用两个指针，指针ahead先出发k-1步，当ahead到达链表尾节点的时候，behind指针刚好到达倒数第k个节点
但是要注意以下情况：

  • 输入的head为空指针
  • 输入的链表长度小于k
  • k = 0
    class Solution:
 def FindKthToTail(self, head, k):
     # write code here          
     if not head or k == 0:
         return None #注意这里要返回None，而不是False，才能通过
     ahead = head
     behind = head 
     for i in range(k-1):
         if not ahead.next:
             return None 
         else:
             ahead = ahead.next
     while ahead.next:
         ahead = ahead.next
         behind = behind.next
     return behind