想象一下，我们有像
这样的矩阵a11 a12 a13
a21 a22 a23
a31 a32 a33
我想要做的是，从文本框值旋转此矩阵，以便，例如，如果我写2并按 rotate ，程序必须保持矩阵的两个对角值（在这种情况下为a11， a22，a33，a13，a31）并顺时针旋转2次其他值。所以结果必须像
a11 a32 a13
a23 a22 a21
a31 a12 a33
它必须适用于所有N x N大小的矩阵，正如您所看到的，每4次旋转将矩阵置于默认状态。
所以想法就是这样，我有两种形式。首先获取矩阵的大小（1值，例如，如果它是5，则生成5x5矩阵）。当我按确定时，它会生成第二个表单文本框矩阵，如
表格1代码
private void button1_Click(object sender, EventArgs e)
{
int matrixSize;
matrixSize = int.Parse(textBox1.Text);
Form2 form2 = new Form2(matrixSize);
form2.Width = matrixSize * 50 + 100;
form2.Height = matrixSize *60 + 200;
form2.Show();
//this.Hide();
}
表单2代码从给定值生成文本框矩阵，并将随机值放入此字段
public Form2(int matrSize)
{
int counter = 0;
InitializeComponent();
TextBox[] MatrixNodes = new TextBox[matrSize*matrSize];
Random r = new Random();
for (int i = 1; i <= matrSize; i++)
{
for (int j = 1; j <= matrSize; j++)
{
var tb = new TextBox();
int num = r.Next(1, 1000);
MatrixNodes[counter] = tb;
tb.Name = string.Format("Node_{0}{1}", i, j);
Debug.Write(string.Format("Node_{0}{1}", i, j));
tb.Text = num.ToString();
tb.Location = new Point(j * 50, i * 50);
tb.Width = 30;
tb.Visible = true;
this.splitContainer1.Panel2.Controls.Add(tb);
counter++;
}
}
}
表单2有1个用于控制旋转的文本框（其他文件框是以编程方式生成的）。我想要做的是，当我输入旋转计数并在此文本框中按 Enter 时，我想旋转文本框矩阵，如上所述。无法弄清楚该怎么做。
答案 0 :(得分：9)
将两个对角线复制到单独的阵列，然后旋转矩阵并替换对角线。下面的代码显示了每个步骤：
class Program
{
static void Main(string[] args)
{
int matrixSize = 3;
string[,] matrix = new string[matrixSize,matrixSize];
//create square matrix
for (int x = 0; x < matrixSize; x++)
{
for (int y = 0; y < matrixSize; y++)
{
matrix[x, y] = "a" + (x + 1).ToString() + (y + 1).ToString();
}
}
Console.WriteLine(Environment.NewLine + "Base square matrix");
for (int x = 0; x < matrixSize; x++)
{
for (int y = 0; y < matrixSize; y++)
{
Console.Write(matrix[x, y] + " ");
}
Console.Write(Environment.NewLine);
}
Console.ReadKey();
//copy diagonals
string[] leftDiagonal = new string[matrixSize];
string[] rightDiagonal = new string[matrixSize];
for (int x = 0; x < matrixSize; x++)
{
leftDiagonal[x] = matrix[x, x];
rightDiagonal[x] = matrix[matrixSize - 1 - x, x];
}
Console.WriteLine(Environment.NewLine + "Diagonals");
for (int x = 0; x < matrixSize; ++x)
{
Console.Write(leftDiagonal[x] + " " + rightDiagonal[x] + Environment.NewLine);
}
Console.ReadKey();
//rotate matrix
string[,] rotatedMatrix = new string[matrixSize, matrixSize];
for (int x = 0; x < matrixSize; x++)
{
for (int y = 0; y < matrixSize; y++)
{
rotatedMatrix[x, y] = matrix[matrixSize - y - 1, x];
}
}
Console.WriteLine(Environment.NewLine + "Rotated");
for (int x = 0; x < matrixSize; x++)
{
for (int y = 0; y < matrixSize; y++)
{
Console.Write(rotatedMatrix[x, y] + " ");
}
Console.Write(Environment.NewLine);
}
Console.ReadKey();
//rotate matrix again
string[,] rotatedMatrixAgain = new string[matrixSize, matrixSize];
for (int x = 0; x < matrixSize; x++)
{
for (int y = 0; y < matrixSize; y++)
{
rotatedMatrixAgain[x, y] = rotatedMatrix[matrixSize - y - 1, x];
}
}
Console.WriteLine(Environment.NewLine + "Rotated again");
for (int x = 0; x < matrixSize; x++)
{
for (int y = 0; y < matrixSize; y++)
{
Console.Write(rotatedMatrixAgain[x, y] + " ");
}
Console.Write(Environment.NewLine);
}
Console.ReadKey();
//replace diagonals
for (int x = 0; x < matrixSize; x++)
{
rotatedMatrixAgain[x, x] = leftDiagonal[x];
rotatedMatrixAgain[matrixSize - 1 - x, x] = rightDiagonal[x];
}
Console.WriteLine(Environment.NewLine + "Completed" + Environment.NewLine);
for (int x = 0; x < matrixSize; x++)
{
for (int y = 0; y < matrixSize; y++)
{
Console.Write(rotatedMatrixAgain[x, y] + " ");
}
Console.Write(Environment.NewLine);
}
Console.ReadKey();
}
}
答案 1 :(得分：2)
我不知道C＃，所以我只能提出伪代码的建议：
输入： N个N矩阵in
输出：按照OP out
for i = 1 to N
for j = 1 to N
if N - j != i and i != j // Do not change values on either diagonal
out[j][N-i] = in[i][j]
else
out[i][j] = in[i][j]
免责声明：此算法未经测试。我建议你使用调试器检查它是否正常工作。
答案 2 :(得分：0)
这似乎是一个非常不正统的UI演示文稿，但是在能够实现您的功能方面你并没有太远。矩形阵列不是线性阵列，而是使您的工作更容易。实际旋转可以通过for循环重复单个旋转来实现（这将在case 1代码中实现），但我决定将它们组合成四种可能的情况。这实际上允许您输入旋转次数的负数。这提醒我，你真的应该做更多的错误检查。至少要防止int.Parse在它使用的地方抛出异常（使用try catch块或切换到int.TryParse）并确保它返回一个有意义的数字（大于0，可能为int.MaxValue中的matrixSize设置button1_Click以外的合理最大值。
namespace RotatingMatrices
{
public class Form2 : Form
{
// note these class fields
private TextBox[,] matrixNodes;
private int matrixSize;
public Form2(int matrSize)
{
InitializeComponent();
// note these inits
matrixSize = matrSize;
matrixNodes = new TextBox[matrixSize, matrixSize];
Random r = new Random();
// note the new loop limits
for (int i = 0; i < matrixSize; i++)
{
for (int j = 0; j < matrixSize; j++)
{
var tb = new TextBox();
int num = r.Next(1, 1000);
// note the change in indexing
matrixNodes[i,j] = tb;
tb.Name = string.Format("Node_{0}_{1}", i, j);
Debug.Write(string.Format("Node_{0}_{1}", i, j));
tb.Text = num.ToString();
tb.Location = new Point(j * 50, i * 50);
tb.Width = 30;
tb.Visible = true;
this.splitContainer1.Panel2.Controls.Add(tb);
}
}
}
private void buttonRotate_Click(object sender, EventArgs e)
{
string[,] matrix = new string[matrixSize, matrixSize];
int rotations = (4 + int.Parse(textBoxRotations.Text)) % 4; // note the addition of and mod by 4
switch(rotations)
{
case 1: // rotate clockwise
for (int i = 0; i < matrixSize; i++)
{
for (int j = 0; j < matrixSize; j++)
{
matrix[j, matrixSize - i - 1] = matrixNodes[i, j].Text;
}
}
break;
case 2: // rotate 180 degrees
for (int i = 0; i < matrixSize; i++)
{
for (int j = 0; j < matrixSize; j++)
{
matrix[i, j] = matrixNodes[matrixSize - i - 1, matrixSize - j - 1].Text;
}
}
break;
case 3: // rotate counter-clockwise
for (int i = 0; i < matrixSize; i++)
{
for (int j = 0; j < matrixSize; j++)
{
matrix[i, j] = matrixNodes[j, matrixSize - i - 1].Text;
}
}
break;
default: // do nothing
return;
}
// restore diagonals
for(int i = 0; i < matrixSize; i++)
{
matrix[i, i] = matrixNodes[i, i].Text;
matrix[i, matrixSize - i - 1] = matrixNodes[i, matrixSize - i - 1].Text;
}
// write strings back to text boxes
for (int i = 0; i < matrixSize; i++)
{
for (int j = 0; j < matrixSize; j++)
{
matrixNodes[i, j].Text = matrix[i, j];
}
}
}
}
}
答案 3 :(得分：0)
我决定使用listView而不是文本框解决问题，这使我的逻辑变得更容易。使用这种方法，我能够将矩阵视为连续的框。我从外面开始向中间移动，每次都改变我的盒子的大小。
另外，我使用的不是两种形式。在顶部，我有一个文本框，用户输入他们想要数组的大小，以及标有＆＃34; Fill＆＃34; （按钮2）。在底部我有一个文本框，用户输入旋转度。当他们点击＆＃34;旋转时，＆＃34;它启动了一个向链表添加值，组合和移动列表，然后写回矩阵的过程。我确信我让它变得比以前更复杂，但这是一次很棒的学习练习。
在查看上面的杰里代码后，我想我会研究矩形阵列。 ：）
using System;
using System.Collections.Generic;
using System.ComponentModel;
using System.Data;
using System.Drawing;
using System.Linq;
using System.Text;
using System.Windows.Forms;
namespace Recycle
{
public partial class Form1 : Form
{
public int size;
public LinkedList<string> topRight = new LinkedList<string>();
public LinkedList<string> bottomLeft = new LinkedList<string>();
public LinkedList<string> myMatrix = new LinkedList<string>();
public LinkedList<string> shiftMatrix = new LinkedList<string>();
public Form1()
{
InitializeComponent();
}
private void button2_Click(object sender, EventArgs e)
{
listView1.Clear();
size = int.Parse(textBox2.Text);
int c = 0;
int q = 0;
int w = 0;
string[] content = new string[size];
Random rnd = new Random();
for (c = 0; c < size; c++)
{
listView1.Columns.Add("", 25);
}
for (q = 0; q < size; q++)
{
for (w = 0; w < size; w++)
{
content[w] = rnd.Next(9,100).ToString();
}
ListViewItem lvi = new ListViewItem(content);
listView1.Items.Add(lvi);
}
}
public bool iseven(int size)
{
if (size % 2 == 0)
{
return true;
}
else
{
return false;
}
}
public void button1_Click(object sender, EventArgs e)
{
if (listView1.Items.Count < 3)
{
MessageBox.Show("Matrix cannot be rotated.");
return;
}
bool even = false;
int shift = int.Parse(textBox1.Text); //amount to shift by
int box = listView1.Items.Count - 1; //size of box
int half = Convert.ToInt32(listView1.Items.Count / 2);
int corner = 0; //inside corner of box
if (shift > listView1.Items.Count)
{
shift = shift % ((listView1.Items.Count - 2) * 4);
}
do
{
eachPass(shift, box, corner);
++corner;
--box;
} while (box >= half + 1);
}
public void eachPass(int shift, int box, int corner)
{
int x;
int i;
//Read each non-diagonal value into one of two lists
for (x = corner + 1; x < box; x++)
{
topRight.AddLast(listView1.Items[corner].SubItems[x].Text);
}
x = box;
for (i = corner + 1; i < box; i++)
{
topRight.AddLast(listView1.Items[i].SubItems[x].Text);
}
for (x = box - 1; x > corner; x--)
{
bottomLeft.AddLast(listView1.Items[box].SubItems[x].Text);
}
x = corner;
for (i = box - 1; i > corner; i--)
{
bottomLeft.AddLast(listView1.Items[i].SubItems[x].Text);
}
string myTest = "";
//join the two lists, so they can be shifted
foreach (string tR in topRight)
{
myMatrix.AddLast(tR);
}
foreach (string bL in bottomLeft)
{
myMatrix.AddLast(bL);
}
int sh;
//shift the list using another list
for (sh = shift; sh < myMatrix.Count; sh++)
{
shiftMatrix.AddLast(myMatrix.ElementAt(sh));
}
for (sh = 0; sh < shift; sh++)
{
shiftMatrix.AddLast(myMatrix.ElementAt(sh));
}
//we need the sizes of the current lists
int trCnt = topRight.Count;
int blCnt = bottomLeft.Count;
//clear them for reuse
topRight.Clear();
bottomLeft.Clear();
int s;
//put the shifted values back
for (s = 0; s < trCnt; s++)
{
topRight.AddLast(shiftMatrix.ElementAt(s));
}
for (s = blCnt; s < shiftMatrix.Count; s++)
{
bottomLeft.AddLast(shiftMatrix.ElementAt(s));
}
int tRn = 0;
int bLn = 0;
//write each non-diagonal value from one of two lists
for (x = corner + 1; x < box; x++)
{
listView1.Items[corner].SubItems[x].Text = topRight.ElementAt(tRn);
++tRn;
}
x = box;
for (i = corner + 1; i < box; i++)
{
listView1.Items[i].SubItems[x].Text = topRight.ElementAt(tRn);
++tRn;
}
for (x = box - 1; x > corner; x--)
{
listView1.Items[box].SubItems[x].Text = bottomLeft.ElementAt(bLn);
++bLn;
}
x = corner;
for (i = box - 1; i > corner; i--)
{
listView1.Items[i].SubItems[x].Text = bottomLeft.ElementAt(bLn);
++bLn;
}
myMatrix.Clear();
shiftMatrix.Clear();
topRight.Clear();
bottomLeft.Clear();
}
}
}