Polish kth to last element challenge and solution

Update algorithm discussion and solution.
This commit is contained in:
Donne Martin 2016-06-12 23:27:36 -04:00
parent 801e583762
commit 90f7505018
2 changed files with 28 additions and 31 deletions

View File

@ -175,21 +175,21 @@
], ],
"metadata": { "metadata": {
"kernelspec": { "kernelspec": {
"display_name": "Python 2", "display_name": "Python 3",
"language": "python", "language": "python",
"name": "python2" "name": "python3"
}, },
"language_info": { "language_info": {
"codemirror_mode": { "codemirror_mode": {
"name": "ipython", "name": "ipython",
"version": 2 "version": 3
}, },
"file_extension": ".py", "file_extension": ".py",
"mimetype": "text/x-python", "mimetype": "text/x-python",
"name": "python", "name": "python",
"nbconvert_exporter": "python", "nbconvert_exporter": "python",
"pygments_lexer": "ipython2", "pygments_lexer": "ipython3",
"version": "2.7.10" "version": "3.5.0"
} }
}, },
"nbformat": 4, "nbformat": 4,

View File

@ -63,10 +63,10 @@
"source": [ "source": [
"## Algorithm\n", "## Algorithm\n",
"\n", "\n",
"* Setup two pointers, current and previous\n", "* Setup two pointers, fast and slow\n",
"* Give current a headstart, incrementing it once if k = 1, twice if k = 2, ...\n", "* Give fast a headstart, incrementing it once if k = 1, twice if k = 2, ...\n",
"* Increment both pointers until current reaches the end\n", "* Increment both pointers until fast reaches the end\n",
"* Return the value of previous\n", "* Return the value of slow\n",
"\n", "\n",
"Complexity:\n", "Complexity:\n",
"* Time: O(n)\n", "* Time: O(n)\n",
@ -103,26 +103,23 @@
"\n", "\n",
" def kth_to_last_elem(self, k):\n", " def kth_to_last_elem(self, k):\n",
" if self.head is None:\n", " if self.head is None:\n",
" return\n", " return None\n",
" if k >= len(self):\n", " fast = self.head\n",
" return\n", " slow = self.head\n",
" curr = self.head\n",
" prev = self.head\n",
" counter = 0\n",
"\n", "\n",
" # Give current a headstart, incrementing it\n", " # Give fast a headstart, incrementing it\n",
" # once for k = 1, twice for k = 2, etc\n", " # once for k = 1, twice for k = 2, etc\n",
" while counter < k:\n", " for _ in range(k):\n",
" curr = curr.next\n", " fast = fast.next\n",
" counter += 1\n", " # If k >= num elements, return None\n",
" if curr is None:\n", " if fast is None:\n",
" return\n", " return None\n",
"\n", "\n",
" # Increment both pointers until current reaches the end\n", " # Increment both pointers until fast reaches the end\n",
" while curr.next is not None:\n", " while fast.next is not None:\n",
" curr = curr.next\n", " fast = fast.next\n",
" prev = prev.next\n", " slow = slow.next\n",
" return prev.data" " return slow.data"
] ]
}, },
{ {
@ -212,21 +209,21 @@
], ],
"metadata": { "metadata": {
"kernelspec": { "kernelspec": {
"display_name": "Python 2", "display_name": "Python 3",
"language": "python", "language": "python",
"name": "python2" "name": "python3"
}, },
"language_info": { "language_info": {
"codemirror_mode": { "codemirror_mode": {
"name": "ipython", "name": "ipython",
"version": 2 "version": 3
}, },
"file_extension": ".py", "file_extension": ".py",
"mimetype": "text/x-python", "mimetype": "text/x-python",
"name": "python", "name": "python",
"nbconvert_exporter": "python", "nbconvert_exporter": "python",
"pygments_lexer": "ipython2", "pygments_lexer": "ipython3",
"version": "2.7.10" "version": "3.5.0"
} }
}, },
"nbformat": 4, "nbformat": 4,