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

一直兩個很大的數用連結串列逆序表示。
比如：342，表示為：(2 -> 4 -> 3)
465，表示為：(5 -> 6 -> 4)
計算342+465=807，結果也是逆序表示為：7 -> 0 -> 8

在例如，1314+520=1834被表示為：
4->1->3->1, 0->2->5， 結果為：4->3->8->1

# Definition for singly-linked list.
 

# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def addTwoNumbers(self, l1, l2):
        """
        :type l1: ListNode
        :type l2: ListNode
        :rtype: ListNode
        """
        if l1 is None:
            return
 
 l2
        elif l2 is None:
            return l1
        else:          
            carry = 0
            ret =ListNode(0) #頭結點
            ret_Last = ret 

            while(l1 or l2):
                sum = 0
                if(l1):
                    sum = l1.val
                    l1 = l1.next
                if(l2):
                    sum += l2.val
                    l2 = l2.next
                sum += carry  #如果有進位則加True(1)，否則加False(0)
                ret_Last.next = ListNode(sum%10)
                ret_Last = ret_Last.next
                carry = (sum >= 10) #如果sum>10，則carry表示進位，True被當作1來累加
            if(carry):
                ret_Last.next =ListNode(1)
            ret_Last = ret.next
            del ret #刪除頭結點
            return ret_Last