在python中反轉鏈接列表時無法訪問鏈接列表的下一個節點

Question

我對python有點陌生，我已經看到了解決反向鏈接列表問題的正確解決方案，但是我想知道為什么我的解決方案不起作用。 特別地，由於“ new_list.head.next = prev”這一行，反向函數停留在以下代碼的while循環內

class Node:
    def __init__(self, value):
        self.value = value
        self.next = None

class LinkedList:
    def __init__(self):
        self.head = None

    def append(self, value):
        if self.head is None:
            self.head = Node(value)
            return

        node = self.head
        while node.next:
            node = node.next

        node.next = Node(value)

    def __iter__(self):
        node = self.head
        while node:
            yield node.value
            node = node.next

    def __repr__(self):
        return str([v for v in self])

def reverse(linked_list):
    new_list = LinkedList()
    if linked_list is None:
        return new_list
    node = linked_list.head
    new_list.head = node 
    while node.next:
        prev = node
        node = node.next
        new_list.head = node
        new_list.head.next = prev
    return new_list   

if __name__ == "__main__":
    a = LinkedList()
    b = [1,2,3,4,5]
    for item in b:
        a.append(item)
    print a
    c = reverse(a) 
    print c

Answer 1

如果您使用Python3標記問題，請確保它在python 3中運行。

原因是因為您混淆了點並創建了無限循環。 打印該value ，它可以幫助您發現錯誤。 我將使用這些值指出問題。

    while node.next:
        # First node that comes in value = 1
        print(node.value) #
        prev = node # prev = 1
        node = node.next # node = 2
        new_list.head = node # You are setting the head = 2 (i.e. head = node.next)
        new_list.head.next = prev # You are setting next of head = 1 (i.e. head.next = node.next.next)
        # however this also set node.next.next = 1
        # So going into the next loop: node.value = 2 and node.next.value = 1

由於指針混亂，您將永遠在第一個節點和第二個節點之間循環。

Answer 2

這是您的reverse情況：

def reverse(linked_list):
    new_list = LinkedList()
    if linked_list is None:
        return new_list
    node = linked_list.head
    new_list.head = Node(node.value)
    while node.next:
        node = node.next
        prev_head = new_list.head
        new_list.head = Node(node.value)
        new_list.head.next = prev_head
    return new_list

有了它，我得到了所需的print c輸出print c ： [5, 4, 3, 2, 1]

一般建議：創建新節點，而不是分配給初始列表中的節點。

Answer 3

如果您想到兩個引用，那么對此（至少對我來說）要容易一點：

•您未曾看到的原始清單的其余部分之一
•新清單的首位

在每次迭代時，將剩余部分向上移動，並將舊的剩余部分設置為新列表的開頭。 順序在這里很重要-如您所見，如果不小心，很容易意外地在指向同一節點的兩個不同變量上進行下一步更改：

def reverse(linked_list):
    new_list = LinkedList()

    if linked_list is None:
        return new_list

    remaining = linked_list.head

    while remaining:
        prev_head = new_list.head       # the old head becomes the first link
        new_list.head = remaining       # new head becomese the first remaining
        remaining = remaining.next      # move remaing one up the chain
        new_list.head.next = prev_head  # point the new head to the previous   
    return new_list