1、题目描述:
但实际上,这道题还有更优的解法,下面来看。
2、方法一:一次遍历,反转指针指向
思路:
代码:
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def reverseList(self, head: ListNode) -> ListNode:
pre = None # 前驱节点
cur = head # 当前要遍历的节点
while cur != None:
temp = cur.next # 保存当前节点的下一个节点,防止丢失
cur.next = pre # 当前指针指向前一个
pre = cur # 后移
cur = temp # 后移
return pre
3、方法二:保存栈,再取栈顶,串起来
思路:
细节都在代码中。
代码:
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def reverseList(self, head: ListNode) -> ListNode:
if head == None: return head # 为空直接返回
stack = [] # 栈
while head != None:
stack.append(head) # 保存对象
head = head.next
head = stack.pop() # 保存第一个节点
cur = head # 当前要遍历的节点
while stack:
cur.next = stack.pop() # 当前节点指向出栈的元素
cur = cur.next # 当前节点向后移动一位
cur.next = None # 后移动后,让next指向空
return head