Linked Lists

Agenda

1. The LinkedList and Node classes
2. Implementing append
3. Implementing deletion
4. Bidirectional links
5. Run-time analysis
6. Closing remarks

1. The LinkedList and Node classes

class LinkedList:
    class Node:
        def __init__(self, val, next=None):
            self.val = val
            self.next = next
    
    def __init__(self):
        self.head = None
        self.count = 0
        
    def prepend(self, value):
        self.head = LinkedList.Node(value, self.head)
        self.count += 1
        
    def __len__(self):
        return self.count
        
    def __iter__(self):
        n = self.head
        while n:
            yield n.val
            n = n.next
    
    def __repr__(self):
        return '[' + ', '.join(str(x) for x in self) + ']'

lst = LinkedList()
for i in range(10):
    lst.prepend(i)
lst

[9, 8, 7, 6, 5, 4, 3, 2, 1, 0]

2. Implementing append

Option 1

class LinkedList (LinkedList):
    def append(self, value):
        if len(self) == 0:
            self.prepend(value)
        else:
            n = self.head
            while n.next:
                n = n.next
            n.next = LinkedList.Node(value)
            self.count += 1

lst = LinkedList()
for i in range(10):
    lst.append(i)
lst

[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

Option 2

class LinkedList (LinkedList):
    def __init__(self):
        self.head = self.tail = None
        self.count = 0
        
    def prepend(self, value):
        self.head = LinkedList.Node(value, self.head)
        if not self.tail:
            self.tail = self.head
        self.count += 1

    def append(self, value):
        if len(self) == 0:
            self.prepend(value)
        else:
            self.tail.next = LinkedList.Node(value)
            self.tail = self.tail.next
            self.count += 1

lst = LinkedList()
for i in range(10):
    lst.append(i)
lst

[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

3. Implementing deletion

Deleting the head

class LinkedList (LinkedList):
    def del_head(self):
        assert(len(self) > 0)
        if self.head:
            self.head = self.head.next
        self.count -= 1
        if len(self) == 0:
            self.tail = None

lst = LinkedList()
for i in range(10):
    lst.append(i)
lst.del_head()
lst.del_head()
lst

[2, 3, 4, 5, 6, 7, 8, 9]

Deleting the tail

class LinkedList (LinkedList):
    def del_tail(self):
        assert(len(self) > 0)
        # YUCK!!!
        if len(self) == 1:
            self.head = self.tail = None
        else:
            n = self.head
            while n.next is not self.tail:
                n = n.next
            n.next = None
            self.tail = n
        self.count -= 1

lst = LinkedList()
for i in range(10):
    lst.append(i)
lst.del_tail()
lst.del_tail()
lst

[0, 1, 2, 3, 4, 5, 6, 7]

4. Bidirectional links (Doubly-linked list)

class LinkedList:
    class Node:
        def __init__(self, val, prior=None, next=None):
            self.val = val
            self.prior = prior
            self.next  = next
    
    def __init__(self):
        self.head = LinkedList.Node(None) # *sentinel* node!
        self.head.prior = self.head.next = self.head
        self.count = 0
        
    def prepend(self, value):
        n = LinkedList.Node(value, prior=self.head, next=self.head.next)
        # important: Python carries out multiple assignments from *left to right*!
        self.head.next.prior = self.head.next = n
        self.count += 1
        
    def append(self, value):
        n = LinkedList.Node(value, prior=self.head.prior, next=self.head)
        n.prior.next = n.next.prior = n
        self.count += 1
        
    def del_head(self):
        assert(len(self) > 0)
        to_del = self.head.next
        # note: following is location agnostic!
        to_del.prior.next = to_del.next
        to_del.next.prior = to_del.prior
        self.count -= 1
        
    def __len__(self):
        return self.count
        
    def __iter__(self):
        n = self.head.next
        while n is not self.head:
            yield n.val
            n = n.next
    
    def __repr__(self):
        return '[' + ', '.join(str(x) for x in self) + ']'

lst = LinkedList()
for i in range(10):
    lst.prepend(i)
for i in range(10):
    lst.append(i)
lst.del_head()
lst

[8, 7, 6, 5, 4, 3, 2, 1, 0, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

5. Run-time analysis

Run-time complexities for circular, doubly-linked list of $N$ elements:

• indexing (position-based access) = $O(N)$
• search (unsorted) = $O(N)$
• search (sorted) = $O(N)$ --- binary search isn't possible!
• prepend = $O(1)$
• append = $O(1)$
• insertion at arbitrary position: traversal = $O(N)$ + insertion = $O(1)$
• deletion of arbitrary element: traversal = $O(N)$ + deletion = $O(1)$

6. Closing remarks