题解 | #输出单向链表中倒数第k个结点#

class Node:
    def __init__(self,val=None):
        self.val = val
        self.next = None

while True:
    try:
        l,s,k,head = int(input()),list(map(int,input().split())),int(input()),Node()
        while k:
            head.next = Node(s.pop())
            head = head.next
            k -= 1
        print(head.val)
    except:
        break

定义一个节点类，初始化val&next都为None。，之后将新节点的地址赋值给head.next,表示头节点指向新节点，更新头指针，利用循环while k:控制循环结束，当k=0时，输出头节点所在的值