题解 | 输出单向链表中倒数第k个结点

class ListNode():
    def __init__(self,val=0):
        self.val = val 
        self.next = None

def fun(num_list:list):
    head = ListNode() #初始化一个虚拟头结点 val=0 next=None
    current = head
    for val in num_list:
        current.next = ListNode(val)
        current = current.next
    
    fast = head.next
    slow = head.next

    for _ in range(k):  #fast指针先往前遍历k步
        if not fast:
            print(None) #判断fast没有超出链表边界，如果超出边界，会返回head.next为None
            break
        fast = fast.next
    else:
        while fast:
            slow = slow.next
            fast = fast.next
        print(slow.val)       

while True:
    try:
        n = int(input())
        num_list = list(map(int,input().split()))
        k = int(input())
        fun(num_list)
    except Exception as error:
        break