题解 | #输出单向链表中倒数第k个结点#
输出单向链表中倒数第k个结点
https://www.nowcoder.com/practice/54404a78aec1435a81150f15f899417d
import sys class ListNode: def __init__(self,Key): self.Key = Key self.next=None class ListLink: def __init__(self): self.rear = None def add(self,Key): t = ListNode(Key) t.next = self.rear self.rear = t while True: try: a = int(input()) b = input().split() c = int(input()) y = ListLink() for i in range(len(b)): y.add(b[i]) d = y.rear if c == 1: print(d.Key) else: for i in range(c-1): d = d.next print(d.Key) if c == 0: print(0) except: break