python实现单链表中删除倒数第K个节点的方法

class LinkedListAlgorithms(object):
 def __init__(self):
  pass

 def rm_last_kth_node(self, k, linked_list): # 删除倒数第 K 个节点，针对单链表的
  if linked_list.is_empty():
   print 'The given linked_list is empty.'
  if k < 1 or k > linked_list.get_length():
   print 'Wrong kth number out of index.'
  k = linked_list.get_length() - k
  if k == 0:
   p = linked_list.head
   linked_list.head = p.next
  else:
   p = linked_list.head
   for i in xrange(k-1):
    p = p.next
   p.next = p.next.next