【LeetCode】每日一题：反转链表

题解思路

循环的方法需要注意prev应该是None开始，然后到结束的时候prev是tail，递归的思路很难绕过弯来，主要在于很难想清楚为什么可以返回尾节点，需要多做递归题，以及递归过程中，可以不使用尾节点来找当前递归位置，用head结点即可，多用边界情况推理。

AC代码

# 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: Optional[ListNode]) -> Optional[ListNode]:
        # if not head or head.next is None:
        #     return head
        # newhead = self.reverseList(head.next)
        # head.next.next = head
        # head.next = None
        # return newhead

        prev = None
        curr = head
        while curr:
            temp = curr.next
            curr.next = prev
            prev = curr
            curr = temp
        return prev