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【Python】实现单链表

作者:彧福 / 发布于2015/5/11/ 852


	#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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　单链表　

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