# 有环单链表

class Node(object):
   def __init__(self, data, next=None):
      self.data = data
      self.next = next


class CircleLinklist(object):
   def __init__(self):
      self.head = None
      self.length = 0

   def insert(self, node):
      if self.length == 0:
         self.head = node
         self.length += 1
         return

      cur = self.head
      while cur.next:
         cur = cur.next

      cur.next = node
      self.length += 1

   def traverse(self):
      cur = self.head
      while cur is not None:
         print(cur.data)
         cur = cur.next

   def judge_circle_quick_slow_pointer(self):
      """
      判断单链表是否有环
      如果在某个时候,快指针追上了慢指针,那么便说明有环
      """
      # 快指针,每次走两个节点
      quick = self.head
      # 慢指针,每次走一个节点
      slow = self.head

      # 标志,判断是否有环
      flag = False
      # 每个指针的步数
      count = 0
      # 加上try-except，因为在没环的情况下，快指针会出错
      try:
         # 循环终止条件,如果某个节点的next为空,就说明没有环了
         while quick.next.next is not None and slow.next is not None:
            count += 1
            # 排除第一次相等的情况
            # 用后面那个条件做判断不是个好想法，万一环的起点就是head，然后节点个数又是偶数的话，不会停止了吧
            if quick.data == slow.data and count != 1:  # quick != self.head
               print('快指针追上慢指针时的节点的值: ',quick.data)
               flag = True
               break
            # 一次跨两个节点
            quick = quick.next.next
            # 一次跨一个节点
            slow = slow.next
      except Exception as e:
         print(e.args)
         print('quick条件报错，说明没有环')
      finally:
         print('步数: ', count)
         print('是否有环: ', flag)

   def judge_circle_by_traversing(self):
      """
      将每个遍历过的节点保存到集合中，如果下一次遍历的节点在集合中了，就说明有环
      这个方法很傻，但却可以得到环的点是哪个
      快慢指针只能判断是否有环，不能得到环的起点
      """
      nodes = []
      cur = self.head
      while cur is not None:
         if cur not in nodes:
            nodes.append(cur)
         else:
            print('有环，环的值是: ', cur.data)
            break
         cur = cur.next
      else:
         print('没有环')


if __name__ == '__main__':
   cl = CircleLinklist()
   nodes = []
   for i in 'ABCDEFGHIJKLMN':
      nodes.append(Node(i))
   nodes[len(nodes) - 1].next = nodes[5]  # 将最后一个节点的next指向F

   for node in nodes:
      cl.insert(node)

   # cl.traverse()

   cl.judge_circle_quick_slow_pointer()
   print('-' * 80)
   cl.judge_circle_by_traversing()