第一种方法使用了stack，但是超时了，
第二种是循环，仿照前面某一题目的写法
# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution(object):
def reverseList(self, head):
"""
:type head: ListNode
:rtype: ListNode
"""
if not head or not head.next:
return head
dummy = ListNode(0)
dummy.next = head
pre = dummy
precurr = head
while precurr.next:
curr = precurr.next
precurr.next = curr.next
curr.next = pre.next
pre.next = curr
return dummy.next
比较简单的写法，嗯我被前面的影响了
class Solution:
# @param {ListNode} head
# @return {ListNode}
def reverseList(self, head):
prev = None
while head:
curr = head
head = head.next
curr.next = prev
prev = curr
return prev
然后是递归：这个我没写出来很好的
# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution(object):
def reverseList(self, head):
"""
:type head: ListNode
:rtype: ListNode
"""
return self._reverse(head)
def _reverse(self, node, prev=None):
if not node:
return prev
n = node.next
node.next = prev
return self._reverse(n, node)
