此文转载自：https://blog.csdn.net/qq_36556893/article/details/110231369
目录
一、题目内容
给你两个 非空 链表来代表两个非负整数。数字最高位位于链表开始位置。它们的每个节点只存储一位数字。将这两数相加会返回一个新的链表。
你可以假设除了数字 0 之外，这两个数字都不会以零开头。
进阶：
如果输入链表不能修改该如何处理？换句话说，你不能对列表中的节点进行翻转。
示例：
输入：(7 -> 2 -> 4 -> 3) + (5 -> 6 -> 4)
输出：7 -> 8 -> 0 -> 7
二、解题思路
哎，太难了，还是两个函数来回套吧：
1.两数相加：leetcode_2. 两数相加
2.反转链表：leetcode_206. 反转链表
三、代码
# Definition for singly-linked list.
class ListNode:
def __init__(self, x):
self.val = x
self.next = None
class Solution:
def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
head = self._addTwoNumbers(self.reverseList(l1), self.reverseList(l2))
return self.reverseList(head)
def _addTwoNumbers(self, l1, l2):
"""
:type l1: ListNode
:type l2: ListNode
:rtype: ListNode
"""
head = ListNode(0)
# print(head.val, head.next)
p = head
q = p
carry = 0
while l1 or l2:
if l1 is not None:
l1_val = l1.val
# print("l1:", l1_val)
else:
l1_val = 0
if l2 is not None:
l2_val = l2.val
# print("l2:", l2_val)
else:
l2_val = 0
temp = l1_val + l2_val + carry
if temp >= 10:
carry = 1
p.val = temp - 10
p.next = ListNode(1)
else:
carry = 0
p.val = temp
p.next = ListNode(0)
q = p
p = p.next
if l1 is not None:
l1 = l1.next
if l2 is not None:
l2 = l2.next
if q.next.val == 0:
q.next = None
del p
return head
def reverseList(self, head: ListNode):
# null 1 --> 2 --> 3 --> 4 --> null
# null <-- 1 <-- 2 <-- 3 <-- 4 null
if not head:
return None
cur, cur_new_next = head, None
while cur:
# save original next node
origin_next = cur.next
# link new next node
cur.next = cur_new_next
# current node turns to new next node
cur_new_next = cur
# original next node turns to current node
cur = origin_next
return cur_new_next