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.

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
  
class Solution:
    def addTwoNumbers(self, l1, l2):
        """
        :type l1: ListNode
        :type l2: ListNode
        :rtype: ListNode
        """   
        carry = 0
        root = n = ListNode(0)
        while l1 is not None or l2 is not None or carry != 0:            
            v1 = v2 = 0
            if l1 is not None:
                v1 = l1.val
                l1 = l1.next
            if l2 is not None:
                v2 = l2.val
                l2 = l2.next
            carry, val = divmod(v1+v2+carry, 10)
            n.next = ListNode(val)
            n = n.next
        return root.next

思路：
其实很简单，遍历两个链表，从低位开始相加。
注意进位的情况，l1和l2都遍历完了，要检查carry是不是零。

要自己构造一个新的链表。

Complexity Analysis:

Time complexity : O(max(m, n)). Assume that m and n represents the length of l1 and l2 respectively, the algorithm above iterates at most max(m, n) times.

Space complexity : O(max(m, n)). The length of the new list is at most max(m,n)+1.