题解 | #输出单向链表中倒数第k个结点#

构造链表

• 定义一个链表结构
• 尾部添加链表
• for循环链接，题目要求是倒数第k个，那我们正向for循环，取 链表长度 - k 的那个节点值，就是我们要的答案
• 题目要求需要多次输入，所以要用while去保证可以多次输入样例

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


def append_link(head: ListNode, val: int):
    cur = head
    while cur.next is not None:
        cur = cur.next

    cur.next = ListNode(val)

    return head


def get_nth_from_the_end_in_link(head: ListNode, nth: int):
    link_length = 0
    cur = head
    while cur is not None:
        cur = cur.next
        link_length += 1

    count = link_length - nth
    res = head
    while count > 0 and res:
        count -= 1
        res = res.next
    
    if res:
        print(res.val)


while True:
    try:
        scan = int(input())
        numbers = list(map(int, input().split()))
        k = int(input())
        head = ListNode(numbers[0])
        for n in numbers[1:]:
            append_link(head, n)

        get_nth_from_the_end_in_link(head, k)
    except:
        break