{ "cells": [ { "cell_type": "markdown", "metadata": {}, "source": [ "## Problem: Determine if a linked list is a palindrome.\n", "\n", "* [Clarifying Questions](#Clarifying-Questions)\n", "* [Test Cases](#Test-Cases)\n", "* [Algorithm](#Algorithm)\n", "* [Code](#Code)\n", "* [Pythonic-Code](#Pythonic-Code)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Clarifying Questions\n", "\n", "* Is a single character or number a palindrome?\n", " * No" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Test Cases\n", "\n", "* Empty list\n", "* Single element list\n", "* Two element list, not a palindrome\n", "* Three element list, not a palindrome\n", "* General case: Palindrome with even length\n", "* General case: Palindrome with odd length" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Algorithm\n", "\n", "* Reverse the linked list\n", "* Compare the reversed list with the original list\n", " * Only need to compare the first half\n", "\n", "Complexity:\n", "* Time: O(1)\n", "* Space: O(n)\n", "\n", "Note:\n", "* We could also do this iteratively, using a stack to effectively reverse the first half of the string." ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Code" ] }, { "cell_type": "code", "execution_count": null, "metadata": { "collapsed": true }, "outputs": [], "source": [ "%run linked_list.py" ] }, { "cell_type": "code", "execution_count": null, "metadata": { "collapsed": false }, "outputs": [], "source": [ "class MyLinkedList(LinkedList):\n", " def is_palindrome(self):\n", " if self.head is None or self.head.next is None:\n", " return False\n", " curr = self.head\n", " reversed_list = MyLinkedList()\n", " length = 0\n", " while curr is not None:\n", " reversed_list.insert_to_front(curr.data)\n", " length += 1\n", " curr = curr.next\n", " iterations_to_compare_half = length / 2\n", " curr = self.head\n", " curr_reversed = reversed_list.head\n", " for _ in xrange(0, iterations_to_compare_half):\n", " if curr.data != curr_reversed.data:\n", " return False\n", " curr = curr.next\n", " curr_reversed = curr_reversed.next\n", " return True" ] }, { "cell_type": "code", "execution_count": null, "metadata": { "collapsed": false }, "outputs": [], "source": [ "print('Empty list')\n", "linked_list = MyLinkedList()\n", "print(linked_list.is_palindrome())\n", "print('Single element list')\n", "head = Node(1)\n", "linked_list = MyLinkedList(head)\n", "print(linked_list.is_palindrome())\n", "print('Two element list, not a palindrome')\n", "linked_list.append(2)\n", "print(linked_list.is_palindrome())\n", "print('Three element list, not a palindrome')\n", "linked_list.append(3)\n", "print(linked_list.is_palindrome())\n", "print('General case: Palindrome with even length')\n", "head = Node('a')\n", "linked_list = MyLinkedList(head)\n", "linked_list.append('b')\n", "linked_list.append('b')\n", "linked_list.append('a')\n", "print(linked_list.is_palindrome())\n", "print('General case: Palindrome with odd length')\n", "head = Node(1)\n", "linked_list = MyLinkedList(head)\n", "linked_list.append(2)\n", "linked_list.append(3)\n", "linked_list.append(2)\n", "linked_list.append(1)\n", "print(linked_list.is_palindrome())" ] } ], "metadata": { "kernelspec": { "display_name": "Python 2", "language": "python", "name": "python2" }, "language_info": { "codemirror_mode": { "name": "ipython", "version": 2 }, "file_extension": ".py", "mimetype": "text/x-python", "name": "python", "nbconvert_exporter": "python", "pygments_lexer": "ipython2", "version": "2.7.9" } }, "nbformat": 4, "nbformat_minor": 0 }