相当简单了,直接得到链表长度,然后长度小于k时,返回None,长度大于等于k时,链表移动到长度-k个node,然后返回就行。

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
        cur = pHead
        size=0
        while cur:
            size+=1
            cur=cur.next
        # print(size)
        if size<k:
            return None
        else:
            ind = size - k 
            cur=pHead
            for i in range(ind):
                cur = cur.next
            return cur

head=ListNode(1)
head.next = ListNode(2)
head.next.next = ListNode(3)
print(Solution().FindKthToTail(head,2).val)