206.反转链表

题目链接：206.反转链表
难度：简单

描述：
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。

提示：
链表中的结点数是[0, 5000]。

题解

思路：
没什么好说的。

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def reverseList(self, head: ListNode) -> ListNode:
        prev = None
        cur = head
        while cur is not None:
            temp = cur.next
            cur.next = prev
            prev = cur
            cur = temp
        return prev

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def reverseList(self, head: ListNode) -> ListNode:
        if head is None or head.next is None:
            return head
        # 找到最后一个有效节点
        ret = self.reverseList(head.next)
        head.next.next = head
        head.next = None
        return ret