python环形单链表的约瑟夫问题详解-创新互联
题目:
创新互联是一家成都网站设计、网站制作,提供网页设计,网站设计,网站制作,建网站,定制设计,网站开发公司,自2013年创立以来是互联行业建设者,服务者。以提升客户品牌价值为核心业务,全程参与项目的网站策划设计制作,前端开发,后台程序制作以及后期项目运营并提出专业建议和思路。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
分享文章:python环形单链表的约瑟夫问题详解-创新互联
本文地址:http://scyanting.com/article/icjsj.html