Python单链表简单实现代码
本文实例讲述了Python单链表简单实现代码。分享给大家供大家参考,具体如下:
用Python模拟一下单链表,比较简单,初学者可以参考参考
#coding:utf-8
class Node(object):
def __init__(self, data):
self.data = data
self.next = None
class NodeList(object):
def __init__(self, node):
self.head = node
self.head.next = None
self.end = self.head
def add_node(self, node):
self.end.next = node
self.end = self.end.next
def length(self):
node = self.head
count = 1
while node.next is not None:
count += 1
node = node.next
return count
# delete node and return it's value
def delete_node(self, index):
if index+1 > self.length():
raise IndexError('index out of bounds')
i = 0
node = self.head
while True:
if i==index-1:
break
node = node.next
i += 1
tmp_node = node.next
node.next = node.next.next
return tmp_node.data
def show(self):
node = self.head
node_str = ''
while node is not None:
if node.next is not None:
node_str += str(node.data) + '->'
else:
node_str += str(node.data)
node = node.next
print node_str
# Modify the original position value and return the old value
def change(self, index, data):
if index+1 > self.length():
raise IndexError('index out of bounds')
i = 0
node = self.head
while True:
if i == index:
break
node = node.next
i += 1
tmp_data = node.data
node.data = data
return tmp_data
# To find the location of index value
def find(self, index):
if index+1 > self.length():
raise IndexError('index out of bounds')
i = 0
node = self.head
while True:
if i == index:
break
node = node.next
i += 1
return node.data
#test case
n1 = Node(0)
n2 = Node(1)
n3 = Node(2)
n4 = Node(3)
n5 = Node(4)
node_list = NodeList(n1)
node_list.add_node(n2)
node_list.add_node(n3)
node_list.add_node(n4)
node_list.add_node(n5)
#node = node_list.delete_node(3)
#print node
#d = node_list.change(0,88)
data = node_list.find(5)
print data
node_list.show()
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希望本文所述对大家Python程序设计有所帮助。
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