python环形单链表的约瑟夫问题详解
题目：
一个环形单链表，从头结点开始向后，指针每移动一个结点，就计数加1，当数到第m个节点时，就把该结点删除，然后继续从下一个节点开始从1计数，循环往复，直到环形单链表中只剩下了一个结点，返回该结点。
这个问题就是著名的约瑟夫问题。
代码：
首先给出环形单链表的数据结构：
class Node(object):
def __init__(self, value, next=0):
self.value = value
self.next = next # 指针
class RingLinkedList(object):
# 链表的数据结构
def __init__(self):
self.head = 0 # 头部
def __getitem__(self, key):
if self.is_empty():
print 'Linked list is empty.'
return
elif key < 0 or key > self.get_length():
print 'The given key is wrong.'
return
else:
return self.get_elem(key)
def __setitem__(self, key, value):
if self.is_empty():
print 'Linked list is empty.'
return
elif key < 0 or key > self.get_length():
print 'The given key is wrong.'
return
else:
return self.set_elem(key, value)
def init_list(self, data): # 按列表给出 data
self.head = Node(data[0])
p = self.head # 指针指向头结点
for i in data[1:]:
p.next = Node(i) # 确定指针指向下一个结点
p = p.next # 指针滑动向下一个位置
p.next = self.head
def get_length(self):
p, length = self.head, 0
while p != 0:
length += 1
p = p.next
if p == self.head:
break
return length
def is_empty(self):
if self.head == 0:
return True
else:
return False
def insert_node(self, index, value):
length = self.get_length()
if index < 0 or index > length:
print 'Can not insert node into the linked list.'
elif index == 0:
temp = self.head
self.head = Node(value, temp)
p = self.head
for _ in xrange(0, length):
p = p.next
print "p.value", p.value
p.next = self.head
elif index == length:
elem = self.get_elem(length-1)
elem.next = Node(value)
elem.next.next = self.head
else:
p, post = self.head, self.head
for i in xrange(index):
post = p
p = p.next
temp = p
post.next = Node(value, temp)
def delete_node(self, index):
if index < 0 or index > self.get_length()-1:
print "Wrong index number to delete any node."
elif self.is_empty():
print "No node can be deleted."
elif index == 0:
tail = self.get_elem(self.get_length()-1)
temp = self.head
self.head = temp.next
tail.next = self.head
elif index == self.get_length()-1:
p = self.head
for i in xrange(self.get_length()-2):
p = p.next
p.next = self.head
else:
p = self.head
for i in xrange(index-1):
p = p.next
p.next = p.next.next
def show_linked_list(self): # 打印链表中的所有元素
if self.is_empty():
print 'This is an empty linked list.'
else:
p, container = self.head, []
for _ in xrange(self.get_length()-1): #
container.append(p.value)
p = p.next
container.append(p.value)
print container
def clear_linked_list(self): # 将链表置空
p = self.head
for _ in xrange(0, self.get_length()-1):
post = p
p = p.next
del post
self.head = 0
def get_elem(self, index):
if self.is_empty():
print "The linked list is empty. Can not get element."
elif index < 0 or index > self.get_length()-1:
print "Wrong index number to get any element."
else:
p = self.head
for _ in xrange(index):
p = p.next
return p
def set_elem(self, index, value):
if self.is_empty():
print "The linked list is empty. Can not set element."
elif index < 0 or index > self.get_length()-1:
print "Wrong index number to set element."
else:
p = self.head
for _ in xrange(index):
p = p.next
p.value = value
def get_index(self, value):
p = self.head
for i in xrange(self.get_length()):
if p.value == value:
return i
else:
p = p.next
return -1
然后给出约瑟夫算法：
def josephus_kill_1(head, m):
'''
环形单链表，使用 RingLinkedList 数据结构，约瑟夫问题。
:param head:给定一个环形单链表的头结点，和第m个节点被杀死
:return:返回最终剩下的那个结点
本方法比较笨拙，就是按照规定的路子进行寻找，时间复杂度为o(m*len(ringlinkedlist))
'''
if head == 0:
print "This is an empty ring linked list."
return head
if m < 2:
print "Wrong m number to play this game."
return head
p = head
while p.next != p:
for _ in xrange(0, m-1):
post = p
p = p.next
#print post.next.value
post.next = post.next.next
p = post.next
return p
分析：
我采用了最原始的方法来解决这个问题，时间复杂度为o(m*len(ringlinkedlist))。
但是实际上，如果确定了链表的长度以及要删除的步长，那么最终剩余的结点一定是固定的，所以这就是一个固定的函数，我们只需要根剧M和N确定索引就可以了，这个函数涉及到了数论，具体我就不细写了。
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