swap all elements from opposite lines in a C ++ matrix - c++

Question

I would like to change each element of the matrix with its "opposite" (element 0,0 becomes element n, n element 0,1 becomes n, n-1 and etc ..).
below an example:
1 2 3 9 8 7
4 5 6 --> 6 5 4
7 8 9 3 2 1
or:
4 5 4 7 22 14 12 2
5 2 6 8 ---> 2 4 6 2
2 6 4 2 8 6 2 5
2 12 14 22 7 4 5 4
anyone have any ideas on how to do it?

Answer 1

It is enough to use the standard algorithm std::reverse to perform the task.
Here is a demonstration program.
#include <iostream>
#include <iterator>
#include <algorithm>
int main()
{
const size_t N1 = 3;
int a[N1][N1] =
{
{ 1, 2, 3 },
{ 4, 5, 6 },
{ 7, 8, 9 }
};
std::reverse( std::begin( a ), std::end( a ) );
for ( auto &row : a )
{
std::reverse( std::begin( row ), std::end( row ) );
}
for (const auto &row : a)
{
for (const auto &item : row)
{
std::cout << item << ' ';
}
std::cout << '\n';
}
std::cout << '\n';
const size_t N2 = 4;
int b[N2][N2] =
{
{ 4, 5, 4, 7 },
{ 5, 2, 6, 8,},
{ 2, 6, 4, 2 },
{ 2, 12, 14, 22 }
};
std::reverse( std::begin( b ), std::end( b ) );
for (auto &row : b)
{
std::reverse( std::begin( row ), std::end( row ) );
}
for (const auto &row : b)
{
for (const auto &item : row)
{
std::cout << item << ' ';
}
std::cout << '\n';
}
std::cout << '\n';
}
The program output is
9 8 7
6 5 4
3 2 1
22 14 12 2
2 4 6 2
8 6 2 5
7 4 5 4
You could write a separate function to perform such an operation like
template <typename T, size_t N>
void reverse_matrix( T ( &a )[N][N] )
{
std::reverse( std::begin( a ), std::end( a ) );
for (auto &row : a)
{
std::reverse( std::begin( row ), std::end( row ) );
}
}
Another approach is to reinterpret the two-dimensional arrays as ine-dimensional arrays like
std::reverse( std::begin( reinterpret_cast< int ( & )[N1 * N1] > ( a ) ),
std::end( reinterpret_cast< int ( & )[N1 * N1] >( a ) ) );
and
std::reverse( std::begin( reinterpret_cast< int ( & )[N2 * N2] > ( b ) ),
std::end( reinterpret_cast< int ( & )[N2 * N2] >( b ) ) );
Without using the standard algorithm std::reverse you will need to use loops as for example
for ( size_t i = 0; i < N1 / 2; i++ )
{
std::swap( a[i], a[N1-i-1] );
}
for ( size_t i = 0; i < N1; i++ )
{
for ( size_t j = 0; j < N1 / 2; j++ )
{
std::swap( a[i][j], a[i][N1-j-1] );
}
}

Answer 2

You can do it simply with for loop. This is code:
#include <iostream>
using namespace std;
int main(){
int n;
cin>>n;
int arr[n][n], arr1[n][n];
for(int i = 0; i < n; i++){
for(int j = 0; j < n; j++)
cin>>arr[i][j];
}
for(int i = 0; i < n; i++){
for(int j = 0; j < n; j++)
arr1[i][j] = arr[n-1-i][n-1-j];
}
for(int i = 0; i < n; i++){
for(int j = 0; j < n; j++){
cout<<arr1[i][j]<<" ";
}
cout<<endl;
}
}