python算法题 链表反转详解

# encoding: utf-8
class Node(object):
def __init__(self):
self.value = None
self.next = None
def __str__(self):
return str(self.value)
def reverse_loop(head):
if not head or not head.next:
return head
pre = None 
while head:
next = head.next # 缓存当前节点的向后指针，待下次迭代用
head.next = pre # 这一步是反转的关键，相当于把当前的向前指针作为当前节点的向后指针
pre = head # 作为下次迭代时的（当前节点的）向前指针
head = next # 作为下次迭代时的（当前）节点
return pre # 返回头指针，头指针就是迭代到最后一次时的head变量（赋值给了pre）

if __name__ == '__main__':
three = Node()
three.value = 3
two = Node()
two.value = 2
two.next = three
one = Node()
one.value = 1
one.next = two
head = Node()
head.value = 0
head.next = one
newhead = reverse_loop(head)
while newhead:
print(newhead.value, )
newhead = newhead.next

3
2
1
0
2

head.next = None
head.next.next = head.next
head.next.next.next = head.next.next
...
...

def reverse_recursion(head):
if not head or not head.next:
return head
new_head = reverse_recursion(head.next)
head.next.next = head
head.next = None
return new_head