题解 | #输出单向链表中倒数第k个结点#

class Node:  # 节点
    def __init__(self, v):
        self.v = v
        self.next = None


class SingleLinkList:  # 链表
    def __init__(self):  # 初始化空
        self.head = None

    def append(self, i):
        n = Node(i)
        # 空
        if not self.head:
            self.head = n
            return
        cur = self.head
        while cur.next:
            cur = cur.next
        cur.next = n
        n.next = None

    def pop(self):
        # 1. 只有1个节点
        cur = self.head
        if cur.next == None:
            r= self.head.v
            self.head = self.head.next
            return r
        # 2. 非头部
        while 1:
            cur_next = cur.next
            if cur_next.next:
                cur = cur.next
            else:
                r  = cur.next.v
                cur.next = None
                break
        return r



if __name__ == '__main__':
    while True:
        try:
            # input
            n = int(input())
            lst = list(map(int, input().split()))
            k = int(input())
            # in SingleLinkList
            L = SingleLinkList()
            # get answer
            for i in lst:
                L.append(i)
            lst01 = []
            for j in range(k):
                lst01.append(L.pop())
            print(lst01[-1])
        except:
            break