【Leetcode刷题Python】206.反转链表

1 题目

给你单链表的头节点 head ，请你反转链表，并返回反转后的链表。

示例 1：
输入：head = [1,2,3,4,5]
输出：[5,4,3,2,1]
示例 2：
输入：head = [1,2]
输出：[2,1]
示例 3：

输入：head = []
输出：[]

提示：

链表中节点的数目范围是 [0, 5000]
-5000 <= Node.val <= 5000
进阶：链表可以选用迭代或递归方式完成反转。你能否用两种方法解决这道题？

2 Python代码

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    # 方案一：递归
    # def reverseList(self, head: ListNode) -> ListNode:
    #     if head ==None or head.next==None:
    #         return head
    #     curr = self.reverseList(head.next)
    #     head.next.next = head
    #     head.next = None
    #     return curr
    # 方案二：迭代
    def reverseList(self, head: ListNode) -> ListNode:
        prev,curr = None,head
        while curr:
            # 小心这一步，别忘了
            next = curr.next
            curr.next = prev
            prev = next
            curr = curr.next
        return prev