如何填充数组：
1 2 3 4 5 6 7 8
20 21 22 23 24 9
19 30 31 32 25 10
18 29 28 27 26 11
17 16 15 14 13 12
螺旋 C＃ 感谢
答案 0 :(得分：10)
从元素（0,0）（左上角）开始遍历数组，向右移动（增加列索引）。保持一个运行的计数器，每次填充元素时都会递增，以及您尚未填充的行和列的上限和下限。对于M列乘N列矩阵，行边界应为0和（M-1），并且列的边界为0和（N-1）。向右走，直到你达到上限列，递减你的上行界限，向下直到你达到你的上行界限，递减你的上行界限，向左走直到你达到你的下列界限，增加你的下列界限，向上直到你达到你的下行界限，增加你的下限，并重复，直到你的上行和下行或列边界相等（或直到你的运行计数是M * N）。
答案 1 :(得分：2)
我亲自制作了一个节目。检查一下。
using System;
using System.Collections.Generic;
using System.Text;
namespace SpiralMatrix
{
class Program
{
static void Main(string[] args)
{
int m = 0, n = 0, start = 0, step = 0;
bool errorOcured = false;
Console.WriteLine("====Spiral Matrix====\n");
try
{
Console.WriteLine("Enter size of the matrix:");
Console.Write("Row (m)? ");
m = Convert.ToInt32(Console.ReadLine());
Console.Write("Column (n)? ");
n = Convert.ToInt32(Console.ReadLine());
Console.Write("Enter the starting number: ");
start = Convert.ToInt32(Console.ReadLine());
Console.Write("Enter step: ");
step = Convert.ToInt32(Console.ReadLine());
if (m < 0 || n < 0 || start < 0 || step < 0) throw new FormatException();
}
catch (FormatException e)
{
Console.WriteLine("Wrong input. [Details: {0}]", e.Message);
Console.WriteLine("Program will now exit...");
errorOcured = true;
}
if (!errorOcured)
{
int[,] mat = new int[m, n];
mat = initMatrix(m, n, start, step);
Console.WriteLine("\nIntial matrix generated is:");
displayMatrix(mat, m, n);
Console.WriteLine("\nSpiral Matrix generated is:");
mat = calculateSpider(mat, m, n);
displayMatrix(mat, m, n);
}
Console.Write("\nPress enter to exit...");
Console.Read();
}
private static int[,] initMatrix(int m, int n, int start, int step)
{
int[,] ret = new int[m, n];
for (int i = 0; i < m; i++)
{
for (int j = 0; j < n; j++)
{
ret[i, j] = start;
start += step;
}
}
return ret;
}
private static void displayMatrix(int[,] mat, int m, int n)
{
for (int i = 0; i < m; i++)
{
for (int j = 0; j < n; j++)
{
Console.Write("\t{0}", mat[i, j]);
}
Console.WriteLine();
}
}
private static int[,] calculateSpider(int[,] mat, int m, int n)
{
int[,] intMat;
if (m <= 2 || n <= 2)
{
if (m == 2 && n == 2)
{
int[,] t = new int[m, n];
t[0, 0] = mat[0, 0];
t[0, 1] = mat[0, 1];
t[1, 0] = mat[1, 1];
t[1, 1] = mat[1, 0];
return t;
}
else if (m == 2)
{
int[,] t = new int[m, n];
for (int i = 0; i < n; i++)
{
t[0, i] = mat[0, i];
t[1, n - 1 - i] = mat[1, i];
}
return t;
}
else if (n == 2)
{
int[,] t = new int[m, n];
int[] stMat = new int[m * n];
int c = 0;
for (int i = 0; i < m; i++)
{
for (int j = 0; j < n; j++)
{
stMat[c] = mat[i, j];
c++;
}
}
c = 0;
for (int i = 0; i < n; i++)
{
t[0, i] = stMat[c];
c++;
}
for (int i = 1; i < m; i++)
{
t[i, 1] = stMat[c];
c++;
}
if(m>1) t[m - 1, 0] = stMat[c];
c++;
for (int i = m - 2; i >= 1; i--)
{
t[i, 0] = stMat[c];
c++;
}
return t;
}
else return mat;
}
intMat = new int[m - 2, n - 2];
int[,] internalMatrix = new int[m - 2, n - 2]; //internal matrix
for (int i = 0; i < ((m - 2) * (n - 2)); i++)
{
internalMatrix[(m - 2) - 1 - i / (n - 2), (n - 2) - 1 - i % (n - 2)] = mat[m - 1 - (i / n), n - 1 - (i % n)];
}
intMat = calculateSpider(internalMatrix, m - 2, n - 2);
int[,] retMat = new int[m, n]; //return matrix
//copy some characters to a single dimentional array
int[] tempMat = new int[(m * n) - ((m - 2) * (n - 2))];
for (int i = 0; i < (m * n) - ((m - 2) * (n - 2)); i++)
{
tempMat[i] = mat[i / n, i % n];
}
int count = 0;
//copy fist row
for (int i = 0; i < n; i++)
{
retMat[0, i] = tempMat[count];
count++;
}
//copy last column
for (int i = 1; i < m; i++)
{
retMat[i, n - 1] = tempMat[count];
count++;
}
//copy last row
for (int i = n - 2; i >= 0; i--)
{
retMat[m - 1, i] = tempMat[count];
count++;
}
//copy first column
for (int i = m - 2; i >= 1; i--)
{
retMat[i, 0] = tempMat[count];
count++;
}
//copy others
for (int i = 1; i < m - 1; i++)
{
for (int j = 1; j < n - 1; j++)
{
retMat[i, j] = intMat[i - 1, j - 1];
}
}
return retMat;
}
}
}
答案 2 :(得分：1)
Web上有几种解决方案，包括StackOverflow和其他地方：
答案 3 :(得分：0)
好吧，我不会给你代码。你不会从我身上学到任何东西，但是我会给你一个提示。
如果您要使用此螺旋图案填充N x M矩形，请注意其中的N-1 x M-1矩形也是螺旋图案。同样，内部的N-2 x M-2矩形也是一个螺旋图案，依此类推，直到你有一个1x1的矩形。
因此很可能是一种递归解决方案。
答案 4 :(得分：0)
您可以这样做：
class Spiral
{
int[,] matrix;
int m_size;
int currentCount;
static void Main(string[] args)
{
Spiral s = new Spiral(2);
s.DrawSpiral();
Console.ReadLine();
}
public Spiral(int size)
{
this.m_size = size;
matrix = new int[size, size];
currentCount = 1;
}
public void DrawSpiral()
{
//x,y x, y+size-1
//x+1, y+size-1 x+size-1, y+size-1
//x+size-1, y+size-2 x+size-1, y
//x+size-2, y x+1, y
int x = 0, y = 0, size = m_size;
while (size > 0)
{
for (int i = y; i <= y + size - 1; i++)
{
matrix[x, i] = currentCount++;
}
for (int j = x + 1; j <= x + size - 1; j++)
{
matrix[j, y + size - 1] = currentCount++;
}
for (int i = y + size - 2; i >= y; i--)
{
matrix[x + size - 1, i] = currentCount++;
}
for (int i = x + size - 2; i >= x + 1; i--)
{
matrix[i, y] = currentCount++;
}
x = x + 1;
y = y + 1;
size = size - 2;
}
PrintMatrix();
}
private void PrintMatrix()
{
for (int i = 0; i < m_size; i++)
{
for (int j = 0; j < m_size; j++)
{
Console.Write(matrix[i, j]);
Console.Write(" ");
}
Console.WriteLine();
}
}
}
答案 5 :(得分：0)
只需在2循环中输入x和y（对于你的，-2，-2到2,2）
while(y <= 2)<br>
{<br>
while(x <= 2)<br>
{<br>
// PRINT RESULTS OF (25-spiral_get_value(x,y))<br>
x += 1;<br>
}<br>
x = -2;<br>
y += 1;<br>
}
逻辑就在这里。您将不得不操纵它以与C＃一起使用。下面的算法立即计算（x，y）处的螺旋数，从from [0,0] == 1开始，顺时针旋转，where [0,-1] == 2.
在将x / y值放入算法之前简单地否定它们，或者将x替换为y并取消其中一个或两个，以改变方向（顺时针/逆时针），输出是否被翻转（水平，垂直或两者）等等。
// spiral_get_value(x,y);<br>
sx = argument0;<br>
sy = argument1;<br>
a = max(sqrt(sqr(sx)),sqrt(sqr(sy)));<br>
c = -b;<br>
d = (b*2)+1;<br>
us = (sy==c and sx !=c);<br>
rs = (sx==b and sy !=c);<br>
bs = (sy==b and sx !=b);<br>
ls = (sx==c and sy !=b);<br>
ra = rs*((b)*2);<br>
ba = bs*((b)*4);<br>
la = ls*((b)*6);<br>
ax = (us*sx)+(bs*-sx);<br>
ay = (rs*sy)+(ls*-sy);<br>
add = ra+ba+la+ax+ay;<br>
value = add+sqr(d-2)+b;<br>
return(value);