题解 | #输出单向链表中倒数第k个结点#

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

class LNode:
    def __init__(self, head):
        self.head = Node(head)
        
    def link(self, vl):
        cur = self.head
        while cur.next:
            cur = cur.next
        for v in vl:
            cur.next = Node(v)
            cur = cur.next
    def findk(self, k):
        cur = self.head
        pre = self.head
        k1 = 0
        k2 = 0
        while cur:
            cur = cur.next
            k1+=1
            if k1 - k2 == k+1:
                k2+=1
                pre = pre.next
        return pre.val
            
while True:
    try:
        n= int(input())
        a = list(map(int, input().split()))
        k = int(input())
        ln = LNode(a[0])
        ln.link(a[1:n])
        v = ln.findk(k)
        print(str(v))
    except:
        break