1. 方法一：利用数组

# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
# 
# @param pHead ListNode类 
# @param k int整型 
# @return ListNode类
# 空间复杂度 n
class Solution:
    def FindKthToTail(self , pHead: ListNode, k: int) -> ListNode:
        # write code here
        if k<=0:
            return None
        a = []
        cur = pHead
        count = 0
        while cur :
            count+=1
            a.append(cur)
            cur = cur.next
        if count>=k:
            return a[count-k]
        else:
            return None

2. 方法二：双指针 先后指针

倒数第k个节点，就先让先指针先走K步，最后停下的时候，输出后指针指向的节点就行

# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
# 
# @param pHead ListNode类 
# @param k int整型 
# @return ListNode类
#
class Solution:
    def FindKthToTail(self , pHead: ListNode, k: int) -> ListNode:
        # write code here
        fast = slow = pHead
        count = k
		# 快指针先行k步
        while fast and count :
            count-=1
            fast = fast.next
            if count ==0:
                break
		# 注意是否走了 k 步 没走到就说明链子太短了
        if count > 0:
            return None
		# 快慢指针同步，快指针先到底，慢指针指向倒数第k个
        while fast:
            fast = fast.next
            slow = slow.next
        return slow