方法一思路：
用现有节点存放结果（用l1存的，l1短则把l2的多余节点拼到l1上）。
class Solution(object):
def addTwoNumbers(self, l1, l2):
"""
:type l1: ListNode
:type l2: ListNode
:rtype: ListNode
"""
# 进位
extra = 0
# 记录l1和l2的前一个节点
pre1 = ListNode(0)
pre2 = ListNode(0)
ans = ListNode(0)
ans.next = l1
while l1 and l2:
sum = l1.val + l2.val + extra
temp = sum % 10
l1.val = temp
# 计算进位
extra = int(sum / 10)
# 指针后移
pre1 = l1
l1 = l1.next
pre2 = l2
l2 = l2.next
# 若l1遍历完，l2没完
if not l1 and l2:
# 先把l2剩余部分接到l1后，pre1再跟着l2走
pre1.next = l2
while l2:
sum = l2.val + extra
temp = sum % 10
extra = int(sum / 10)
l2.val = temp
pre1 = l2
l2 = l2.next
# 最后一位有进位且l2已遍历完
if not l2 and extra:
node = ListNode(0)
node.val = extra
pre1.next = node
# 若l2遍历完，l1没完
if not l2 and l1:
while l1:
sum = l1.val + extra
temp = sum % 10
extra = int(sum / 10)
l1.val = temp
pre1 = l1
l1 = l1.next
if not l1 and extra:
node = ListNode(0)
node.val = extra
pre1.next = node
# 若l1和l2都遍历完，进位不为0
if not l1 and not l2 and extra:
node = ListNode(0)
node.val = extra
pre1.next = node
return ans.next
方法二思路：
新建链表来存放结果。
class Solution(object):
def addTwoNumbers(self, l1, l2):
"""
:type l1: ListNode
:type l2: ListNode
:rtype: ListNode
"""
pre = node = ListNode(0)
# 进位
extra = 0
while l1 and l2:
node.next = ListNode(0)
# 求当前位的总和
sum = l1.val + l2.val + extra
# 计算当前位的值
node.next.val = sum % 10
# 计算进位
extra = int(sum / 10)
# 顺移
node = node.next
l1 = l1.next
l2 = l2.next
# 取未遍历完的
res = l1 or l2
while res:
node.next = ListNode(0)
sum = res.val + extra
node.next.val = sum % 10
extra = int(sum / 10)
node = node.next
res = res.next
# 最后一步还有进位
if extra:
node.next = ListNode(0)
node.next.val = extra
return pre.next