Python Linked Lists

python_linked_lists.py

import bisect
from heapq import *

"""
Linked Lists

A linked list is an ordered collection of values.
Linked lists are similar to arrays in the sense that they contain objects in a linear order.
However they differ from arrays in their memory layout.

Arrays are contiguous data structures and they’re composed of fixed-size data records stored in adjoining blocks of memory.
Linked lists, however, are made up of data records linked together by pointers.

https://dbader.org/blog/python-linked-list
"""
class SListNode():

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

    def __repr__(self):
        return repr(self.data)


class DListNode():
    """
    A node in a doubly-linked list.
    """
    def __init__(self, data=None, prev=None, next=None):
        self.data = data
        self.prev = prev
        self.next = next

    def __repr__(self):
        return repr(self.data)


class SinglyLinkedList:

    def __init__(self):
        """
        Create a new singly-linked list.
        Takes O(1) time.
        """
        self.head = None

    def __repr__(self):
        """
        Return a string representation of the list.
        Takes O(n) time.
        """
        nodes = []
        curr = self.head
        while curr:
            nodes.append(repr(curr))
            curr = curr.next
        return '[' + ', '.join(nodes) + ']'

    def prepend(self, data):
        """
        Insert a new element at the beginning of the list.
        Takes O(1) time.
        """
        self.head = SListNode(data=data, next=self.head)

    def append(self, data):
        """
        Insert a new element at the end of the list.
        Takes O(n) time.

        https://www.hackerrank.com/challenges/30-linked-list/problem
        """
        if not self.head:
            self.head = SListNode(data=data)
            return
        curr = self.head
        while curr.next:
            curr = curr.next
        curr.next = SListNode(data=data)

    def insert(self, data):
        """
        Insert a new element at the end of the list.
        Takes O(n) time.
        """
        if not self.head:
            self.head = SListNode(data=data)
            return
        curr = self.head
        while curr.next:
            curr = curr.next
        curr.next = SListNode(data=data)

    def find(self, key):
        """
        Search for the first element with `data` matching
        `key`. Return the element or `None` if not found.
        Takes O(n) time.
        """
        curr = self.head
        while curr and curr.data != key:
            curr = curr.next
        return curr  # Will be None if not found

    def remove(self, key):
        """
        Remove the first occurrence of `key` in the list.
        Takes O(n) time.
        """
        # Find the element and keep a
        # reference to the element preceding it
        curr = self.head
        prev = None
        while curr and curr.data != key:
            prev = curr
            curr = curr.next
        # Unlink it from the list
        if prev is None:
            self.head = curr.next
        elif curr:
            prev.next = curr.next
            curr.next = None

    def reverse(self):
        """
        Reverse the list in-place.
        Takes O(n) time.
        """
        curr = self.head
        prev_node = None
        next_node = None
        while curr:
            next_node = curr.next
            curr.next = prev_node
            prev_node = curr
            curr = next_node
        self.head = prev_node

    def remove_duplicates(self):
        """
        Your remove_duplicates function should return the head of the updated linked list. after removing duplicates

        https://www.hackerrank.com/challenges/30-linked-list-deletion/problem
        """
        if not self.head:
            return None
        n = self.head
        while n.next:
            if n.data == n.next.data:
                n.next = n.next.next
            else:
                n = n.next
        return self.head


class DoublyLinkedList():

    def __init__(self):
        """
        Create a new doubly linked list.
        Takes O(1) time.
        """
        self.head = None

    def __repr__(self):
        """
        Return a string representation of the list.
        Takes O(n) time.
        """
        nodes = []
        curr = self.head
        while curr:
            nodes.append(repr(curr))
            curr = curr.next
        return '[' + ', '.join(nodes) + ']'

    def prepend(self, data):
        """
        Insert a new element at the beginning of the list.
        Takes O(1) time.
        """
        new_head = DListNode(data=data, next=self.head)
        if self.head:
            self.head.prev = new_head
        self.head = new_head

    def append(self, data):
        """
        Insert a new element at the end of the list.
        Takes O(n) time.
        """
        if not self.head:
            self.head = DListNode(data=data)
            return
        curr = self.head
        while curr.next:
            curr = curr.next
        curr.next = DListNode(data=data, prev=curr)

    def find(self, key):
        """
        Search for the first element with `data` matching
        `key`. Return the element or `None` if not found.
        Takes O(n) time.
        """
        curr = self.head
        while curr and curr.data != key:
            curr = curr.next
        return curr  # Will be None if not found

    def remove_elem(self, node):
        """
        Unlink an element from the list.
        Takes O(1) time.
        """
        if node.prev:
            node.prev.next = node.next
        if node.next:
            node.next.prev = node.prev
        if node is self.head:
            self.head = node.next
        node.prev = None
        node.next = None

    def remove(self, key):
        """
        Remove the first occurrence of `key` in the list.
        Takes O(n) time.
        """
        elem = self.find(key)
        if not elem:
            return
        self.remove_elem(elem)

    def reverse(self):
        """
        Reverse the list in-place.
        Takes O(n) time.
        """
        curr = self.head
        prev_node = None
        while curr:
            prev_node = curr.prev
            curr.prev = curr.next
            curr.next = prev_node
            curr = curr.prev
        self.head = prev_node.prev