Question

基本上，我要实现一个LinkedList类，然后实现各种方法来使用它。下面是LinkedList类（及其相关的Node类）的代码

# A simple linked list implementation

class Node:
# attributes:
#   data (can be anything)
#   next (another Node)

    def __init__(self, data):
        self.data = data
        self.next = None


class LinkedList:
# attributes:
#   head (a Node)
# ****************
# methods:
#   insert
#   find
#   delete

    def __init__(self):
        self.head = None

    def __str__(self):
        output = []
        current_node = self.head
        while current_node:
            output.append(str(current_node.data))
            current_node = current_node.next
        return(", ".join(output))


    # Creates a Node with the given data and inserts into the front of the list.
    def insert(self, data):
        new_node = Node(data)
        new_node.next = self.head
        self.head = new_node

    # Finds the first node with the given data. Returns None if there is no such node.
    def find(self, data):
        current_node = self.head
        while(current_node):
            if (current_node.data == data):
                return current_node
            current_node = current_node.next
        return None # We reached the end of the list without finding anything.

    # Deletes given node. Can be used in conjunction with find.
    def delete(self, deleted_node):
        if deleted_node == self.head:
            self.head = self.head.next
            return
        current_node = self.head
        while(current_node.next != deleted_node):
            current_node = current_node.next
        current_node.next = deleted_node.next

然后，我正在尝试实现一个rotate（my_list，N）函数，该函数将把my_list旋转N。以下是我的代码：

import linkedlists as ll
from ErrorHandler import sanitize
import random, math, time, copy

def length(my_list):
    sanitize(my_list, ll.LinkedList)
    if my_list.head == None: return 0
    count = 1 #First item! Ah, ah, ah
    current_node = my_list.head
    while current_node.next != None:
        current_node = current_node.next
        count += 1 #One more item! Ah, ah, ah
    return count

def get_Nth(my_list, N):
    sanitize(my_list, ll.LinkedList)
    if my_list.head == None: return None
    current_node = my_list.head
    count = 0
    while current_node.next != None:
        if count == N:
            return current_node
        count +=1
        current_node = current_node.next
    return current_node

def rotate(my_list, N):
    sanitize(my_list, ll.LinkedList)
    if my_list.head == None: return None
    N = N % length(my_list)
    lent = length(my_list)
    get_Nth(my_list, lent-1).next = my_list.head
    my_list.head = get_Nth(my_list, lent-N)
    get_Nth(my_list, lent-N-1).next = None

但是，在包含从0到9的数字升序的LinkedList上调用rotate（）会返回8,9,0,1,2,3,4,5。为什么？我很确定这与倒数第二行有关，因为在将get_Nth（my_list，lent-1）.next分配给my_list.head时，它仅指向my_list.head，而不是节点对象my_list.head当时指向。

我该如何解决？

Answer 1

您的错误就在这里：get_Nth(my_list, lent-N-1).next = None

我假设您打电话给rotate(my_list, 2)，因此此时列表看起来像这样[8, 9, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, ...]。因此，当您调用get_Nth(my_list, lent-N-1)时，lent-N-1为7，那么，索引7处的元素实际上为5。

您应该只使用lent-1。

如何通过Python中的变量直接引用对象

1 个答案: