leecode206（反转链表）

输入：head = [1,2,3,4,5]
输出：[5,4,3,2,1]

输入：head = [1,2]
输出：[2,1]

输入：head = []
输出：[]

def reverseList(self, head: ListNode) -> ListNode:
    if head is None or head.next is None:
        return head
    
    p = self.reverseList(head.next)
    head.next.next = head
    head.next = None

    return p

# 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, curr = None, head     #prev是previous前一个，curr是current当前
        while curr is not None:
            next = curr.next
            curr.next = prev
            prev = curr
            curr = next
        return prev