}->{0,9}

class Solution {

public:

	int Length(ListNode *l)

	{

		int len=0;

		while(l)

		{

			l=l->next;

			len++;

		}

		return len;

	}

	ListNode *func(ListNode *l1,ListNode *l2)

	{

		ListNode *p=l1;

		ListNode *q=l2;

		ListNode *r=p;

		bool flag=false;

		while(q)

		{

			r=p;

			if(flag)

			{

				p->val+=q->val+1;

				flag=false;

			}

			else

				p->val+=q->val;

			if(p->val>9)

			{

				flag=true;

				p->val=p->val%10;

			}

			p=p->next;

			q=q->next;

		}

		while(flag)

		{

			if(!p)

			{

				ListNode *newnode=new ListNode(1);

				r->next=newnode;

				flag=false;

			}

			else

			{

				p->val+=1;

				flag=false;

				if(p->val>9)

				{

					flag=true;

					p->val=p->val%10;

					r=p;

					p=p->next;

				}

			}

		}

		return l1;

	}

    ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {

        // IMPORTANT: Please reset any member data you declared, as

        // the same Solution instance will be reused for each test case.

		int len1=Length(l1);

		int len2=Length(l2);

		if(len1==0) return l2;

		else if (len2==0) return l1;

		else if(len1<len2) return func(l2,l1);

		else return func(l1,l2);

    }

};