在Python中迭代链接列表的问题

Posted 2021-05-02

我试图在Python 3.7中从头开始实现链接列表，经过多次尝试，我似乎无法使用print_values（）方法按预期的顺序打印所有节点值（顺序）。此时我不确定问题是使用append（）方法还是print_values（）方法。

    class Node:
        def __init__(self, node_value):
            self.node_value = node_value
            self.nextNode = None


    class SinglyLinkedList:
        # methods that should be available: get_size, insert_at, append, remove, 
        # update_node_value
        def __init__(self):
            self.head_node = None
            self.tail_node = None
            self.size = 0

        def get_list_size(self):
            """This returns the value for the size variable which get incremented 
               every time a new node is added.
               This implementation is better because it has a running time of O(1) 
               as opposed to iterating through the
               whole list which has a running time of O(n)"""
            return self.size

        def append(self, value):
            new_node = Node(value)
            if self.head_node is None:
                self.head_node = new_node
                self.size += 1
            else:
                while self.head_node.nextNode is not None:
                    self.head_node = self.head_node.nextNode
                self.head_node.nextNode = new_node
                self.size += 1

        def print_values(self):
            current_node = self.head_node
            list_values = []
            while current_node.nextNode is not None:
                list_values.append(current_node.node_value)
                if current_node.nextNode.nextNode is None:
                    list_values.append(current_node.nextNode.node_value)
                current_node = current_node.nextNode
            if list_values is not None:
                print("Linked list: " + str(list_values))
            else:
                print("Linked List is currently empty.")


# Helper code below.
new_ll = SinglyLinkedList()
new_ll.append("alpha")
print(new_ll.get_list_size())
new_ll.append("beta")
print(new_ll.get_list_size())
new_ll.append("gamma")
print(new_ll.get_list_size())
new_ll.append("delta")
print(new_ll.get_list_size())
new_ll.append("epsilon")
print(new_ll.get_list_size())
new_ll.append("zeta")
print(new_ll.get_list_size())
new_ll.print_values()

我输出的所有内容都是这样的：

1
2
3
4
5
6
Linked list: ['epsilon', 'zeta']

答案

通常，单链表仅跟踪头部。 （也不是尾巴）。因此通常不使用self.tail_node = None。

当使用linkedlist或tree时，它将使您的生活更容易使用递归而不是使用循环。如果您只想查看列表，循环工作正常，但如果您想更改它，那么我建议使用递归解决方案。

据说问题不在于你的print与你的append。

你永远不能移动头节点。你必须总是做一个指针，所以这导致了问题： self.head_node = self.head_node.nextNode

固定：

def append(self, value):
    new_node = Node(value)
    if self.head_node is None:
        self.head_node = new_node
        self.size += 1
    else:
        temp_head = self.head_node
        while temp_head.nextNode is not None:
            temp_head = temp_head.nextNode
        temp_head.nextNode = new_node
        self.size += 1

递归解决方案：

def append(self, value):
    new_node = Node(value)
    self.size += 1
    self.head_node = self.__recursive_append(self.head_node, new_node)
def __recursive_append(self, node, new_node):
    if not node:
        node = new_node
    elif not node.nextNode:
        node.nextNode = new_node
    else:
        node.nextNode = self.__recursive_append(node.nextNode, new_node)
    return node

话虽如此，直到我重新打印你的打印之后我才意识到这一点，所以这里是一个更干净的打印方法，使用python生成器可以帮助你。

生成器是你可以使用python的东西，你通常不能与其他编程语言一起使用，它使得链接列表变成一个非常容易做到的值列表：

def print_values(self, reverse=False):
    values = [val for val in self.__list_generator()]
    if values:
        print("Linked list: " + str(values))
    else:
        print("Linked List is currently empty.")

def __list_generator(self):
    '''
    A Generator remembers its state.
    When `yield` is hit it will return like a `return` statement would
    however the next time the method is called it will
    start at the yield statment instead of starting at the beginning
    of the method.
    '''
    cur_node = self.head_node
    while cur_node != None:
        yield cur_node.node_value # return the current node value

        # Next time this method is called
        # Go to the next node
        cur_node = cur_node.nextNode

免责声明：生成器很好，但我只是这样做，以匹配你的方式（即从链表中获取一个列表）。如果列表不重要但你只想输出链表中的每个元素，那么我就是这样做的：

def print_values(self, reverse=False):
    cur_node = self.head_node
    if cur_node:
        print('Linked list: ', end='')
        while cur_node:
            print("'{}' ".format(cur_node.node_value), end='')
            cur_node = cur_node.nextNode
    else:
        print("Linked List is currently empty.")

另一答案

我同意Error - Syntactical Remorse的答案，因为问题是附加的，而while循环的主体......这里是伪代码中的例子：

append 0:
  head = alpha

append 1:
  //skip the while loop
  head = alpha
  next = beta

append 2:
  //one time through the while loop because next = beta
  head = beta
  //beta beta just got assigned to head, and beta doesn't have next yet...
  next = gamma

append 3:
  //head is beta, next is gamma...the linked list can only store 2 nodes
  head = gamma //head gets next from itself
  //then has no next
  next = delta

...etc.