# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
ans = l1
while l1 or l2:
if not l1:
l1 = ListNode(val = 0)
if not l2:
l2 = ListNode(val = 0)
l1.val += l2.val
if l1.val >= 10:
l1.val %= 10
if l1.next:
l1.next.val += 1
else:
l1.next = ListNode(val = 1)
l1 = l1.next
l2 = l2.next
return ans
这是力扣第二题,因为ans返回的是最开始的l1头节点,而这个头节点后面会因为l1长度不够提前指向none,那么问题来了,怎么返回遍历相加后的头节点