难度：简单
一、题目描述：
二、解题分析：
1、剑指解析
2、测试用例
3、代码实现
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def getKthFromEnd(self, head: ListNode, k: int) -> ListNode:
result = head
temp = he
