啊要命了,很快就写出了用迭代实现的方法,但是卡在了用递归反转,然后就跑去玩了很久,赶快跑过来把别人的代码先抄一下否则今天就废了。这个算法的时间复杂度是O(N),空间复杂度是O(1)。
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def reverseList1(self, head: ListNode) -> ListNode:
prev = None
while head is not None:
tmp = head.next
head.next = prev
prev = head
head = tmp
return prev
def reverseList2(self, head: ListNode) -> ListNode:
if not head or not head.next:
return head
p = self.reverseList(head.next)
head.next.next = head
head.next = None
return p
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