Converting a list to a linked list

问题

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() +'>')

回答1:

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

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).

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()

回答2:

This is what you want.

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

class LinkedList(object):
    def __init__(self, sequence):
        self.head = Node(sequence[0])
        current = self.head
        for item in sequence[1:]:
            current.reference = Node(item)
            current = current.reference
a = range(10)
li = LinkedList(li)
current = li.head
while current is not None:
    print current.value
    current = current.reference

回答3:

I have an idea using dummy ListNode. This makes code simple and neat.

class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None


def lst2link(lst):
    cur = dummy = ListNode(0)
    for e in lst:
        cur.next = ListNode(e)
        cur = cur.next
    return dummy.next

来源：https://stackoverflow.com/questions/31553576/converting-a-list-to-a-linked-list