在C ++中转置矩阵 [英] Transposing a matrix in C++

问题描述

我正在编写一个程序，使用分配的内存来转置给定的矩阵。该函数完美的与方阵NxN（行== cols），但它崩溃与MxN矩阵（行！= cols）。请帮助

I'm writing a program to transpose a given matrix using allocated memory. The function works perfect with square matrix NxN (rows==cols) but it crashes with MxN matrix (rows != cols). Please help

void transpose(int **matrix, int *row, int *col)
{
    // dynamically allocate an array
    int **result;
    result = new int *[*col]; //creates a new array of pointers to int objects
    // check for error
    if (result == NULL)
    {
        cout << "Error allocating array";
        exit(1);
    }
    for (int count = 0; count < *col; count++)
    {
        *(result + count) = new int[*row];
    }

    // transposing
    for (int i = 0; i<*row; i++)
    {
       for (int j = i+1; j<*col; j++)
       {
        int temp = *(*(matrix + i) + j);
        *(*(matrix + i) + j) = *(*(matrix + j) + i);
        *(*(matrix + j) + i) = temp;
       }
    }

    for (int i = 0; i<*row; i++)
    {
       for (int j = 0; j<*col; j++)
       {
          *(*(result + i) + j) = *(*(matrix + i) + j);
          cout << *(*(result + i) + j) << "\t";
       }
       cout << endl;
    }
}

推荐答案

行：

for (int i = 0; i<*row; i++)
{
   for (int j = i+1; j<*col; j++)
   {
    int temp = *(*(matrix + i) + j);
    *(*(matrix + i) + j) = *(*(matrix + j) + i);
    *(*(matrix + j) + i) = temp;
   }
}

问题是矩阵的索引由i，然后j，而不是j，然后我喜欢你在第二和第三行在while循环。该矩阵是一个2×3矩阵，然后你试图执行矩阵[2] [3] =矩阵[3] [2]，但矩阵[3] [2]不存在。

are the issue. The problem is that matrix is indexed by i then j, not j then i like you are doing in the second and third line in the while loop. Image that matrix is a 2x3 matrix, then you try to perform matrix[2][3] = matrix[3][2], but matrix[3][2] does not exist.

最好直接在这个循环中初始化结果：

It is best to go about simply initializing result directly in this loop:

for (int i = 0; i<*row; i++)
   for (int j = 0; j<*col; j++)
     result[j][i] = matrix[i][j];

然后你可以像下面输出，或者删除矩阵和重新分配矩阵，我的整个转置函数成为下面的代码（row和col不需要指向int传递的值是正确的。也访问矩阵应该使用数组下标，因为它是更好的风格）：

Then you can output like below, or delete matrix and reassign matrix to be result as you wish. My entire transpose function became the following code (row and col need not be pointers to int pass by value is just fine. Also accessing matrices should use array subscripts as it is nicer style):

void transpose(int **matrix, int row, int col)
{
  // dynamically allocate an array
  int **result;
  result = new int *[col]; //creates a new array of pointers to int objects
  for (int i = 0; i < col; i++)
    result[i] = new int[row];

  // transposing
  for (int i = 0; i<row; i++)
   for (int j = 0; j<col; j++)
     result[j][i] = matrix[i][j];

  //output resulting matrix
  for (int i = 0; i<col; i++) {
   for (int j = 0; j<row; j++)
    cout << result[i][j] << "\t";
   cout << endl;
  }
}

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