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