我试图在python中创建一个简单的单链接列表。 (我知道不需要在python中实现list,但这不是重点)
这是我的代码:
class Node:
def __init__(self,data):
self.data = data
self.next= None
class List:
def __init__(self):
self.firstNode = Node(None)
def inserthead(self,newnode):
if not self.firstNode.next:
newnode.next = None
self.firstNode.next = newnode
else:
newnode.next = self.firstNode.next
self.firstNode.next= newnode
def __show(self,start):
if start.next:
print start.data
self.__show(start.next)
def printlist(self):
self.__show(self.firstNode)
def __reverte_recursive(self,node):
temp = None
if not node.next: return node
else:
temp = self.__reverte_recursive(node.next)
node.next.next= node
node.next = None
return temp
def reverte_list1(self):
self.firstNode=self.__reverte_recursive(self.firstNode)
def __reverte_iterative(self,node):
temp = None
previous = None
while node and node.next:
temp = node.next
node.next= previous
previous = node
node = temp
return previous
def reverte_iterative(self):
self.firstNode=self.__reverte_iterative(self.firstNode)
nodeA = Node("A")
nodeB = Node("B")
nodeC = Node("C")
nodeD = Node("D")
nodeE = Node("E")
list1= List()
list1.inserthead(nodeA)
list1.inserthead(nodeB)
class Node:
def __init__(self,data):
self.data = data
self.next= None
class List:
def __init__(self):
self.firstNode = Node(None)
def inserthead(self,newnode):
if not self.firstNode.next:
newnode.next = None
self.firstNode.next = newnode
else:
newnode.next = self.firstNode.next
self.firstNode.next= newnode
def __show(self,start):
if start.next:
print start.data
self.__show(start.next)
def printlist(self):
self.__show(self.firstNode)
def __reverte_recursive(self,node):
temp = None
if not node.next: return node
else:
temp = self.__reverte_recursive(node.next)
node.next.next= node
node.next = None
return temp
def reverte_list1(self):
self.firstNode=self.__reverte_recursive(self.firstNode)
def __reverte_iterative(self,node):
temp = None
previous = None
while node and node.next:
temp = node.next
node.next= previous
previous = node
node = temp
return previous
def reverte_iterative(self):
self.firstNode=self.__reverte_iterative(self.firstNode)
nodeA = Node("A")
nodeB = Node("B")
nodeC = Node("C")
nodeD = Node("D")
nodeE = Node("E")
list1= List()
list1.inserthead(nodeA)
list1.inserthead(nodeB)
list1.inserthead(nodeC)
list1.inserthead(nodeD)
list1.inserthead(nodeE)
print "list"
list1.printlist()
print "list reverse"
list1.reverte_list1()
list1.printlist()
list1.reverte_iterative()
print "list reverse reverse"
list1.printlist()
结果是:
None
E
D
C
B
list reverse
A
B
C
D
E
list reverse reverse
E
D
C
B
出于某种原因,我无法打印所有列表,在第一种情况下,也无法打印“A”节点 但是打印第一个节点(但是我选中了,B节点指向A) 第一个倒车可以 但是第三个节点即使被B节点指向也不会打印A节点。 可能打印的问题在于uu show函数。 但我想我是个概念错误。
谢谢
它只在当前节点有下一个节点时打印当前节点,这就是为什么从未打印最后一个节点的原因。 应该是:
在整个代码中,检查/分配节点而不是下一个节点时也会犯类似的错误,反之亦然(例如在
inserthead()
中-这会导致None
被打印)相关问题 更多 >
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