mirror of
https://github.com/donnemartin/interactive-coding-challenges.git
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170 lines
4.0 KiB
Plaintext
170 lines
4.0 KiB
Plaintext
{
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"cells": [
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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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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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"## Problem: Implement the function void Reverse(char* str)\n",
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"\n",
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"* [Clarifying Questions](#Clarifying-Questions)\n",
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"* [Test Cases](#Test-Cases)\n",
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"* [Algorithm](#Algorithm)\n",
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"* [Code](#Code)\n",
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"* [Pythonic-Code: Not In-Place](#Pythonic-Code:-Not-In-Place)"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"## Clarifying Questions\n",
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"\n",
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"* Based on the function signature, it seems you have to implement this in C/C++?\n",
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" * Yes\n",
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"* Can you use additional data structures?\n",
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" * No, do the operation in-place"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"## Test Cases\n",
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"\n",
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"* NULL input\n",
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"* '' -> ''\n",
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"* 'foo bar' -> 'rab oof'"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"## Algorithm\n",
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"\n",
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"We'll want to keep two pointers:\n",
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"* i is a pointer to the first char\n",
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"* j is a pointer to the last char\n",
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"\n",
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"To get a pointer to the last char, we need to loop through all characters, take note of null terminator.\n",
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"\n",
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"* while i < j\n",
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" * swap i and j\n",
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"\n",
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"Complexity:\n",
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"* Time: O(n)\n",
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"* Space: In-place\n",
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"\n",
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"Note:\n",
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"* Instead of using i, you can use str instead, although this might not be as intuitive."
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"## Code"
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"metadata": {
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"collapsed": false
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},
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"outputs": [],
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"source": [
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"# %load reverse_string.cpp\n",
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"#include <stdio.h>\n",
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"\n",
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"void Reverse(char* str) {\n",
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" if (str) {\n",
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" char* i = str;\t// first letter\n",
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" char* j = str;\t// last letter\n",
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" \n",
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" // find the end of the string\n",
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" while (*j) {\n",
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" j++;\n",
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" }\n",
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" \n",
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" // don't point to the null terminator\n",
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" j--;\n",
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" \n",
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" char tmp;\n",
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" \n",
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" // swap chars to reverse the string\n",
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" while (i < j) {\n",
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" tmp = *i;\n",
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" *i++ = *j;\n",
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" *j-- = tmp;\n",
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" }\n",
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" }\n",
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"}\n",
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"\n",
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"int main() {\n",
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" char test0[] = \"\";\n",
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" char test1[] = \"foo\";\n",
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" Reverse(NULL);\n",
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" Reverse(test0);\n",
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" Reverse(test1);\n",
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" printf(\"%s \\n\", test0);\n",
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" printf(\"%s \\n\", test1);\n",
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" return 0;\n",
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"}"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"## Pythonic-Code: Not In-Place\n",
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"\n",
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"The following code is Pythonic, but requires using additional data structures as Python strings are immutable. You could use a bytearray or a list instead of a string to simulate manipulating an array of characters."
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"metadata": {
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"collapsed": false
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},
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"outputs": [],
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"source": [
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"def reverse_string_alt(string):\n",
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" return string[::-1]\n",
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"\n",
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"def reverse_string_alt2(string):\n",
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" return ''.join(reversed(string))"
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]
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}
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],
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"metadata": {
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"kernelspec": {
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"display_name": "Python 2",
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"language": "python",
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"name": "python2"
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},
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"language_info": {
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"codemirror_mode": {
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"name": "ipython",
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"version": 2
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},
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"file_extension": ".py",
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"mimetype": "text/x-python",
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"name": "python",
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"nbconvert_exporter": "python",
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"pygments_lexer": "ipython2",
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"version": "2.7.10"
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}
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},
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"nbformat": 4,
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"nbformat_minor": 0
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}
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