C++实现矩阵的转置

矩阵的转置

转置用一维数组描述的二维矩阵，类中实现，可拆解出来。如若是二维数组的转置。可以考虑二维转一维再转置。。。
当然，你愿意的话。。。（数据结构、算法与应用C++描述第七章习题）

习题汇总

matrix类

//matrix.h
template<class T>
class matrix
{
    friend std::ostream& operator<<
		(std::ostream& out, const matrix<T>& theMatrix);
public:
    matrix(int theRows = 0, int theColumns = 0);
    matrix(const matrix<T>& theMatrix);
    ...
    matrix<T> tranpose();
    T& operator() (int i, int j) const;//类似数学中的访问
    ...

private:
    int rows, columns;
    T* element;
};

template<class T>
matrix<T>::matrix(int theRows, int theColumns)
{//构造函数
	if (theRows < 0 || theColumns < 0)
		throw illegalParameterValue("Rows and columns must be >= 0");
	if ((theRows == 0 || theColumns == 0)
		&& (theRows != 0 || theColumns != 0))
		throw illegalParameterValue
		("Either both or neither rows and columns should be zero");

	rows = theRows;
	columns = theColumns;
	element = new T[rows * columns];
}

template<class T>
matrix<T>::matrix(const matrix<T>& theMatrix)
{//复制构造
	rows = theMatrix.rows;
	columns = theMatrix.columns;
	element = new T[rows * columns];
	std::copy(theMatrix.element, theMatrix.element + rows * columns, element);
}

转置实现

template<class T>
matrix<T> matrix<T>::tranpose()
{//矩阵转置
    matrix<T> w(columns, rows);
    //用i遍历完一维数组,利用除法和求余锁定步长
    for (int i = 0; i < rows * columns; ++i)
          w.element[i / columns + rows * (i % columns)] = element[i];
    return w;
}

//官方答案
template<class T>
void matrix<T>::transpose(matrix<T>& b)
{// Set b to be the transpose of *this.
   // create result matrix
   b.theRows = theColumns;
   b.theColumns = theRows;
   delete [] b.element;
   b.element = new T [theRows * theColumns];

   // copy from this->element to b.element
   int ct = 0;  // next element to move into the transpose
   for (int i = 1; i <= theRows; i++)
      for (int j = 1; j <= theColumns; j++)
         b.element[(j - 1) * theRows + i - 1] = element[ct++];

}

测试

#include <iostream>
#include "matrix.h"

using namespace std;
int main()
{
    try
    {
        matrix<int> x(3, 2), y, z;
        int i, j;
        for (i = 1; i <= 3; i++)
            for (j = 1; j <= 2; j++)
                x(i, j) = 2 * i + j;
        cout << "Initialized x(i,j) = 2*i + j" << endl;
        cout << "x(3,1) = " << x(3, 1) << endl;
        cout << "x is" << endl;;
        cout << x << endl;

        y = x.tranpose();
        cout << y << endl;
    }
    catch (...) {
        cerr << "An exception has occurred" << endl;
    }

    return 0;
}

输出

Initialized x(i,j) = 2*i + j
x(3,1) = 7
x is
3  4
5  6
7  8

3  5  7
4  6  8