Dada una array 2D, imprímala en forma de espiral inversa. Ya hemos discutido Imprimir una array dada en forma de espiral . Este artículo explica cómo hacer la impresión inversa. Vea los siguientes ejemplos.
Input:
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
Output:
10 11 7 6 5 9 13 14 15 16 12 8 4 3 2 1
Input:
1 2 3 4 5 6
7 8 9 10 11 12
13 14 15 16 17 18
Output:
11 10 9 8 7 13 14 15 16 17 18 12 6 5 4 3 2 1
Implementación:
C++
// This is a modified code of
// https://www.geeksforgeeks.org/print-a-given-matrix-in-spiral-form/
#include <iostream>
#define R 3
#define C 6
using namespace std;
// Function that print matrix in reverse spiral form.
void ReversespiralPrint(int m, int n, int a[R][C])
{
// Large array to initialize it
// with elements of matrix
long int b[100];
/* k - starting row index
l - starting column index*/
int i, k = 0, l = 0;
// Counter for single dimension array
//in which elements will be stored
int z = 0;
// Total elements in matrix
int size = m*n;
while (k < m && l < n)
{
// Variable to store value of matrix.
int val;
/* Print the first row from the remaining rows */
for (i = l; i < n; ++i)
{
// printf("%d ", a[k][i]);
val = a[k][i];
b[z] = val;
++z;
}
k++;
/* Print the last column from the remaining columns */
for (i = k; i < m; ++i)
{
// printf("%d ", a[i][n-1]);
val = a[i][n-1];
b[z] = val;
++z;
}
n--;
/* Print the last row from the remaining rows */
if ( k < m)
{
for (i = n-1; i >= l; --i)
{
// printf("%d ", a[m-1][i]);
val = a[m-1][i];
b[z] = val;
++z;
}
m--;
}
/* Print the first column from the remaining columns */
if (l < n)
{
for (i = m-1; i >= k; --i)
{
// printf("%d ", a[i][l]);
val = a[i][l];
b[z] = val;
++z;
}
l++;
}
}
for (int i=size-1 ; i>=0 ; --i)
{
cout<<b[i]<<" ";
}
}
/* Driver program to test above functions */
int main()
{
int a[R][C] = { {1, 2, 3, 4, 5, 6},
{7, 8, 9, 10, 11, 12},
{13, 14, 15, 16, 17, 18}};
ReversespiralPrint(R, C, a);
return 0;
}
Java
// JAVA Code for Print a given matrix in
// reverse spiral form
class GFG {
public static int R = 3, C = 6;
// Function that print matrix in reverse spiral form.
public static void ReversespiralPrint(int m, int n,
int a[][])
{
// Large array to initialize it
// with elements of matrix
long b[] = new long[100];
/* k - starting row index
l - starting column index*/
int i, k = 0, l = 0;
// Counter for single dimension array
//in which elements will be stored
int z = 0;
// Total elements in matrix
int size = m * n;
while (k < m && l < n)
{
// Variable to store value of matrix.
int val;
/* Print the first row from the remaining
rows */
for (i = l; i < n; ++i)
{
val = a[k][i];
b[z] = val;
++z;
}
k++;
/* Print the last column from the remaining
columns */
for (i = k; i < m; ++i)
{
val = a[i][n-1];
b[z] = val;
++z;
}
n--;
/* Print the last row from the remaining
rows */
if ( k < m)
{
for (i = n-1; i >= l; --i)
{
val = a[m-1][i];
b[z] = val;
++z;
}
m--;
}
/* Print the first column from the remaining
columns */
if (l < n)
{
for (i = m-1; i >= k; --i)
{
val = a[i][l];
b[z] = val;
++z;
}
l++;
}
}
for (int x = size-1 ; x>=0 ; --x)
{
System.out.print(b[x]+" ");
}
}
/* Driver program to test above function */
public static void main(String[] args)
{
int a[][] = { {1, 2, 3, 4, 5, 6},
{7, 8, 9, 10, 11, 12},
{13, 14, 15, 16, 17, 18}};
ReversespiralPrint(R, C, a);
}
}
// This code is contributed by Arnav Kr. Mandal.
Python3
# Python3 Code to Print a given
# matrix in reverse spiral form
# This is a modified code of
# https:#www.geeksforgeeks.org/print-a-given-matrix-in-spiral-form/
R, C = 3, 6
def ReversespiralPrint(m, n, a):
# Large array to initialize it
# with elements of matrix
b = [0 for i in range(100)]
#/* k - starting row index
#l - starting column index*/
i, k, l = 0, 0, 0
# Counter for single dimension array
# in which elements will be stored
z = 0
# Total elements in matrix
size = m * n
while (k < m and l < n):
# Variable to store value of matrix.
val = 0
# Print the first row
# from the remaining rows
for i in range(l, n):
# printf("%d ", a[k][i])
val = a[k][i]
b[z] = val
z += 1
k += 1
# Print the last column
# from the remaining columns
for i in range(k, m):
# printf("%d ", a[i][n-1])
val = a[i][n - 1]
b[z] = val
z += 1
n -= 1
# Print the last row
# from the remaining rows
if (k < m):
for i in range(n - 1, l - 1, -1):
# printf("%d ", a[m-1][i])
val = a[m - 1][i]
b[z] = val
z += 1
m -= 1
# Print the first column
# from the remaining columns
if (l < n):
for i in range(m - 1, k - 1, -1):
# printf("%d ", a[i][l])
val = a[i][l]
b[z] = val
z += 1
l += 1
for i in range(size - 1, -1, -1):
print(b[i], end = " ")
# Driver Code
a = [[1, 2, 3, 4, 5, 6],
[7, 8, 9, 10, 11, 12],
[13, 14, 15, 16, 17, 18]]
ReversespiralPrint(R, C, a)
# This code is contributed by mohit kumar
C#
// C# Code for Print a given matrix in
// reverse spiral form
using System;
class GFG {
public static int R = 3, C = 6;
// Function that print matrix in reverse spiral form.
public static void ReversespiralPrint(int m, int n,
int [,]a)
{
// Large array to initialize it
// with elements of matrix
long []b = new long[100];
/* k - starting row index
l - starting column index*/
int i, k = 0, l = 0;
// Counter for single dimension array
//in which elements will be stored
int z = 0;
// Total elements in matrix
int size = m * n;
while (k < m && l < n)
{
// Variable to store value of matrix.
int val;
/* Print the first row from the remaining
rows */
for (i = l; i < n; ++i)
{
val = a[k,i];
b[z] = val;
++z;
}
k++;
/* Print the last column from the remaining
columns */
for (i = k; i < m; ++i)
{
val = a[i,n-1];
b[z] = val;
++z;
}
n--;
/* Print the last row from the remaining
rows */
if ( k < m)
{
for (i = n-1; i >= l; --i)
{
val = a[m-1,i];
b[z] = val;
++z;
}
m--;
}
/* Print the first column from the remaining
columns */
if (l < n)
{
for (i = m-1; i >= k; --i)
{
val = a[i,l];
b[z] = val;
++z;
}
l++;
}
}
for (int x = size-1 ; x>=0 ; --x)
{
Console.Write(b[x]+" ");
}
}
/* Driver program to test above function */
public static void Main()
{
int [ ,]a = { {1, 2, 3, 4, 5, 6},
{7, 8, 9, 10, 11, 12},
{13, 14, 15, 16, 17, 18}};
ReversespiralPrint(R, C, a);
}
}
// This code is contributed by vt_m.
PHP
<?php
// PHP Code for Print a given
// matrix in reverse spiral form
$R=3;
$C=6;
// Function that print matrix
// in reverse spiral form.
function ReversespiralPrint($m, $n, array $a)
{
// Large array to initialize it
// with elements of matrix
$b;
// k - starting row index
// l - starting column index
$k = 0;
$l = 0;
// Counter for single dimension array
// in which elements will be stored
$z = 0;
// Total elements in matrix
$size = $m*$n;
while ($k < $m && $l < $n)
{
// Variable to store
// value of matrix.
$val;
// Print the first row from
// the remaining rows
for ($i = $l; $i < $n; ++$i)
{
$val = $a[$k][$i];
$b[$z] = $val;
++$z;
}
$k++;
// Print the last column from
// the remaining columns
for ($i = $k; $i < $m; ++$i)
{
// printf("%d ", a[i][n-1]);
$val = $a[$i][$n-1];
$b[$z] = $val;
++$z;
}
$n--;
// Print the last row from
// the remaining rows
if ( $k < $m)
{
for ($i = $n-1; $i >= $l; --$i)
{
// printf("%d ", a[m-1][i]);
$val = $a[$m-1][$i];
$b[$z] = $val;
++$z;
}
$m--;
}
// Print the first column
// from the remaining columns
if ($l < $n)
{
for ($i = $m - 1; $i >= $k; --$i)
{
$val = $a[$i][$l];
$b[$z] = $val;
++$z;
}
$l++;
}
}
for ($i = $size - 1; $i >= 0; --$i)
{
echo $b[$i]." ";
}
}
// Driver Code
$a= array(array(1, 2, 3, 4, 5, 6),
array(7, 8, 9, 10, 11, 12),
array(13, 14, 15, 16, 17, 18));
ReversespiralPrint($R, $C, $a);
// This Code is contributed by mits
?>
Javascript
<script>
// This is a modified code of
// https://www.geeksforgeeks.org/
// print-a-given-matrix-in-spiral-form/
let R = 3;
let C = 6;
// Function that print matrix in
// reverse spiral form.
function ReversespiralPrint(m, n, a)
{
// Large array to initialize it
// with elements of matrix
let b = new Array(100);
/* k - starting row index
l - starting column index*/
let i, k = 0, l = 0;
// Counter for single dimension array
//in which elements will be stored
let z = 0;
// Total elements in matrix
let size = m*n;
while (k < m && l < n)
{
// Variable to store value of matrix.
let val;
/* Print the first row from
the remaining rows */
for (i = l; i < n; ++i)
{
// printf("%d ", a[k][i]);
val = a[k][i];
b[z] = val;
++z;
}
k++;
/* Print the last column from
the remaining columns */
for (i = k; i < m; ++i)
{
// printf("%d ", a[i][n-1]);
val = a[i][n-1];
b[z] = val;
++z;
}
n--;
/* Print the last row from the
remaining rows */
if ( k < m)
{
for (i = n-1; i >= l; --i)
{
// printf("%d ", a[m-1][i]);
val = a[m-1][i];
b[z] = val;
++z;
}
m--;
}
/* Print the first column from the
remaining columns */
if (l < n)
{
for (i = m-1; i >= k; --i)
{
// printf("%d ", a[i][l]);
val = a[i][l];
b[z] = val;
++z;
}
l++;
}
}
for (let i=size-1 ; i>=0 ; --i)
{
document.write(b[i] + " ");
}
}
/* Driver program to test above functions */
let a = [ [1, 2, 3, 4, 5, 6],
[7, 8, 9, 10, 11, 12],
[13, 14, 15, 16, 17, 18]];
ReversespiralPrint(R, C, a);
</script>
Producción
11 10 9 8 7 13 14 15 16 17 18 12 6 5 4 3 2 1
Complejidad del Tiempo: O(m*n), Para recorrer la array O(m*n) se requiere tiempo.
Espacio Auxiliar: O(1), No se requiere espacio adicional.
Este artículo es una contribución de Sahil Rajput . Si te gusta GeeksforGeeks y te gustaría contribuir, también puedes escribir un artículo usando write.geeksforgeeks.org o enviar tu artículo por correo a review-team@geeksforgeeks.org. Vea su artículo que aparece en la página principal de GeeksforGeeks y ayude a otros Geeks.
Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA