C#-将字符串数组转换为多维字符串数组

我试图将一个36个字符的字符串数组放入一个6x6多维字符串数组中，但在如何实现这一点上，我遇到了很多困难

这是我的字符串：

public static readonly string[] Supported = { "0", "1", "2", "3", "4", "5", "6", "7", "8", "9", "a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k", "l", "m", "n", "o", "p", "q", "r", "s", "t", "u", "v", "w", "x", "y", "z" };

我想将其放入这个数组，但要以编程方式：

string[,] grid = new string[,] 
        {
            { "0", "1", "2", "3", "4", "5" },
            { "6", "7", "8", "9", "a", "b" },
            { "c", "d", "e", "f", "g", "h" },
            { "i", "j", "k", "l", "m", "n" },
            { "o", "p", "q", "r", "s", "t" },
            { "u", "v", "w", "x", "y", "z" }
        };

---

这样做现实可行/简单吗？

试试这样的方法：

int width = 6;
string[,] grid = new string[width,width];
for(int i = 0; i < Supported.Length; i++)
{
    grid[i / width, i % width] = Supported[i];
}

class Program
{
    static void Main(string[] args)
    {
        string[] strsArray = { "0", "1", "2", "3", "4", "5", "6", "7", "8", "9", "a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k", "l", "m", "n", "o", "p", "q", "r", "s", "t", "u", "v", "w", "x", "y", "z" };

        string[,] strsMatrix = ConvertStrsArrayToStrsMatrix(strsArray, 6);
    }

    private static string[,] ConvertStrsArrayToStrsMatrix(string[] strsArray, int maxAmountRow)
    {
        int max = strsArray.Length / maxAmountRow;
        string[,] strsMatrix = new string[max, maxAmountRow];
        int counter = 0;
        for(int row = 0; row < max; row++)
        {
            for(int column = 0; column < maxAmountRow; column++)
            {
                strsMatrix[row, column] = strsArray[counter];
                counter++;
            }
        }

        return strsMatrix;
    }
}

int-width=6；
字符串[，]网格=新字符串[宽度，宽度]；
for（int i=0；i

---

正如@DragAndDrop所指出的，

intwidth=6

仅适用于您给出的特定示例。我把它作为一个变量，而不是在

网格

初始化中硬编码

6

，以显示这些值可以调整（如果矩阵不是平方的，则留出空间）。您应该找到一种查找值的方法（即

MathSqrt（Supported.Length）

）。此外，网格上没有边界检查，请确保在计算网格的长度和宽度时，我会这样做：

int width = 6;
string[,] grid = new string[width,width];
for(int i = 0; i < Supported.Length; i++)
{
    grid[i / width, i % width] = Supported[i];
}

class Program
{
    static void Main(string[] args)
    {
        string[] strsArray = { "0", "1", "2", "3", "4", "5", "6", "7", "8", "9", "a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k", "l", "m", "n", "o", "p", "q", "r", "s", "t", "u", "v", "w", "x", "y", "z" };

        string[,] strsMatrix = ConvertStrsArrayToStrsMatrix(strsArray, 6);
    }

    private static string[,] ConvertStrsArrayToStrsMatrix(string[] strsArray, int maxAmountRow)
    {
        int max = strsArray.Length / maxAmountRow;
        string[,] strsMatrix = new string[max, maxAmountRow];
        int counter = 0;
        for(int row = 0; row < max; row++)
        {
            for(int column = 0; column < maxAmountRow; column++)
            {
                strsMatrix[row, column] = strsArray[counter];
                counter++;
            }
        }

        return strsMatrix;
    }
}

类程序
{
静态void Main（字符串[]参数）
{
字符串[]strArray={“0”、“1”、“2”、“3”、“4”、“5”、“6”、“7”、“8”、“9”、“a”、“b”、“c”、“d”、“e”、“f”、“g”、“h”、“i”、“j”、“k”、“l”、“m”、“n”、“o”、“p”、“q”、“r”、“s”、“t”、“u”、“v”、“w”、“x”、“y”、“z”}；
字符串[，]strsMatrix=convertstrarraytostrsmatrix（strarray，6）；
}
私有静态字符串[，]convertstrsarytostrsmatrix（字符串[]strsArray，int-maxamontraw）
{
int max=strArray.Length/maxamontraw；
字符串[，]strsMatrix=新字符串[max，maxAmountRow]；
int计数器=0；
对于（int row=0；row

如果您想支持可能的过度行为：

class Program
{
    static void Main(string[] args)
    {
        string[] strsArray = { "0", "1", "2", "3", "4", "5", "6", "7", "8", "9", "a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k", "l", "m", "n", "o", "p", "q", "r", "s", "t", "u", "v", "w", "x", "y", "z", "mustafa" };

        string[,] strsMatrix = ConvertStrsArrayToStrsMatrix(strsArray, 6);
    }

    private static string[,] ConvertStrsArrayToStrsMatrix(string[] strsArray, int maxAmountRow)
    {
        int max = strsArray.Length / maxAmountRow;
        if(max * maxAmountRow < strsArray.Length)
        {
            max++;
        }
        string[,] strsMatrix = new string[max, maxAmountRow];
        int counter = 0;
        for(int row = 0; row < max; row++)
        {
            for(int column = 0; column < maxAmountRow; column++)
            {
                if (counter < strsArray.Length)
                {
                    strsMatrix[row, column] = strsArray[counter];
                    counter++;
                }
                else
                {
                    break;
                }
            }
        }

        return strsMatrix;
    }
}

类程序
{
静态void Main（字符串[]参数）
{
字符串[]strArray={“0”、“1”、“2”、“3”、“4”、“5”、“6”、“7”、“8”、“9”、“a”、“b”、“c”、“d”、“e”、“f”、“g”、“h”、“i”、“j”、“k”、“l”、“m”、“n”、“o”、“p”、“q”、“r”、“s”、“t”、“u”、“v”、“w”、“x”、“y”、“z”、“mustafa”}；
字符串[，]strsMatrix=convertstrarraytostrsmatrix（strarray，6）；
}
私有静态字符串[，]convertstrsarytostrsmatrix（字符串[]strsArray，int-maxamontraw）
{
int max=strArray.Length/maxamontraw；
if（最大*最大长度

如果您

Math.天花（Math.Sqrt（Supported.Length））

，这将给出矩阵的大小（均匀或不均匀矩阵）

从支持的文件中删除第一个元素：

1 2 3 4 5 6
7 8 9 a b c
d e f g h i
j k l m n o
p q r s t u
v w x y z

将“A”添加到原始数组的开头

A 0 1 2 3 4 5
6 7 8 9 a b c
d e f g h i j
k l m n o p q
r s t u v w x
y z

方法（1）：

0,1,2,3,4,5
6,7,8,9,a,b
c,d,e,f,g,h
i,j,k,l,m,n
o,p,q,r,s,t
u,v,w,x,y,z

以下代码还可用于将支持的数组转换为二维数组并打印二维数组

    public static readonly string[] Supported = { "0", "1", "2", "3", "4", "5", "6", "7", "8", "9", "a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k", "l", "m", "n", "o", "p", "q", "r", "s", "t", "u", "v", "w", "x", "y", "z" };
    static void Main(string[] args)
    {
        int noOfCols = 6;//your own choice to split w.r.t columns
        int noOfRows=Supported.Length/noOfCols;
        string[,] array2D=new string[noOfRows,noOfCols];
        int n = 0, m = 0;
        for(int i=0;i<Supported.Length;i++)
        {
            array2D[n, m] = Supported[i];
            m++;
            if((m=m%noOfCols)==0)
            {
                n++;
            }
        }
    }

这是thread1和thread2的两种方法

static void FillFromZeroToMidIndex(string[,] array2D,int midIndex,int noOfCols)
    {
        int n = 0, m = 0;
        for (int i = 0; i<=midIndex; i++)
        {
            array2D[n, m] = Supported[i];
            m++;
            if ((m = m % noOfCols) == 0)
            {
                n++;
            }
        }
    }
    static void FillFromLastToMidUpIndex(string[,] array2D, int midUpIndex, int lastIndex, int noOfCols, int noOfRows)
    {
        int n = noOfRows-1, m = noOfCols-1;
        for (int i = lastIndex; i >= midUpIndex; i--)
        {
            array2D[n, m] = Supported[i];
            m--;
            if (m<0)
            {
                m = noOfCols-1;
                n--;
            }
        }
    }

---

很有可能，先试：-）可能的重复和相关。您的问题只是一个特定的案例，将列表拆分成n个大小的束。回答为两种可能的行为处理方式您定义了magic int 6，在数学中使用它，这样改变一个会改变另一个。并且要小心越界。@拖放：我们必须使用显式

6

（除非矩阵是平方一）：

36=2*18=3*12=…=6*6

@dragandrop correct，

int 6

是这样定义的，只是因为OP的矩阵是6x6，我将编辑我的答案以解决您的一些评论：-）@DmitryBychenko，

I/width

相当于

I/6

。和

支持。长度
为
width²

        for(int i=0;i<noOfRows;i++)
        {
            for(int j=0;j<noOfCols;j++)
            {
                Console.Write(array2D[i,j]);
                if(j<noOfCols-1)
                {
                    Console.Write(",");
                }
            }
            Console.WriteLine();
        }

0,1,2,3,4,5
6,7,8,9,a,b
c,d,e,f,g,h
i,j,k,l,m,n
o,p,q,r,s,t
u,v,w,x,y,z