interactive-coding-challenges/linked_lists/palindrome/palindrome.ipynb
2015-06-30 05:55:58 -04:00

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"<small><i>This notebook was prepared by [Donne Martin](http://donnemartin.com). Source and license info is on [GitHub](https://bit.ly/code-notes).</i></small>"
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"## Problem: Determine if a linked list is a palindrome.\n",
"\n",
"* [Constraints](#Constraints)\n",
"* [Test Cases](#Test-Cases)\n",
"* [Algorithm](#Algorithm)\n",
"* [Code](#Code)\n",
"* [Unit Test](#Unit-Test)"
]
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"source": [
"## Constraints\n",
"\n",
"*Problem statements are often intentionally ambiguous. Identifying constraints and stating assumptions can help to ensure you code the intended solution.*\n",
"\n",
"* Is a single character or number a palindrome?\n",
" * No\n",
"* Can we assume we already have a linked list class that can be used for this problem?\n",
" * Yes"
]
},
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"cell_type": "markdown",
"metadata": {},
"source": [
"## Test Cases\n",
"\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"
]
},
{
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"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."
]
},
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"source": [
"## Code"
]
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"%run linked_list.py"
]
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"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",
" \n",
" # Reverse the linked list\n",
" while curr is not None:\n",
" reversed_list.insert_to_front(curr.data)\n",
" length += 1\n",
" curr = curr.next\n",
" \n",
" # Compare the reversed list with the original list\n",
" # Only need to compare the first half \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": "markdown",
"metadata": {},
"source": [
"## Unit Test"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"*It is important to identify and run through general and edge cases from the [Test Cases](#Test-Cases) section by hand. You generally will not be asked to write a unit test like what is shown below.*"
]
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"text": [
"Test: Empty list\n",
"Test: Single element list\n",
"Test: Two element list, not a palindrome\n",
"Test: Three element list, not a palindrome\n",
"Test: General case: Palindrome with even length\n",
"Test: General case: Palindrome with odd length\n",
"Success: test_is_palindrome\n"
]
}
],
"source": [
"from nose.tools import assert_equal\n",
"\n",
"class Test(object):\n",
" def test_is_palindrome(self):\n",
" print('Test: Empty list')\n",
" linked_list = MyLinkedList()\n",
" assert_equal(linked_list.is_palindrome(), False)\n",
"\n",
" print('Test: Single element list')\n",
" head = Node(1)\n",
" linked_list = MyLinkedList(head)\n",
" assert_equal(linked_list.is_palindrome(), False)\n",
"\n",
" print('Test: Two element list, not a palindrome')\n",
" linked_list.append(2)\n",
" assert_equal(linked_list.is_palindrome(), False)\n",
"\n",
" print('Test: Three element list, not a palindrome')\n",
" linked_list.append(3)\n",
" assert_equal(linked_list.is_palindrome(), False)\n",
"\n",
" print('Test: 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",
" assert_equal(linked_list.is_palindrome(), True)\n",
"\n",
" print('Test: 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",
" assert_equal(linked_list.is_palindrome(), True)\n",
" \n",
" print('Success: test_is_palindrome')\n",
"\n",
"if __name__ == '__main__':\n",
" test = Test()\n",
" test.test_is_palindrome()"
]
}
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