你有两个用链表代表的整数，其中每个节点包含一个数字。数字存储按照在原来整数中相反的顺序，使得第一个数字位于链表的开头。写出一个函数将两个整数相加，用链表形式返回和。
您在真实的面试中是否遇到过这个题？
样例
给出两个链表 3->1->5->null 和 5->9->2->null，返回 8->0->8->null
"""
Definition for singly-linked list.
class ListNode:
def __init__(self, x):
self.val = x
self.next = None
"""
class Solution:
"""
@param: l1: the first list
@param: l2: the second list
@return: the sum list of l1 and l2
"""
def addLists(self, l1, l2):
# write your code here
if l1 is None and l2 is None:
return None
cur1 = l1.next
cur2 = l2.next
head = ListNode((l1.val + l2.val)%10)
cur = head
carry = (l1.val + l2.val) // 10
#当其中一条链表为空的时候,为该链表添加一个值为0的节点
#直到两条都为None,这种方式会修改原先的列表.
while cur1 is not None or cur2 is not None:
if cur1 is None:cur1 = ListNode(0)
if cur2 is None:cur2 = ListNode(0)
val = cur1.val + cur2.val + carry
carry = val // 10
cur.next = ListNode(val%10)
cur = cur.next
cur1 = cur1.next
cur2 = cur2.next
if carry != 0:
cur.next = ListNode(carry)
return head
