refactor: refactor project structure

legacy/498 FindDiagonalOrder/C#/C#.sln

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+Microsoft Visual Studio Solution File, Format Version 12.00
+# Visual Studio Version 17
+VisualStudioVersion = 17.5.2.0
+MinimumVisualStudioVersion = 10.0.40219.1
+Project("{FAE04EC0-301F-11D3-BF4B-00C04F79EFBC}") = "FindDiagonalOrder", "FindDiagonalOrder.csproj", "{B3151728-C057-4305-DD4C-9A2AFBFC189A}"
+EndProject
+Global
+	GlobalSection(SolutionConfigurationPlatforms) = preSolution
+		Debug|Any CPU = Debug|Any CPU
+		Release|Any CPU = Release|Any CPU
+	EndGlobalSection
+	GlobalSection(ProjectConfigurationPlatforms) = postSolution
+		{B3151728-C057-4305-DD4C-9A2AFBFC189A}.Debug|Any CPU.ActiveCfg = Debug|Any CPU
+		{B3151728-C057-4305-DD4C-9A2AFBFC189A}.Debug|Any CPU.Build.0 = Debug|Any CPU
+		{B3151728-C057-4305-DD4C-9A2AFBFC189A}.Release|Any CPU.ActiveCfg = Release|Any CPU
+		{B3151728-C057-4305-DD4C-9A2AFBFC189A}.Release|Any CPU.Build.0 = Release|Any CPU
+	EndGlobalSection
+	GlobalSection(SolutionProperties) = preSolution
+		HideSolutionNode = FALSE
+	EndGlobalSection
+	GlobalSection(ExtensibilityGlobals) = postSolution
+		SolutionGuid = {DB56539A-6F32-4D28-B91F-BB1DCF46F4D0}
+	EndGlobalSection
+EndGlobal

legacy/498 FindDiagonalOrder/C#/FindDiagonalOrder.csproj

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+<Project Sdk="Microsoft.NET.Sdk">
+
+  <PropertyGroup>
+    <OutputType>Exe</OutputType>
+    <TargetFramework>net8.0</TargetFramework>
+    <ImplicitUsings>enable</ImplicitUsings>
+    <Nullable>enable</Nullable>
+  </PropertyGroup>
+
+</Project>

legacy/498 FindDiagonalOrder/C#/Program.cs

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+public class Solution {
+    public int[] FindDiagonalOrder(int[][] mat)
+    {
+        if (mat == null || mat.Length == 0 || mat[0].Length == 0)
+        {
+            return new int[0];
+        }
+
+        var m = mat.Length;
+        var n = mat[0].Length;
+
+        var diagonals = new Dictionary<int, List<int>>();
+
+        for (var i = 0; i < m; i++)
+        {
+            for (var j = 0; j < n; j++)
+            {
+                int diagonalIndex = i + j;
+                if (!diagonals.ContainsKey(diagonalIndex))
+                {
+                    diagonals[diagonalIndex] = new List<int>();
+                }
+                diagonals[diagonalIndex].Add(mat[i][j]);
+            }
+        }
+
+        var result = new List<int>();
+        foreach (var key in diagonals.Keys)
+        {
+            var diagonal = diagonals[key];
+            if (key % 2 == 0)
+            {
+                result.AddRange(diagonal);
+            }
+            else
+            {
+                diagonal.Reverse();
+                result.AddRange(diagonal);
+            }
+        }
+
+        return result.ToArray();
+    }
+}
+
+class Program
+{
+    static void Main()
+    {
+        var Solution = new Solution();
+
+        int[][] mat1 = new int[][]{
+            new int[]{1,2,3},
+            new int[]{4,5,6},
+            new int[]{7,8,9}
+        };
+
+        Console.WriteLine(string.Join(", ", Solution.FindDiagonalOrder(mat1)));
+
+    }
+}

legacy/498 FindDiagonalOrder/Go/main.go

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+package main
+
+import "fmt"
+
+func findDiagonalOrder(mat [][]int) []int {
+	if len(mat) == 0 || len(mat[0]) == 0{
+		return []int{}
+	}
+
+	m := len(mat)
+	n := len(mat[0])
+	
+	// maps
+	diagonals := make(map[int][]int)
+
+	for i:=0; i< m;i++{
+		for j:=0; j < n; j++{
+			index := i+j
+			diagonals[index] = append(diagonals[index], mat[i][j])
+		}
+	}
+
+	result := []int{}
+
+	for i:= 0; i< m+n-1; i++{
+		if i% 2 == 0{
+			reverse(diagonals[i])
+			result = append(result, diagonals[i]...)
+		}else{
+			result = append(result, diagonals[i]...)
+		}
+	}
+
+	return result
+}
+
+func reverse(slice []int){
+	left, right :=0, len(slice)-1
+
+	for left < right{
+		slice[left], slice[right] = slice[right], slice[left]
+		left++
+		right--
+	}
+}

legacy/498 FindDiagonalOrder/solution.md

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+1. 觀察矩陣對角線模式
+   1. (0,0) -> 1 -> i+j = 0
+   2. (1, 0), (0, 1) -> 2 -> i+j = 1
+   3. (2, 0), (1, 1), (0, 2) -> 3 -> i+j = 2
+2. how to deal with diagonal sorting?
+   1. while (i+j) % 2 == 1 
+   2. while (i+j) % 2 == 0 reversed