Python单链表简单实现代码
(编辑:jimmy 日期: 2024/11/19 浏览:3 次 )
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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