反转一个单链表。

示例:

迭代1笨办法：

class Solution:
    def reverseList(self, head: ListNode) -> ListNode:
        if not head:
            return
        p = head
        res = []
        while(p):
            res.append(p.val)
            p = p.next
        head1 = ListNode(res[-1])
        q = head1
        res = res[:-1]
        while res:
            val = res.pop()
            q.next=ListNode(val)
            q = q.next
        return head1

迭代2:代码搬运工，思路是用引入一个指针pre，先让它指向空，迭代中，先把cur.next的值用tmp存下来，然后把cur.next这个指针断开用pre替代，然后pre指针指向cur本身，相当于把往后指的指针往前指惹，最后用后一个节点做cur.

class Solution:
    def reverseList(self, head: ListNode) -> ListNode:
        cur = head
        pre = None
        while(cur):
            tmp = cur.next
            cur.next = pre
            pre = cur
            cur = tmp
        return pre

递归1：代码搬运工，思路是先把cur.next的节点用tmp存，tmp的next接前面反转过的子链表

这个代码思路上行得通，但是会超出时间限制

输入: 1->2->3->4->5->NULL
输出: 5->4->3->2->1->NULL

[1,None]   

[2,1,None]

[3,2,1,None]

[4,3,2,1,None]

[5,4,3,2,1,None]

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
 
class Solution:
    def reverseList(self, head: ListNode) -> ListNode:
        #停止条件:cur遍历到节点最后一个节点(None),返回头节点pre
        #返回头节点
        def loop(pre,cur):
            if not cur:
                return pre
            tmp = cur.next
            cur.next = pre
            return loop(cur,tmp)
        return loop(None,head)