我有这个程序,无论大小如何,都应该转置一个矩阵。但是,它没有按预期的方式工作,我也不知道为什么,我确实从中获取了输出,没有编译错误,但是{1,2,3},{4,5,6}的输出是{ 1,0,0},{0,0,4},这对我来说绝对没有意义。我已经在纸上多次处理“内存快照”,但是找不到我所缺少的东西,在这一点上,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;
}
//这样就可以了,如果COLS很小并且说ROWS很大,这是由于中间的2d平方向量,那不是很有效,但是对于其他所有事情,如果它工作正常,应该还不错。
最佳答案
您的代码有几个主要问题,此行:
if (ROWS=COLS)
本来应该是:
if (ROWS == COLS)
第一种情况是将
COLS
的值分配给ROWS
,第二种情况是要检查它们是否相等。在所有for
循环中,当您应使用<=
时,您都在使用<
,否则您将访问数组边界之外的一个。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,最简单的方法将是反转坐标,因此您无需访问
(i,j)
即可访问(j,i)
。另一方面,如果性能是您最关心的问题,那么该previous thread很好地涵盖了我的解决方案主题,并提出了我的解决方案(可能适合您)。