顺时针打印阵列

本文介绍了顺时针打印阵列的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我有这个代码是从TextBox接收数组然后顺时针打印它，但是当我跟踪这段代码时没有以正确的方式打印我没有错误但是在运行时它没有显示顺时针的结果我不知道错误在哪里

I Have this code is receive the array from TextBox and then print it in the clockwise ,but doesn't print in correct way when I make trace to this code I Don't have error but when run it don't show the result in a clockwise I Don't know where the error

int i, k = 0, l = 0; int m=4; int n=4; /* k - starting row index m - ending row index l - starting column index n - ending column index i - iterator */ string [,]a = new string[4,4]; string s = null; for(int q=0;q<=4;q++){ for(int j=0;j<=4;j++){ a[q,j] = textBox1.Text[q].ToString(); } } while (k < m && l < n) { /* Print the first row from the remaining rows */ for (i = l; i < n; i++) { s += a[k, i].ToString(); } k++; /* Print the last column from the remaining columns */ for (i = k; i < m; i++) { s += a[i, n - 1].ToString(); } n--; /* Print the last row from the remaining rows */ if (k < m) { for (i = n - 1; i >= l; i--) { s += a[m - 1, i].ToString(); } m--; } /* Print the first column from the remaining columns */ if (l < n) { for (i = m - 1; i >= k; i--) { s += a[i, l].ToString(); } l++; } richTextBox1.Text = s.ToString(); }

推荐答案

编辑1 ：添加了另一种方法体 您好。 这是我的解决方案。 你将找到3个重载方法来接受所有字符的二维矩阵（ char [，] ），一个行列表（ string [] ），一个文本（ string ）。 char [，] 重载是唯一进行计算的（其他两个委托给它）。 EDIT 1: Added an alternative method body Hi. Here is my solution. You will find 3 overloaded methods to accept a 2D matrix of all chars (char[,]), a list of lines (string[]), a text (string). The char[,] overload is the only doing the computation (the others two delegates to it). static string RebuildClockwise(string text) { // Delegate computation to "string[]" overload var result = RebuildClockwise(text.Split(new[] { Environment.NewLine }, StringSplitOptions.None)); return result; } static string RebuildClockwise(params string[] lines) { // Fetch total counts of rows/lines and columns (max length of all lines) var nRowCount = lines.Length; var nColCount = lines.Max(line => line.Length); // Loop all lines with lesser length and pad them with spaces for (int ixLine = 0; ixLine < lines.Length; ixLine++) { var line = lines[ixLine]; if (line.Length < nColCount) lines[ixLine] = line.PadRight(nColCount); } // Translate the lines into chars var chars = new char[nRowCount, nColCount]; for (int ixRow = 0; ixRow < nRowCount; ixRow++) { for (int ixCol = 0; ixCol < nColCount; ixCol++) { chars[ixRow, ixCol] = lines[ixRow][ixCol]; } } // Delegate computation to "char[,]" overload var result = RebuildClockwise(chars); return result; } static string RebuildClockwise(char[,] chars) { // Fetch total count of chars for loop safety var nTotalCount = chars.GetLength(0) * chars.GetLength(1); // Store the limits var aLimits = new int[4]; aLimits[Dir_Right] = chars.GetLength(1) - 1; aLimits[Dir_Down] = chars.GetLength(0) - 1; aLimits[Dir_Left] = 0; aLimits[Dir_Up] = 0; var result = string.Empty; // Start at Top Left corner (0,0) pointing Right var ixCol = 0; var ixRow = 0; var nDirection = Dir_Right; // For safety, check the count of processed chars to avoid infinite loop for (var nProcessedCount = 0; nProcessedCount < nTotalCount; nProcessedCount++) { result += chars[ixRow, ixCol]; if (!RebuildClockwise_Move(aLimits, ref nDirection, ref ixCol, ref ixRow)) break; } return result; }

主要功能（ RebuildClockwise（char [，] chars）），首先存储矩阵限制并初始化方向和单元格位置。 然后循环更新结果并调用另一个改变单元格位置和方向的函数（ RebuildClockwise_Move ）。

The main function (RebuildClockwise(char[,] chars)), first stores the matrix limits and initializes the direction and cell location to start with. Then loops updating the result and calling another function (RebuildClockwise_Move) that changes the cell location and direction.

private static bool RebuildClockwise_Move(int[] aLimits, ref int nDirection, ref int ixCol, ref int ixRow) { // Try all 4 directions for (var nTrial = MinDir + 1; nTrial <= MaxDir + 1; nTrial++) { switch (nDirection) { case Dir_Right: // There are still unprocessed columns on the Right if (ixCol < aLimits[nDirection]) { // Select next column to the Right ixCol++; return true; } // If we get here, then we have exhasuted all the Columns for the current Row: discard current Row aLimits[(nDirection + 3) % 4]++; break; case Dir_Down: // There are still unprocessed rows Below if (ixRow < aLimits[nDirection]) { // Select next row to Below ixRow++; return true; } // If we get here, then we have exhasuted all the Rows for the current Column: discard current Column aLimits[(nDirection + 3) % 4]--; break; case Dir_Left: // There are still unprocessed columns on the Left if (ixCol > aLimits[nDirection]) { // Select next column to the Left ixCol--; return true; } // If we get here, then we have exhasuted all the Columns for the current Row: discard current Row aLimits[(nDirection + 3) % 4]--; break; case Dir_Up: // There are still unprocessed rows Above if (ixRow > aLimits[nDirection]) { // Select next row to Above ixRow--; return true; } // If we get here, then we have exhasuted all the Rows for the current Column: discard current Column aLimits[(nDirection + 3) % 4]++; break; } // Pick the next (clockwise) direction nDirection++; if (nDirection > MaxDir) nDirection = MinDir; } // We Tried all directions, then we are at the center of the matrix and should stop return false; }

RebuildClockwise_Move ，根据方向，更改行和/或列索引并更新限制。 当没有行或列可用时，它返回false以停止循环。 这两种方法都使用了一些管理方向的常数：

RebuildClockwise_Move, based on the direction, change the row and/or column index and updates the limits. When no row or column is available, it returns false to stop looping. Both methods make use of some constants to manage the direction:

const int Dir_Right = 0; const int Dir_Down = Dir_Right + 1; const int Dir_Left = Dir_Down + 1; const int Dir_Up = Dir_Left + 1; const int MinDir = 0; const int MaxDir = 3;

以下测试代码可以检查计算是否正确：

The following test code allow to check the computation is correct:

MessageBox.Show(RebuildClockwise("1234", "CDE5", "BGF6", "A987")); MessageBox.Show(RebuildClockwise("12345", "EFGH6", "DKJI7", "CBA98")); MessageBox.Show(RebuildClockwise("1234", "EFG5", "DKH6", "CJI7", "BA98")); MessageBox.Show(RebuildClockwise("12345678")); MessageBox.Show(RebuildClockwise("1234", "8765")); MessageBox.Show(RebuildClockwise("12", "83", "74", "65")); MessageBox.Show(RebuildClockwise("1", "2", "3", "4", "5", "6", "7", "8"));

他们应该在之间显示文字12345678 和 123456789ABCDEFGHIJK （输入值的排列是为了获得可理解的结果）。 总的来说，这个解决方案是比示例代码更通用（和复杂）但可以简化替换 aLimits 数组和使用con的代码切换块stants或plain变量。 问候， Daniele。 编辑1 ：以下函数是 char [，] 重载的替代方法。 这个将使用两个并发变量来移动行和列而不是依赖于 nDirection 变量;此版本还包含移动逻辑。

They should display a text between 12345678 and 123456789ABCDEFGHIJK (the input values are arranged in order to obtain an understandable result). Overall, this solution is more generalized (and complex) than your sample code but it can be simplified replacing the aLimits array and the switch block with code using constants or plain variables. Regards, Daniele. EDIT 1: The following functions is an alternative to the char[,] overload. This one will use two concurrent variables to move row and column instead of relying on the nDirection variable; this version also incorporate the movement logic.

static string RebuildClockwise(char[,] chars) { // Fetch maximum column and row indices var nMaxCol = chars.GetLength(1) - 1; var nMaxRow = chars.GetLength(0) - 1; // Fetch total count of chars for loop safety var nTotalCount = (nMaxRow + 1) * (nMaxCol + 1); var result = string.Empty; // Fixed minimum column and row indices var nMinCol = 0; var nMinRow = 0; // Start at 0,0 var ixCol = 0; var ixRow = 0; // Initial direction is Col++ (toward Right) var nDeltaCol = 1; var nDeltaRow = 0; // For safety, check the count of processed chars to avoid infinite loop for (var nProcessedCount = 0; nProcessedCount < nTotalCount; nProcessedCount++) { result += chars[ixRow, ixCol]; var bMoved = false; // Try all 4 directions for (var nTrial = 1; nTrial <= 4; nTrial++) { // Compute new location var ixColN = ixCol + nDeltaCol; var ixRowN = ixRow + nDeltaRow; // If location is within min and max, then accept it if ((ixColN >= nMinCol) && (ixColN <= nMaxCol) && (ixRowN >= nMinRow) && (ixRowN <= nMaxRow)) { ixCol = ixColN; ixRow = ixRowN; // Tell outside of for loop that the new location has been found bMoved = true; break; } // If location is beyond min column, discard (current) max row if (ixColN < nMinCol) nMaxRow--; // If location is beyond max column, discard (current) min row else if (ixColN > nMaxCol) nMinRow++; // If location is beyond min row, discard (current) min column else if (ixRowN < nMinRow) nMinCol++; // If location is beyond max row, discard (current) max column else if (ixRowN > nMaxRow) nMaxCol--; // Rotate increments (they are like cosine and sine of angles 0, 90, 180, 270) var nTemp = nDeltaCol; nDeltaCol = -nDeltaRow; nDeltaRow = nTemp; } // We Tried all directions, then we are at the center of the matrix and should stop if (!bMoved) break; } return result; }