单链表反转实现
1、递归实现
根据递归，递归到最后一个节点(条件为head3为非空，其下一个指向为空)，将其next指向前一个结点，前一个结点的指向为None。
def recurse(head, newhead): # 递归，head为原链表的头结点，newhead为反转后链表的头结点
if head is None:
return
if head.next is None:
newhead = head
else:
newhead = recurse(head.next, newhead)
head.next.next = head
head.next = None
return newhead
注释：原来链表为{1,2,3,4}
head指向为1，pHead1=head.next pHead2=pHead1.next pHead3=pHead2.next
a、递归最后实现newhead=phead3
b、回到上次递归的结束下方
2、循环实现
def ReverseList(self, pHead):
if pHead is None or pHead.next is None:
return pHead
pre = None
cur = pHead
h = pHead
while cur:
h = cur
tmp = cur.next
cur.next = pre
pre = cur
cur = tmp
return h