每日一题力扣206

反转一个单链表。

class Solution:
    def reverseList(self, head: ListNode) -> ListNode:
        if head is None or head.next is None:
            return head
        # pre 就是那个空链表
        pre, cur = None, head
        # 不断将当前链表移动到空链表上
        while cur:#cur是当前的点
            next_node = cur.next#保存当前节点的后面内容
            cur.next = pre#链接到等于新链表
            pre = cur#赋给新链表
            cur = next_node#再把cur往后移动
        return pre