用 Python 理解链表

Question

我正在使用 Python 对链表进行自学。我正在努力尝试将链表的结构和概念可视化。下面是来自自学的代码，要求我添加缺失的代码。有人可以画出或解释我应该如何想象这个。我熟悉常规的 Python 列表、字典和其他常见的数据结构。但例如对于方法，我的思考过程是

if current:
    return current.value
else:
    return None

但这是不正确的。我是否在检查构造函数是否初始化了一个列表并有一个元素变量 current？以下是完整代码。谢谢。

"""The LinkedList code from before is provided below.
Add three functions to the LinkedList.
"get_position" returns the element at a certain position.
The "insert" function will add an element to a particular
spot in the list.
"delete" will delete the first element with that
particular value.
Then, use "Test Run" and "Submit" to run the test cases
at the bottom."""

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

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

    def append(self, new_element):
        current = self.head
        if self.head:
            while current.next:
                current = current.next
            current.next = new_element
        else:
            self.head = new_element

    def get_position(self, position):
        """Get an element from a particular position.
        Assume the first position is "1".
        Return "None" if position is not in the list."""
        return None

    def insert(self, new_element, position):
        """Insert a new node at the given position.
        Assume the first position is "1".
        Inserting at position 3 means between
        the 2nd and 3rd elements."""
        pass


    def delete(self, value):
        """Delete the first node with a given value."""
        pass

# Test cases
# Set up some Elements
e1 = Element(1)
e2 = Element(2)
e3 = Element(3)
e4 = Element(4)

# Start setting up a LinkedList
ll = LinkedList(e1)
ll.append(e2)
ll.append(e3)

# Test get_position
# Should print 3
print ll.head.next.next.value
# Should also print 3
print ll.get_position(3).value

# Test insert
ll.insert(e4,3)
# Should print 4 now
print ll.get_position(3).value

# Test delete
ll.delete(1)
# Should print 2 now
print ll.get_position(1).value
# Should print 4 now
print ll.get_position(2).value
# Should print 3 now
print ll.get_position(3).value

Answer 1

对于方法，我的思考过程是……

什么方法？ get_position? insert? delete?

正如@JacobIRR 所建议的，添加一种打印链接列表的方式可能会有所帮助。看看：

class Element:

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


class LinkedList:

    def __init__(self):
        self.head = None

    def append(self, value):

        element = Element(value)

        if self.head is None:
            self.head = element
            return

        cursor = self.head
        while cursor.next is not None:
            cursor = cursor.next
        cursor.next = element

    def __str__(self):

        values = []

        cursor = self.head
        while cursor is not None:
            values.append(cursor.value)
            cursor = cursor.next
        return " -> ".join(values)


def main():

    linked_list = LinkedList()

    linked_list.append("Foo")
    linked_list.append("Bar")
    linked_list.append("Fizz")
    linked_list.append("Buzz")

    print(linked_list)

    return 0


if __name__ == "__main__":
    import sys
    sys.exit(main())

输出：

Foo -> Bar -> Fizz -> Buzz

Answer 2

所有你需要做的：

首先，尝试可视化将状态更改为该状态时会发生什么，或者将整个可视化写在纸上甚至任何在线软件中以了解更改。

最后/最后，请确保您了解链表的核心概念，并使用它进行一些棘手的、一堆不同的操作。或者，您可以在 Google 上搜索一些资源。

好吧，这是我为您的问题所做的解决方案：

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

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

    def append(self, new_element):
        current = self.head
        if self.head:
            while current.next:
                current = current.next
            current.next = new_element
        else:
            self.head = new_element

    def get_position(self, position):
        counter = 1
        current = self.head
        if position < 1:
            return None
        while current and counter <= position:
            if counter == position:
                return current
            current = current.next
            counter += 1
        return None

    def insert(self, new_element, position):
        counter = 1
        current = self.head
        if position > 1:
            while current and counter < position:
                if counter == position - 1:
                    new_element.next = current.next
                    current.next = new_element
                current = current.next
                counter += 1
        elif position == 1:
            new_element.next = self.head
            self.head = new_element

    def delete(self, value):
        current = self.head
        previous = None
        while current.value != value and current.next:
            previous = current
            current = current.next
        if current.value == value:
            if previous:
                previous.next = current.next
            else:
                self.head = current.next



# Test cases
# Set up some Elements
e1 = Element(1)
e2 = Element(2)
e3 = Element(3)
e4 = Element(4)

# Start setting up a LinkedList
ll = LinkedList(e1)
ll.append(e2)
ll.append(e3)

# Test get_position
# Should print 3
print(ll.head.next.next.value)
# Should also print 3
print(ll.get_position(3).value)

# Test insert
ll.insert(e4,3)
# Should print 4 now
print(ll.get_position(3).value)

# Test delete
ll.delete(1)
# Should print 2 now
print(ll.get_position(1).value)
# Should print 4 now
print(ll.get_position(2).value)
# Should print 3 now
print(ll.get_position(3).value)

再次，任何进一步的问题；拿一张纸，编写代码并可视化正在发生的事情。