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1 问题 如何利用python实现单向循环链表简化数学问题？ 2 方法 - add方法：向链表头部添加一个节点data
- append方法：向链表尾部添加一个节点，值为data
- remove方法：删除链表中第一个值为data的节点
代码清单 1 class Node:
def __init__(self, data, _next=None):
self.data = data
self.next = _next
class SingleCycleLinkList:
def __init__(self):
self.head = None
self._length = 0
def is_empty(self):
return self._length == 0
def length(self):
return self._length
def nodes_list(self):
ls = []
if self.is_empty():
return ls
else:
ls.append(self.head.data)
cur = self.head.next
while cur != self.head:
ls.append(cur.data)
cur = cur.next
return ls
def add(self, data):
node = Node(data)
if self.is_empty():
self.head = node
node.next = node
else:
node.next = self.head
cur = self.head
while cur.next != self.head:
cur = cur.next
cur.next = node
self.head = node
self._length += 1
def append(self, data):
node = Node(data)
if self.head != None:
cur = self.head
while cur.next != self.head:
cur = cur.next
cur.next = node
node.next = self.head
else:
self.head = node
node.next = node
self._length += 1
def insert(self, pos, data):
if pos <= 0:
self.add(data)
elif pos >= self._length:
self.append(data)
else:
node = Node(data)
cur = self.head
n = 0
while n < pos - 1:
cur = cur.next
n = n + 1
node.next = cur.next
cur.next = node
self._length += 1
def remove(self, data):
if self.is_empty():
return -1
else:
cur = self.head
flag = True
prev = None
while cur != self.head or flag:
flag = False
if cur.data == data:
if prev == None
last_node = self.head
while last_node.next != self.head:
last_node = last_node
last_node.next = self.head.next
self.head = self.head.next
else:
prev.next = cur.next
self._length -= 1
return 0
prev = cur
cur = cur.next
return -1
def modify(self, pos, data
if pos < 0 or pos >= self._length:
print("位置不正确")
else:
cur = self.head
n = 0
while n < pos:
cur = cur.next
n = n + 1
cur.data = data
def search(self, data):
if self.is_empty():
return False
else:
cur = self.head
flag = True
flag = False
if cur.data == data:
return True
cur = cur.next
return False
if __name__ == "__main__":
l1 = SingleCycleLinkList()
print(l1.head, l1.length())
l1.add(1)
print(l1.nodes_list())
l1.add(2)
print(l1.nodes_list())
print(l1.head.data, l1.head.next.data, l1.head.next.next.data)
l1.append(4)
print(l1.nodes_list())
l1.insert(1, 7)
print(l1.nodes_list())
l1.remove(2)
print(l1.nodes_list())
l1.modify(1, 3)
print(l1.nodes_list())
print("查找")
print(l1.search(3)) |
3 结语 运用单向循环链表可以用来解决约瑟夫环问题，但目前通过python来解决此类问题只能停留在最基本的层面上，要想深入解决此类问题，则要通过后续的学习，了解更多的python知识，从来实现对该类问题的完美解决。
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