Dada una array nxn. En la array dada, debe imprimir los elementos de la array en el patrón de serpiente.
Ejemplos:
Input :mat[][] = { {10, 20, 30, 40},
{15, 25, 35, 45},
{27, 29, 37, 48},
{32, 33, 39, 50}};
Output : 10 20 30 40 45 35 25 15 27 29
37 48 50 39 33 32
Input :mat[][] = { {1, 2, 3},
{4, 5, 6},
{7, 8, 9}};
Output : 1 2 3 6 5 4 7 8 9
Recorremos todas las filas. Para cada fila, verificamos si es par o impar. Si es par, imprimimos de izquierda a derecha, de lo contrario imprimimos de derecha a izquierda.
Implementación:
C++
// C++ program to print matrix in snake order
#include <iostream>
#define M 4
#define N 4
using namespace std;
void print(int mat[M][N])
{
// Traverse through all rows
for (int i = 0; i < M; i++) {
// If current row is even, print from
// left to right
if (i % 2 == 0) {
for (int j = 0; j < N; j++)
cout << mat[i][j] << " ";
// If current row is odd, print from
// right to left
} else {
for (int j = N - 1; j >= 0; j--)
cout << mat[i][j] << " ";
}
}
}
// Driver code
int main()
{
int mat[M][N] = { { 10, 20, 30, 40 },
{ 15, 25, 35, 45 },
{ 27, 29, 37, 48 },
{ 32, 33, 39, 50 } };
print(mat);
return 0;
}
Java
// Java program to print matrix in snake order
import java.util.*;
class GFG
{
static void print(int [][] mat)
{
// Traverse through all rows
for (int i = 0; i < mat.length; i++)
{
// If current row is even, print from
// left to right
if (i % 2 == 0)
{
for (int j = 0; j < mat[0].length; j++)
System.out.print(mat[i][j] +" ");
// If current row is odd, print from
// right to left
}
else
{
for (int j = mat[0].length - 1; j >= 0; j--)
System.out.print(mat[i][j] +" ");
}
}
}
// Driver code
public static void main(String[] args)
{
int mat[][] = new int[][]
{
{ 10, 20, 30, 40 },
{ 15, 25, 35, 45 },
{ 27, 29, 37, 48 },
{ 32, 33, 39, 50 }
};
print(mat);
}
}
/* This code is contributed by Mr. Somesh Awasthi */
Python3
# Python 3 program to print # matrix in snake order M = 4 N = 4 def printf(mat): global M, N # Traverse through all rows for i in range(M): # If current row is # even, print from # left to right if i % 2 == 0: for j in range(N): print(str(mat[i][j]), end = " ") # If current row is # odd, print from # right to left else: for j in range(N - 1, -1, -1): print(str(mat[i][j]), end = " ") # Driver code mat = [[ 10, 20, 30, 40 ], [ 15, 25, 35, 45 ], [ 27, 29, 37, 48 ], [ 32, 33, 39, 50 ]] printf(mat) # This code is contributed # by ChitraNayal
C#
// C# program to print
// matrix in snake order
using System;
class GFG
{
static void print(int [,]mat)
{
// Traverse through all rows
for (int i = 0;
i < mat.GetLength(0); i++)
{
// If current row is
// even, print from
// left to right
if (i % 2 == 0)
{
for (int j = 0;
j < mat.GetLength(1); j++)
Console.Write(mat[i, j] + " ");
// If current row is
// odd, print from
// right to left
}
else
{
for (int j = mat.GetLength(1) - 1;
j >= 0; j--)
Console.Write(mat[i, j] + " ");
}
}
}
// Driver code
public static void Main()
{
int [,]mat = {{ 10, 20, 30, 40 },
{ 15, 25, 35, 45 },
{ 27, 29, 37, 48 },
{ 32, 33, 39, 50 }};
print(mat);
}
}
// This code is contributed
// by ChitraNayal
PHP
<?php
// PHP program to print
// matrix in snake order
$M= 4;
$N =4;
function printLN($mat)
{
global $M;
global $N;
// Traverse through all rows
for ($i = 0; $i < $M; $i++)
{
// If current row is even,
// print from left to right
if ($i % 2 == 0)
{
for ($j = 0; $j < $N; $j++)
echo $mat[$i][$j], " ";
// If current row is odd,
// print from right to left
}
else
{
for ($j = $N - 1; $j >= 0; $j--)
echo $mat[$i][$j] , " ";
}
}
}
// Driver code
$mat = array(array(10, 20, 30, 40),
array(15, 25, 35, 45),
array(27, 29, 37, 48),
array(32, 33, 39, 50));
printLN($mat);
// This code is contributed by ajit
?>
Javascript
<script>
// Javascript program to print
// matrix in snake order
let M = 4;
let N = 4;
function print(mat)
{
// Traverse through all rows
for(let i = 0; i < M; i++)
{
// If current row is even, print from
// left to right
if (i % 2 == 0)
{
for(let j = 0; j < N; j++)
document.write(mat[i][j] + " ");
// If current row is odd, print from
// right to left
}
else
{
for(let j = N - 1; j >= 0; j--)
document.write(mat[i][j] + " ");
}
}
}
// Driver code
let mat = [ [ 10, 20, 30, 40 ],
[ 15, 25, 35, 45 ],
[ 27, 29, 37, 48 ],
[ 32, 33, 39, 50 ] ];
print(mat);
// This code is contributed by rameshtravel07
</script>
Producción
10 20 30 40 45 35 25 15 27 29 37 48 50 39 33 32
Complejidad de Tiempo: O(nxm)
Espacio Auxiliar: O(1)
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Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA