2025年单向链表反转代码（编写一个单向链表反转的程序）

编程日记 • 2025-03-10 21:40 • 阅读 8

1 问题

利用python实现链表的反向输出，反转单链表。

2 方法

1.程序分析:无

代码清单 1

class Node:
def __init__(self, data):
self.data = data
self.next = None
def get_data(self):
return self.data
class List:
def __init__(self, head):
self.head = head
def is_empty(self):
return self.get_len() == 0
def get_len(self):
length = 0
temp = self.head
while temp is not None:
length += 1
temp = temp.next
return length
def append(self, node):
temp = self.head
while temp.next is not None:
temp = temp.next
temp.next = node
def delete(self, index):
if index < 1 or index > self.get_len():
print("给定位置不合理")
return
if index == 1:
self.head = self.head.next
return
temp = self.head
cur_pos = 0
while temp is not None:
cur_pos += 1
if cur_pos == index-1:
temp.next = temp.next.next
temp = temp.next
def insert(self, pos, node):
if pos < 1 or pos > self.get_len():
print("插入结点位置不合理")
return
temp = self.head
cur_pos = 0
while temp is not Node:
cur_pos += 1
if cur_pos == pos-1:
node.next = temp.next
temp.next =node
break
temp = temp.next
def reverse(self, head):
if head is None and head.next is None:
return head
pre = head
cur = head.next
while cur is not None:
temp = cur.next
cur.next = pre
pre = cur
cur = temp
head.next = None
return pre
def print_list(self, head):
init_data = []
while head is not None:
init_data.append(head.get_data())
head = head.next
return init_data
if __name__=='__main__':
head=Node('head')
link=List(head)
for i in range(10):
node=Node(i)
link.append(node)
print(link.print_list(head))
print(link.print_list(link.reverse(head)))

3 结语

针对此类边界条件较多的问题，提出定义类，函数的方法，通过实验，证明该方法是有效的。此方法结合熟练运用了所学的基本python知识，知识虽简单，但需要较清晰的逻辑关系来分类讨论。对验证对象条件的充分理解，是解决此题的关键。希望未来能利用更复杂的python知识解决更多问题。

2025年单向链表反转代码（编写一个单向链表反转的程序）

相关推荐