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atwood
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‎.gitignore

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/.vs/
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/bin/
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/obj/

‎LeetCode.csproj

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<Project Sdk="Microsoft.NET.Sdk">
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<PropertyGroup>
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<OutputType>Exe</OutputType>
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<TargetFramework>netcoreapp3.0</TargetFramework>
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</PropertyGroup>
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</Project>

‎LeetCode.sln

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Microsoft Visual Studio Solution File, Format Version 12.00
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# Visual Studio Version 16
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VisualStudioVersion = 16.0.29411.108
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MinimumVisualStudioVersion = 10.0.40219.1
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Project("{FAE04EC0-301F-11D3-BF4B-00C04F79EFBC}") = "LeetCode", "LeetCode.csproj", "{FFEEB876-767D-41EE-ABA9-027E68F2620F}"
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EndProject
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Global
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GlobalSection(SolutionConfigurationPlatforms) = preSolution
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Debug|Any CPU = Debug|Any CPU
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Release|Any CPU = Release|Any CPU
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EndGlobalSection
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GlobalSection(ProjectConfigurationPlatforms) = postSolution
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{FFEEB876-767D-41EE-ABA9-027E68F2620F}.Debug|Any CPU.ActiveCfg = Debug|Any CPU
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{FFEEB876-767D-41EE-ABA9-027E68F2620F}.Debug|Any CPU.Build.0 = Debug|Any CPU
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{FFEEB876-767D-41EE-ABA9-027E68F2620F}.Release|Any CPU.ActiveCfg = Release|Any CPU
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{FFEEB876-767D-41EE-ABA9-027E68F2620F}.Release|Any CPU.Build.0 = Release|Any CPU
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EndGlobalSection
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GlobalSection(SolutionProperties) = preSolution
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HideSolutionNode = FALSE
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EndGlobalSection
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GlobalSection(ExtensibilityGlobals) = postSolution
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SolutionGuid = {C5A6A089-C86A-4CB1-A8A5-DAC22668C201}
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EndGlobalSection
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EndGlobal

‎No13_Math.cs

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using System;
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using System.Collections.Generic;
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using System.Text;
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namespace LeetCode_13
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{
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//static void Main(string[] args)
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//{
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// var solution = new Solution();
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// while (true)
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// {
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// //int input = int.Parse(Console.ReadLine());
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// string input = Console.ReadLine();
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// var res = solution.RomanToInt(input);
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// Console.WriteLine(res);
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// }
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//}
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public class Solution
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{
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/// <summary>
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/// 第一印象，遍历解（没有看到更好的做法，）
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/// </summary>
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/// <param name="s"></param>
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/// <returns></returns>
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public int RomanToInt(string s)
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{
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int sum = 0;
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for (int i = 0; i < s.Length; i++)
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{
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switch (s[i])
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{
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case 'I':
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{
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if (i != s.Length - 1 && s[i + 1] == 'V')
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{
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sum += 4;
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i++;
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}
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else if (i != s.Length - 1 && s[i + 1] == 'X')
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{
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sum += 9;
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i++;
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}
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else
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{
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sum += 1;
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}
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break;
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}
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case 'V':
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sum += 5; break;
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case 'X':
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{
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if (i != s.Length - 1 && s[i + 1] == 'L')
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{
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sum += 40;
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i++;
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}
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else if (i != s.Length - 1 && s[i + 1] == 'C')
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{
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sum += 90;
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i++;
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}
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else
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{
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sum += 10;
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}
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break;
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}
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case 'L':
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sum += 50; break;
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case 'C':
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{
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if (i != s.Length - 1 && s[i + 1] == 'D')
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{
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sum += 400;
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i++;
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}
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else if (i != s.Length - 1 && s[i + 1] == 'M')
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{
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sum += 900;
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i++;
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}
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else
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{
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sum += 100;
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}
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break;
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}
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case 'D':
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sum += 500; break;
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case 'M': sum += 1000; break;
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default: break;
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}
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}
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return sum;
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}
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}
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}

‎No168_Math.cs

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using System;
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using System.Collections.Generic;
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using System.Text;
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namespace LeetCode_168
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{
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//static void Main(string[] args)
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//{
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// var solution = new Solution();
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// while (true)
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// {
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// int input = int.Parse(Console.ReadLine());
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// //string input = Console.ReadLine();
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// //string input2 = Console.ReadLine();
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// var res = solution.ConvertToTitle(input);
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// Console.WriteLine(res);
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// }
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//}
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public class Solution
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{
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public string ConvertToTitle(int n)
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{
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int len = (int)Math.Log(n, 26) + 1;
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char[] res = new char[len];
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//A的ASCII码为65
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for (int i = 0; n > 0; i++)
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{
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int reminder = n % 26;
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n /= 26;
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if (reminder == 0)
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{
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res[i] = (char)(64 + 26);
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//此题重点：因为没有0的表示形态，所以当整除的时候要-1，因为后面的Z已经代表了26
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n -= 1;
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}
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else
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res[i] = (char)(64 + reminder);
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}
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Array.Reverse(res);
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return new string(res).Trim('\0');
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}
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}
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}

‎No171_Math.cs

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using System;
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using System.Collections.Generic;
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using System.Text;
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namespace LeetCode_171
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{
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//static void Main(string[] args)
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//{
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// var solution = new Solution();
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// while (true)
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// {
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// //int input = int.Parse(Console.ReadLine());
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// string input = Console.ReadLine();
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// //string input2 = Console.ReadLine();
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// var res = solution.TitleToNumber(input);
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// Console.WriteLine(res);
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// }
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//}
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public class Solution
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{
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public int TitleToNumber(string s)
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{
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int sum = 0;
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for (int i = s.Length - 1; i >= 0; i--)
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{
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sum += (s[i] - ('A' - 1)) * (int)Math.Pow(26, s.Length - 1 - i);
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}
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return sum;
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}
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}
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}

‎No172_Math.cs

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using System;
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using System.Collections.Generic;
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using System.Text;
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namespace LeetCode_172
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{
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//static void Main(string[] args)
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//{
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// var solution = new Solution();
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// while (true)
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// {
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// int input = int.Parse(Console.ReadLine());
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// //string input = Console.ReadLine();
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// //string input2 = Console.ReadLine();
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// var res = solution.TrailingZeroes(input);
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// Console.WriteLine(res);
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// }
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//}
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public class Solution
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{
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/// <summary>
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/// 更高维度拆解分式，通过数学推出，大概思路就是分解25就是分解两次5，往上也是一样，所以通过不断分解5就可以得到答案
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/// 时间复杂度O(log n),因为是不断除以5，所以是5的幂，即为O(log 5 n),循环内的除法是32位整形内的，所以是O(1),忽略
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/// 空间复杂度O(1),只用了一个zeroCount计算。
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/// </summary>
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/// <param name="n"></param>
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/// <returns></returns>
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public int TrailingZeroes(int n)
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{
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int zeroCount = 0;
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while (n > 0)
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{
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n /= 5;
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zeroCount += n;
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}
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return zeroCount;
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}
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/// <summary>
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/// 尾数的零一定是由2乘以5构成的，所以可以分解成有多少个2和5的组合就能获得答案，因为分解出来的2是远多于5的，
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/// 所以忽略掉2，只考虑5。5还有一种特殊情况就是，例如25这种，可以分解出不只一个5，这种就要特殊考虑。
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/// 时间复杂度O(n)，循环步长为5，所以是1/5n，还有部分进入计算的也可以忽略不计
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/// 空间复杂度O(1)，只用了一个zeroCount计数。
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/// </summary>
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/// <param name="n"></param>
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/// <returns></returns>
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//public int TrailingZeroes(int n)
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//{
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// //写到一半突然觉得不对，笔记本上写了一下，发现了规律。降维打击成就达成（然而并没有，哈哈哈）
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// int zeroCount = 0;
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// for (int i = 5; i <= n; i += 5)
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// {
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// int powerOfFive = 5;
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// while (i % powerOfFive == 0)
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// {
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// zeroCount++;
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// powerOfFive *= 5;
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// }
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// }
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// return zeroCount;
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//}
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}
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}

‎No202_Math.cs

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using System;
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using System.Collections.Generic;
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using System.Text;
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namespace LeetCode_202
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{
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//static void Main(string[] args)
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//{
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// var solution = new Solution();
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// while (true)
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// {
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// int input = int.Parse(Console.ReadLine());
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// //string input = Console.ReadLine();
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// //string input2 = Console.ReadLine();
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// var res = solution.IsHappy(input);
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// Console.WriteLine(res);
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// }
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//}
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public class Solution
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{
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public int GetNextNum(int n)
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{
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int num = 0;
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while (n > 0)
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{
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num += (int)Math.Pow(n % 10, 2);
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n /= 10;
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}
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return num;
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}
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/// <summary>
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/// 快慢指针来查看是否有循环，这样就不用存走过的路途了（弗洛伊德循环查找算法）
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/// 时间复杂度O(log n)
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/// 空间复杂度O(1)，因为不用存走过的路径，只需要不断计算就可以了
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/// </summary>
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/// <param name="n"></param>
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/// <returns></returns>
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public bool IsHappy(int n)
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{
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int rabbit = n;
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int tortoise = n;
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do
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{
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rabbit = GetNextNum(GetNextNum(rabbit));
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tortoise = GetNextNum(tortoise);
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}
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while (rabbit != 1 && rabbit != tortoise);//&& tortoise != 1
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//如果没有循环兔子会最先到达
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return rabbit == 1;// || tortoise == 1;
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}
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/// <summary>
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///hash set记录已经走过的路径，如果有已经走过的，就一定不是happy number
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///时间复杂度:O(log n)
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///空间复杂度:O(log n)
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/// </summary>
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/// <param name="n"></param>
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/// <returns></returns>
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//public bool IsHappy(int n)
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//{
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// HashSet<int> hs = new HashSet<int>();
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// while (n != 1 && !hs.Contains(n))
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// {
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// hs.Add(n);
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// n = GetNextNum(n);
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// }
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// return n == 1;
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//}
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}
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}

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