我有两种算法将随机2d阵列（mхn或mхm）转换为1d阵列。我想知道是否有办法使它们在相反的方向上工作并将结果转换为1d数组，保存数字的顺序。以下是我的程序的完整代码和图片，以了解我的算法是如何工作的。enter image description here 代码：
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Diagnostics;
using static System.Math;
namespace MyProgram
{
public class Program
{
public static void Main(string[] args)
{
Console.Write("Enter number of rows:");
int n = int.Parse(Console.ReadLine());
Console.Write("Enter number of columns:");
int m = int.Parse(Console.ReadLine());
int[] arr1 = new int[n * m];
int[,] arr2 = new int[n, m];
int choice;
do
{
Console.WriteLine("Select option:");
Console.WriteLine("\t1: Diagonal");
Console.WriteLine("\t2: Spiral");
Console.WriteLine("\t3: Exit");
Console.Write("Your selection: ");
choice = int.Parse(Console.ReadLine());
switch (choice)
{
case 1:
{
SetArray(arr2);
PrintArray(arr2);
Diagonal(arr2, arr1);
PrintArray(arr1);
break;
}
case 2:
{
SetArray(arr2);
PrintArray(arr2);
Spiral(arr2, arr1);
PrintArray(arr1);
break;
}
}
Console.WriteLine();
Console.WriteLine();
} while (choice != 5);
}
static void Diagonal(int[,] array2, int[] array1)
{
int k = 0;
int row = 0;
int col = 0;
while (k < array1.Length)
{
array1[k] = array2[row, col];
if ((row + col) % 2 == 0)
{
if ((row == 0) && (col != array2.GetLength(1) - 1)) { col++; }
else
{
if (col == array2.GetLength(1) - 1) { row++; }
else { row--; col++; }
}
}
else
{
if ((col == 0) && (row != array2.GetLength(0) - 1)) { row++; }
else
{
if (row == array2.GetLength(0) - 1) { col++; }
else { row++; col--; }
}
}
k += 1;
}
}
private static void Spiral(int[,] array2, int[] array1)
{
int lengthX = array2.GetLength(0);
int lengthY = array2.GetLength(1);
int Product = lengthX * lengthY;
int CorrectY = 0;
int CorrectX = 0;
int Count = 0;
while (lengthX > 0 && lengthY > 0)
{
for (int j = CorrectY; j < lengthY && Count < Product; j++)
{
array1[Count] = array2[CorrectX, j];
Count++ ;
}
CorrectX++;
for (int i = CorrectX; i < lengthX && Count < Product; i++)
{
array1[Count] = array2[i, lengthY - 1];
Count++ ;
}
if (lengthY > 0 && lengthX > 0) lengthY-- ;
else break;
for (int j = lengthY - 1; j >= CorrectY && Count < Product; j--)
{
array1[Count] = array2[lengthX - 1, j];
Count++ ;
}
if (lengthY > 0 && lengthX > 0) lengthX-- ;
else break;
for (int i = lengthX - 1; i >= CorrectX && Count < Product; i--)
{
array1[Count] = array2[i, CorrectY];
Count++ ;
}
CorrectY++;
}
}
public static void SetArray(int[,] arr)
{
Random r = new Random();
for (int i = 0; i < arr.GetLength(0); i++)
{
for (int j = 0; j < arr.GetLength(1); j++)
{
arr[i, j] = r.Next(11, 99);
}
}
}
public static void SetArray(int[] arr)
{
Random r = new Random();
for (int i = 0; i < arr.Length; i++)
{
arr[i] = r.Next(11, 99);
}
}
public static void PrintArray(int[] arr)
{
Console.Write("print 1d array:");
for (int i = 0; i < arr.Length; i++)
{
Console.Write(arr[i] + " ");
}
Console.WriteLine();
}
public static void PrintArray(int[,] arr)
{
Console.WriteLine("print 2d array:");
for (int i = 0; i < arr.GetLength(0); i++)
{
for (int j = 0; j < arr.GetLength(1); j++)
{
Console.Write(arr[i, j] + " ");
}
Console.WriteLine();
}
}
}
}
答案 0 :(得分：1)
对于我来说，这似乎对我有用Diagonal：
static void BackwardDiagonal(int[,] array2, int[] array1) {
int k = 0;
int row = 0;
int col = 0;
while (k < array1.Length) {
array2[row, col] = array1[k]; // just swap sides of the assignment...
if ((row + col) % 2 == 0) {
if ((row == 0) && (col != array2.GetLength(1) - 1)) { col++; } else {
if (col == array2.GetLength(1) - 1) { row++; } else { row--; col++; }
}
} else {
if ((col == 0) && (row != array2.GetLength(0) - 1)) { row++; } else {
if (row == array2.GetLength(0) - 1) { col++; } else { row++; col--; }
}
}
k += 1;
}
}
对于向后Spiral：
private static void BackwardSpiral(int[,] array2, int[] array1)
{
int lengthX = array2.GetLength(0);
int lengthY = array2.GetLength(1);
int Product = lengthX * lengthY;
int CorrectY = 0;
int CorrectX = 0;
int Count = 0;
while (lengthX > 0 && lengthY > 0)
{
for (int j = CorrectY; j < lengthY && Count < Product; j++)
{
array2[CorrectX, j] = array1[Count]; // just swap sides of the assignment...
Count++ ;
}
CorrectX++;
for (int i = CorrectX; i < lengthX && Count < Product; i++)
{
array2[i, lengthY - 1] = array1[Count];
Count++ ;
}
if (lengthY > 0 && lengthX > 0) lengthY-- ;
else break;
for (int j = lengthY - 1; j >= CorrectY && Count < Product; j--)
{
array2[lengthX - 1, j] = array1[Count];
Count++ ;
}
if (lengthY > 0 && lengthX > 0) lengthX-- ;
else break;
for (int i = lengthX - 1; i >= CorrectX && Count < Product; i--)
{
array2[i, CorrectY] = array1[Count];
Count++ ;
}
CorrectY++;
}
}
我还将此添加到switch语句中以实现它：
case 4:
{
SetArray(arr2);
PrintArray(arr2);
Diagonal(arr2, arr1);
PrintArray(arr1);
int[,] arr3 = new int[n, m]; // new blank array to fill with arr1
BackwardDiagonal(arr3, arr1); // fill arr3 from backward Diagonal algorithm
PrintArray(arr3);
break;
}
case 5:
{
SetArray(arr2);
PrintArray(arr2);
Spiral(arr2, arr1);
PrintArray(arr1);
int[,] arr3 = new int[n, m]; // new blank array to fill with arr1
BackwardSpiral(arr3, arr1); // fill arr3 from backward Spiral algorithm
PrintArray(arr3);
break;
}
当您正在使用它时，还要确保在} while (choice != 3);循环结束时do，以便退出该程序！
答案 1 :(得分：1)
实际上你要问的很简单。只需更改您的算法方法即可接收int rows，int cols并像这样委托
delegate void Apply(int row, int col, int index);
然后将arr2.GetLength(0)替换为rows，将arr2.GetLength(1)替换为cols，将数组元素分配替换为委托调用。
以下是更新后的Diagonal方法（您可以对其他方法执行相同操作）：
static void Diagonal(int rows, int cols, Apply action)
{
int k = 0;
int row = 0;
int col = 0;
int length = rows * cols;
while (k < length)
{
action(row, col, k);
if ((row + col) % 2 == 0)
{
if ((row == 0) && (col != cols - 1)) { col++; }
else
{
if (col == cols - 1) { row++; }
else { row--; col++; }
}
}
else
{
if ((col == 0) && (row != rows - 1)) { row++; }
else
{
if (row == rows - 1) { col++; }
else { row++; col--; }
}
}
k += 1;
}
}
和用法：
SetArray(arr2);
PrintArray(arr2);
Diagonal(n, m, (r, c, i) => arr1[i] = arr2[r, c]);
PrintArray(arr1);
// Inverse
var arr3 = new int[n, m];
Diagonal(n, m, (r, c, i) => arr3[r, c] = arr1[i]);
PrintArray(arr3);