将列表转换为链接列表 [英] Converting a list to a linked list

问题描述

我试图找出将列表转换为链接列表的方法. 我已经有一个用于链接的类，但是我试图找出如何将列表转换为链接列表的方法，例如:

I'm trying to figure out to convert a list to a linked list. I already have a class for the link but I'm trying to figure out how to convert a list to linked list, for example:

def list_to_link(lst):
    """Takes a Python list and returns a Link with the same elements.

    >>> link = list_to_link([1, 2, 3])
    >>> print_link(link)
    <1 2 3>
    """


class Link:

    empty = ()

    def __init__(self, first, rest=empty):
        assert rest is Link.empty or isinstance(rest, Link)
        self.first = first
        self.rest = rest

def print_link(link):
    """Print elements of a linked list link."""

    >>> link = Link(1, Link(2, Link(3)))
    >>> print_link(link)
    <1 2 3>
    >>> link1 = Link(1, Link(Link(2), Link(3)))
    >>> print_link(link1)
    <1 <2> 3>
    >>> link1 = Link(3, Link(Link(4), Link(5, Link(6))))
    >>> print_link(link1)
    <3 <4> 5 6>
    """
    print('<' +helper(link).rstrip() +'>')

推荐答案

Matt的回答很好，但是超出了上面问题中描述的函数原型的约束.

Matt's answer is good, but it's outside the constraint of the function prototype described in the problem above.

阅读抽象/原型，看起来像是问题的创建者想用递归/动态编程方法来解决这个问题.这是一个相当标准的递归算法介绍.与其说是在Python中创建链接列表(不是真正有用或通用)，不如说是更多地了解如何编写优雅的递归代码.

Reading the abstract/prototype, it looks like the creator of the problem wanted to solve this with recursive/dynamic programming methodology. This is a pretty standard recursive algorithm introduction. It's more about understanding how to write elegant recursive code more than creating linked-list in Python (not really useful or common).

这是我想出的一个解决方案.试试看:

Here's a solution I came up with. Try it out:

class Link:
    empty = ()

    def __init__(self, first, rest=empty):
        assert rest is Link.empty or isinstance(rest, Link)
        self.first = first
        self.rest = rest


def print_link(link):
    """Print elements of a linked list link.
    """
    print('<' + helper(link).rstrip() +'>')


def list_to_link(lst):
    """Takes a Python list and returns a Link with the same elements.
    """
    if len(lst) == 1:
        return Link(lst[0])
    return Link(lst[0], list_to_link(lst[1:]))  # <<<< RECURSIVE

def helper(link):
    if isinstance(link.first, Link):
        first = '<' + helper(link.first).rstrip() + '>'  # <<<< RECURSIVE
    else:
        first = str(link.first)

    if link.rest != Link.empty:
        return first + ' ' + helper(link.rest)  # <<<< RECURSIVE
    else:
        return first + ' '

def main():
    """ Below are taken from sample in function prototype comments
    """
    link = list_to_link([1, 2, 3])
    print_link(link)

    link = Link(1, Link(2, Link(3)))
    print_link(link)
    link1 = Link(1, Link(Link(2), Link(3)))
    print_link(link1)
    link1 = Link(3, Link(Link(4), Link(5, Link(6))))
    print_link(link1)


if __name__ == '__main__':
    main()

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