How to traverse a 2D array using recursion - c

Question

I'm looking for someone to help me to traverse and display the 2d array or matrix using recursion.
void display(int** matrix1,int row, int column)
This is what I am doing for the 1D array:
void print_array(int arr[], int size, int i)
{
if (i == size) {
cout << endl;
return;
}
cout << arr[i] << " ";
print_array(arr, size, i+1);
}
I know how to traverse 1D array but unable to do this.
I want to display each element in the matrix using recursion.

Answer 1

First of all, it is not quite clear why you want a recursive solution. Perhaps it is a matter of tase, but I find recursion difficult to write, read and debug. Anyhow, I allowed myself to modify your 1d version:
#include <iostream>
#include <vector>
void print_array(const std::vector<int>& arr,size_t i = 0) {
if (i == arr.size()) {
std::cout << '\n';
return;
}
std::cout << arr[i] << " ";
print_array(arr, i+1);
}
I use size_t instead of int, because thats the type to be used when comparing with a containers size. I supplied a default for i because when you call it you dont want to pass the index, but just print the whole array. std::endl is not only printing a new line, but it also flushes the stream. This is unnecessary in most cases.
For the 2D case all you need to add is a second index and a condition to go to the next row.
That is, if the signature of the function is
void print_array(const std::vector<std::vector<int>>& arr,size_t i=0,size_t j=0)
and it prints arr[i][j] then you need to return without printing anything when i == arr.size() and you have to skip to the next row when j== arr[i].size(). This can be done with a condition along the line of:
if (j == arr[i].size()) {
std::cout << '\n';
print_array( arr, i+1,0); // start with the first element of next row
return;
}
PS is you insist on a int**, it should be straightforward to adapt, but I would strongly suggest to use a vector and perhaps even a std::vector<std::array> if the inner arrays all have same size.

Answer 2

I am sure that you need to write a C program or a C++ program but using functionality of C due to your declaration of "2D array" like int **.:)
The function can be easy implemented if to use one more auxiliary recursive function.
Here is a demonstrative program
#include <stdio.h>
void display_row( int *a, size_t column )
{
column == 0 ? ( void )putchar( '\n' )
: ( void )( printf( "%2d ", *a ), display_row ( ++a, --column ) );
}
void display( int **a, size_t row, size_t column )
{
if ( row != 0 )
{
display_row( *a, column );
display( ++a, --row, column );
}
}
int main(void)
{
enum { M = 3, N = 4 };
int a1[N] = { 1, 2, 3, 4 };
int a2[N] = { 5, 6, 7, 8 };
int a3[N] = { 9, 10, 11, 12 };
int *a[M] = { a1, a2, a3 };
display( a, M, N );
return 0;
}
The program output is
1 2 3 4
5 6 7 8
9 10 11 12
If not to use an auxiliary function then another approach is to use a static variable inside the recursive function.
Here is a demonstrative program that uses a recursive function with a static variable.
#include <stdio.h>
void display( int** a, size_t row, size_t column )
{
static size_t pos = 0;
if ( row != 0 && column == 0 )
{
putchar( '\n' );
size_t tmp = column;
column = pos;
pos = tmp;
--row;
++a;
}
if ( row )
{
printf( "%2d ", a[0][pos++] );
display( a, row, --column );
}
}
int main(void)
{
enum { M = 3, N = 4 };
int a1[N] = { 1, 2, 3, 4 };
int a2[N] = { 5, 6, 7, 8 };
int a3[N] = { 9, 10, 11, 12 };
int *a[M] = { a1, a2, a3 };
display( a, M, N );
return 0;
}
Its output is the same as shown above.
If you need a C++ program then just include headers
#include <iostream>
#include <iomanip>
and instead of for example this call of printf
printf( "%2d ", a[0][pos++] );
use
std::cout << std::setw( 2 ) << a[0][pos++];

Answer 3

Here is a simple recursive solution hope you can understand............
#include <iostream>
using namespace std;
void print(int x,int y,int arr[3][3])
{
cout<<arr[x][y]<<" ";
if(x==2 && y==2)//Its a 3 by 3 matrix.....Index starts from 0 and ends at 2 for both row and colomn.
{
return;
}
else if(y==2)
{
cout<<endl;
print(x+1,0,arr);
}
else
print(x,y+1,arr);
}
int main()
{
int arr[3][3]={{1,2,3},{4,5,6},{7,8,9}};
print(0,0,arr);
return 0;
}