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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