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我有我制作的这个程序，它应该转置一个矩阵，不管它是什么大小。但是它并没有像预期的那样工作，我不知道为什么，我确实从中得到输出，没有编译错误，但 {1,2,3},{4,5,6} 的输出是 { 1,0,0},{0,0,4}，这对我来说完全没有意义。我已经多次在纸上处理“记忆快照”，但找不到我缺少的东西，此时我真的只需要另一双眼睛，谢谢。

#include "stdafx.h"

#include <fstream>

#include <iostream>

#include <iomanip>

#include <string>

#include <cmath>

#include <vector>

using namespace std;

void squaretranspose(int &M, int &MT, int ROWS, int COLS);

int main(void)
{
    int M[2][3]={{1,2,3},{4,5,6}};
    int MT[3][2]={0};
    int ROWS(2),COLS(3);

int i,j;
cout << " The entries of the original matrix " << endl;
for(i=0;i<=ROWS-1;i++)
{
    for(j=0;j<=COLS-1;j++)
{
cout<<M[i][j]<<"\t";
}
cout << endl;
}
squaretranspose(M[0][0],MT[0][0],ROWS,COLS);
cout << " The entries of the transposed non-square matrix " << endl;
for(i=0;i<=COLS-1;i++)
{
    cout << endl;
    for(j=0;j<=ROWS-1;j++)
{
cout<<MT[i][j]<<"\t";
}
}
system ("PAUSE");
return 0;
}
void squaretranspose (int &M, int &MT, int ROWS, int COLS)
{
    // declare pointers to change the input matrice's values
    int *ptr,*ptrT;
// declare indices for a row by row process
    int i,j;
// declare placeholder 2d vectors for swapping the I,j, entries to ,j,i entries  
    vector < vector<int>> temp(ROWS,COLS);
    vector < vector<int>> tempT(COLS,COLS);
    vector < vector<int>> temp_T(ROWS,ROWS);
// set the pointers to point to the first entry of the input and output matrices
    ptr = &M;
    ptrT = &MT;

// if rows=cols we want to use 2d vector temp
    if (ROWS=COLS)
    {
// store all of the input matrice's values in the 2d vector "temp"
    for(i=0;i<=ROWS-1;i++)
    {
        for(j=0;j<=COLS-1;j++)
        {

// set the i,j th entry of the 2d vector "temp" equal to the value currently pointed to by the pointer
        temp[i][j]=*ptr;
// increment the pointer to the address of the next entry of the input matrix unless we are on the last entry
               if ((i!=ROWS-1)&&(j!=COLS-1))
               { 
               ptr++;
               }
        }
        }
    }
// reset pointer address to first entry
    ptr=&M;
// the for loop for swapping the j,i entries of the 2d vector "temp" with the i,j entries of the input matrix
    for(i=0;i<=ROWS-1;i++)
    {
        for(j=0;j<=COLS-1;j++)
        {
// if j is not equal to i swap the value pointed to by the pointer (the i,j entry of the input matrix) for the j,ith entries value of the 2d vector "temp"
        if (j!=i)
            {
            *ptr=temp[j][i];
            }
        // increment the pointer if it is not on the last entry
               if ((i!=ROWS-1)&&(j!=COLS-1))
               { 
               ptr++;
               }
        }
    }*/


// if ROWS<COLS we want to have 2d vector tempT
if (ROWS<COLS)
{
// store all of the input matrice's values in the 2d vector "tempT"
    for(i=0;i<=ROWS-1;i++)
    {
        for(j=0;j<=COLS-1;j++)
        {
// set the j,ith entry of the 2d vector "temp" equal to the value currently pointed to by the pointer
        tempT[j][i]=*ptr;
// increment the pointer to the address of the next entry of the input matrix
        if (((i!=(ROWS-1))&&(j!=(COLS-1))))
        {
            ptr++;
        }
        }
    }
    ptr=&M;
// transport the entries of tempT into the output matrix MT
    for(i=0;i<=COLS-1;i++)
    {
        for(j=0;j<=ROWS-1;j++)
        {
        *ptrT=tempT[i][j];
// increment the pointer
        if (((i!=ROWS-1)&&(j!=COLS-1)))
        {
            ptrT++;
        }
        }
    }
}
ptrT=&MT;

// if ROWS>COLS we want to use the 2d vector temp_T
if (ROWS>COLS)
{
// store all of the input matrice's values in the 2d vector "temp_T"
    for(i=0;i<=ROWS-1;i++)
    {
        for(j=0;j<=COLS-1;j++)
        {

// set the j,i th entry of the 2d vector "temp" equal to the value currently pointed to by the pointer
        temp_T[j][i]=*ptr;
// increment the pointer
        if (((i!=ROWS-1)&&(j!=COLS-1)))
        {
            ptrT++;
        }
    }

// the for loop for swapping the j,i entries of the 2d vector "temp" with the i,j entries of the input matrix
    for(i=0;i<=COLS-1;i++)
    {
        for(j=0;j<=ROWS-1;j++)
        {
// if j is not equal to i swap the value pointed to by the pointer (the i,j entry of the input matrix) for the j,ith entries value of the 2d vector "temp"
            if (j!=i)
            {
            *ptrT=temp_T[j][i];
            }
// increment the pointer
        if (((i!=ROWS-1)&&(j!=COLS-1)))
        {
            ptrT++;
        }
        }
    }
}
return;
}

// 就是这样，如果 COLS 非常小并且说 ROWS 非常大，因为中间的 2d 方形向量，它不是很有效，但是对于其他一切，如果它工作正常，它应该是相当不错的。

4

2 回答 2

0

使用 DIY 索引： ptrT[ index(j,i,COLS,ROWS) ] = ptr[ index(i,j,ROWS,COLS) ]. 现在您所要做的就是编写将 2D indeces 转换为 1D indeces 的函数。

于 2013-07-02T23:11:10.430 回答
0

您的代码有几个主要问题，这一行：

if (ROWS=COLS)

可能是：

if (ROWS == COLS)

在第一种情况下，您将分配 to 的值，COLS在ROWS第二种情况下，您将检查它们是否相等。在您应该使用的所有for循环中，否则您将访问数组边界之外的循环。<=<

除此之外，代码太复杂了，转置函数应该很简单，这是一种可能的方法：

template <int n, int m>
void squaretranspose( int a[n][m], int b[m][n])
{
    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < m; j++)
        {
            b[j][i] = a[i][j];
        }
    }
}

但是转置矩阵的最快和最简单的方法是反转你的坐标，所以不要访问(i,j)你 access (j,i)。另一方面，如果性能是您主要关心的问题，那么这个先前的主题很好地涵盖了我的解决方案，并且我的解决方案与其他可能适合您的解决方案一样。

于 2013-07-02T23:50:19.427 回答