题解 | #反转链表#

# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
#
# @param head ListNode类
# @return ListNode类
#
class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next


class Solution:
    # current.next=pre
    def ReverseList(self, head: ListNode) -> ListNode:
        if not head:
            return None

        pre = None
        current = head
        # 先记录下一个节点，当前节点连接上一个节点，然后断开原本的连接
        while current:
            temp = current.next
            # 注意是节点不是节点的值
            current.next = pre
            pre = current
            current = temp
        return pre


def creat_chain_table(elements):
    if not elements:
        return None
    head = ListNode(elements[0])
    current = head
    for value in elements[1:]:
        current.next = ListNode(value)
        current = current.next
    return head

注意两个指针，分清节点和节点的值即可