题解 | #输出单向链表中倒数第k个结点# 最舒服的构建链表的方法 没有之一(不掉书包)

输出单向链表中倒数第k个结点

http://www.nowcoder.com/practice/54404a78aec1435a81150f15f899417d

class Node:
    def __init__(self,value= None,next_= None):
        self.value = value
        self.next_ = next_
def init_node(li):
    head = Node()
    temp = head

    while li:
        temp.next_ = Node(li.pop(0))
        temp = temp.next_
    return head.next_

while 1: 
    try:
        n = int(input())
        string  = input().strip()
        k = int(input())
        li = string.split(' ')
        head = init_node(li)
        k = n - k
        con = 0
        while head:
            if con == k:
                print(head.value)
                break
            head = head.next_
            con += 1
    except Exception as ex:
        
        break
        









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