Discussion 8: Linked Lists

Linked Lists

A linked list is a Link object or Link.empty.

You can mutate a Link object s in two ways:

  • Change the first element with s.first = ...
  • Change the rest of the elements with s.rest = ...

You can make a new Link object by calling Link:

  • Link(4) makes a linked list of length 1 containing 4.
  • Link(4, s) makes a linked list that starts with 4 followed by the elements of linked list s.
class Link:
    """A linked list is either a Link object or Link.empty

    >>> s = Link(3, Link(4, Link(5)))
    >>> s.rest
    Link(4, Link(5))
    >>> s.rest.rest.rest is Link.empty
    True
    >>> s.rest.first * 2
    8
    >>> print(s)
    (3 4 5)
    """
    empty = ()

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

    def __repr__(self):
        if self.rest:
            rest_repr = ', ' + repr(self.rest)
        else:
            rest_repr = ''
        return 'Link(' + repr(self.first) + rest_repr + ')'

    def __str__(self):
        string = '('
        while self.rest is not Link.empty:
            string += str(self.first) + ' '
            self = self.rest
        return string + str(self.first) + ')'

Q1: Sum Two Ways

Implement both sum_rec and sum_iter. Each one takes a linked list of numbers s and a non-negative integer k and returns the sum of the first k elements of s. If there are fewer than k elements in s, all of them are summed. If k is 0 or s is empty, the sum is 0.

Use recursion to implement sum_rec. Don't use recursion to implement sum_iter; use a while loop instead.

To get started on the recursive implementation, consider the example a = Link(1, Link(6, Link(8))), and the call sum_rec(a, 2). Write down the recursive call to sum_rec that would help compute sum_rec(a, 2). Then, write down what that recursive call should return. Discuss how this return value is useful in computing the return value of sum_rec(a, 2).

Your Answer
Run in 61A Code
Solution 
def sum_rec(s, k):
    """Return the sum of the first k elements in s.

    >>> a = Link(1, Link(6, Link(8)))
    >>> sum_rec(a, 2)
    7
    >>> sum_rec(a, 5)
    15
    >>> sum_rec(Link.empty, 1)
    0
    """
    # Use a recursive call to sum_rec; don't call sum_iter
    if k == 0 or s is Link.empty:
        return 0
    return s.first + sum_rec(s.rest, k - 1)

def sum_iter(s, k):
    """Return the sum of the first k elements in s.

    >>> a = Link(1, Link(6, Link(8)))
    >>> sum_iter(a, 2)
    7
    >>> sum_iter(a, 5)
    15
    >>> sum_iter(Link.empty, 1)
    0
    """
    # Don't call sum_rec or sum_iter
    total = 0
    while k > 0 and s is not Link.empty:
        total, s, k = total + s.first, s.rest, k - 1
    return total
Add s.first to the sum of the first k-1 elements in s.rest. Your base case condition should include s is Link.empty so that you're checking whether s is empty before ever evaluating s.first or s.rest.
Introduce a new name, such as total, then repeatedly (in a while loop) add s.first to total, set s = s.rest to advance through the linked list, and reduce k by one.

Discussion time: When adding up numbers, the intermediate sums depend on the order. (1 + 3) + 5 and 1 + (3 + 5) both equal 9, but the first one makes 4 along the way while the second makes 8 along the way. For the same linked list s and length k, will sum_rec and sum_iter both make the same intermediate sums along the way?

Write a function duplicate_link that takes in a linked list s and a value. duplicate_link will mutate s such that if there is a linked list node that has a first equal to value, that node will be duplicated. Note that you should be mutating the original linked list s; you will need to create new Links, but you should not be returning a new linked list.

Note: In order to insert a link into a linked list, you need to modify the .rest of certain links. We encourage you to draw out a doctest to visualize!

Your Answer Run in 61A Code
Solution 
def duplicate_link(s: Link, val: int) -> None:
    """Mutates s so that each element equal to val is followed by another val.

    >>> x = Link(5, Link(4, Link(5)))
    >>> duplicate_link(x, 5)
    >>> x
    Link(5, Link(5, Link(4, Link(5, Link(5)))))
    >>> y = Link(2, Link(4, Link(6, Link(8))))
    >>> duplicate_link(y, 10)
    >>> y
    Link(2, Link(4, Link(6, Link(8))))
    >>> z = Link(1, Link(2, Link(2, Link(3))))
    >>> duplicate_link(z, 2) # ensures that back to back links with val are both duplicated
    >>> z
    Link(1, Link(2, Link(2, Link(2, Link(2, Link(3))))))
    """
    if s is Link.empty:
        return
    elif s.first == val:
        remaining = s.rest
        s.rest = Link(val, remaining)
        duplicate_link(remaining, val)
    else:
        duplicate_link(s.rest, val)