Dada una array 2D, la tarea es imprimir la array en forma antiespiral:
Ejemplos:
Salida: 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1
Input : arr[][4] = {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 :arr[][6] = {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
La idea es simple, atravesamos la array en forma de espiral y colocamos todos los elementos atravesados en una pila. Finalmente, uno por uno los elementos de la pila e imprímalos.
Implementación:
C++
// C++ program to print matrix in anti-spiral form
#include <bits/stdc++.h>
using namespace std;
#define R 4
#define C 5
void antiSpiralTraversal(int m, int n, int a[R][C])
{
int i, k = 0, l = 0;
/* k - starting row index
m - ending row index
l - starting column index
n - ending column index
i - iterator */
stack<int> stk;
while (k <= m && l <= n)
{
/* Print the first row from the remaining rows */
for (i = l; i <= n; ++i)
stk.push(a[k][i]);
k++;
/* Print the last column from the remaining columns */
for (i = k; i <= m; ++i)
stk.push(a[i][n]);
n--;
/* Print the last row from the remaining rows */
if ( k <= m)
{
for (i = n; i >= l; --i)
stk.push(a[m][i]);
m--;
}
/* Print the first column from the remaining columns */
if (l <= n)
{
for (i = m; i >= k; --i)
stk.push(a[i][l]);
l++;
}
}
while (!stk.empty())
{
cout << stk.top() << " ";
stk.pop();
}
}
/* Driver program to test above functions */
int main()
{
int mat[R][C] =
{
{1, 2, 3, 4, 5},
{6, 7, 8, 9, 10},
{11, 12, 13, 14, 15},
{16, 17, 18, 19, 20}
};
antiSpiralTraversal(R-1, C-1, mat);
return 0;
}
Java
// Java Code for Print matrix in antispiral form
import java.util.*;
class GFG {
public static void antiSpiralTraversal(int m, int n,
int a[][])
{
int i, k = 0, l = 0;
/* k - starting row index
m - ending row index
l - starting column index
n - ending column index
i - iterator */
Stack<Integer> stk=new Stack<Integer>();
while (k <= m && l <= n)
{
/* Print the first row from the remaining
rows */
for (i = l; i <= n; ++i)
stk.push(a[k][i]);
k++;
/* Print the last column from the remaining
columns */
for (i = k; i <= m; ++i)
stk.push(a[i][n]);
n--;
/* Print the last row from the remaining
rows */
if ( k <= m)
{
for (i = n; i >= l; --i)
stk.push(a[m][i]);
m--;
}
/* Print the first column from the remaining
columns */
if (l <= n)
{
for (i = m; i >= k; --i)
stk.push(a[i][l]);
l++;
}
}
while (!stk.empty())
{
System.out.print(stk.peek() + " ");
stk.pop();
}
}
/* Driver program to test above function */
public static void main(String[] args)
{
int mat[][] =
{
{1, 2, 3, 4, 5},
{6, 7, 8, 9, 10},
{11, 12, 13, 14, 15},
{16, 17, 18, 19, 20}
};
antiSpiralTraversal(mat.length - 1, mat[0].length - 1,
mat);
}
}
// This code is contributed by Arnav Kr. Mandal.
Python 3
# Python 3 program to print # matrix in anti-spiral form R = 4 C = 5 def antiSpiralTraversal(m, n, a): k = 0 l = 0 # k - starting row index # m - ending row index # l - starting column index # n - ending column index # i - iterator stk = [] while (k <= m and l <= n): # Print the first row # from the remaining rows for i in range(l, n + 1): stk.append(a[k][i]) k += 1 # Print the last column # from the remaining columns for i in range(k, m + 1): stk.append(a[i][n]) n -= 1 # Print the last row # from the remaining rows if ( k <= m): for i in range(n, l - 1, -1): stk.append(a[m][i]) m -= 1 # Print the first column # from the remaining columns if (l <= n): for i in range(m, k - 1, -1): stk.append(a[i][l]) l += 1 while len(stk) != 0: print(str(stk[-1]), end = " ") stk.pop() # Driver Code mat = [[1, 2, 3, 4, 5], [6, 7, 8, 9, 10], [11, 12, 13, 14, 15], [16, 17, 18, 19, 20]]; antiSpiralTraversal(R - 1, C - 1, mat) # This code is contributed # by ChitraNayal
C#
using System;
using System.Collections.Generic;
// C# Code for Print matrix in antispiral form
public class GFG
{
public static void antiSpiralTraversal(int m, int n, int[][] a)
{
int i, k = 0, l = 0;
/* k - starting row index
m - ending row index
l - starting column index
n - ending column index
i - iterator */
Stack<int> stk = new Stack<int>();
while (k <= m && l <= n)
{
/* Print the first row from the remaining
rows */
for (i = l; i <= n; ++i)
{
stk.Push(a[k][i]);
}
k++;
/* Print the last column from the remaining
columns */
for (i = k; i <= m; ++i)
{
stk.Push(a[i][n]);
}
n--;
/* Print the last row from the remaining
rows */
if (k <= m)
{
for (i = n; i >= l; --i)
{
stk.Push(a[m][i]);
}
m--;
}
/* Print the first column from the remaining
columns */
if (l <= n)
{
for (i = m; i >= k; --i)
{
stk.Push(a[i][l]);
}
l++;
}
}
while (stk.Count > 0)
{
Console.Write(stk.Peek() + " ");
stk.Pop();
}
}
/* Driver program to test above function */
public static void Main(string[] args)
{
int[][] mat = new int[][]
{
new int[] {1, 2, 3, 4, 5},
new int[] {6, 7, 8, 9, 10},
new int[] {11, 12, 13, 14, 15},
new int[] {16, 17, 18, 19, 20}
};
antiSpiralTraversal(mat.Length - 1, mat[0].Length - 1, mat);
}
}
// This code is contributed by Shrikant13
Javascript
<script>
// Javascript Code for Print matrix in antispiral form
function antiSpiralTraversal(m,n,a)
{
let i, k = 0, l = 0;
/* k - starting row index
m - ending row index
l - starting column index
n - ending column index
i - iterator */
let stk=[];
while (k <= m && l <= n)
{
/* Print the first row from the remaining
rows */
for (i = l; i <= n; ++i)
stk.push(a[k][i]);
k++;
/* Print the last column from the remaining
columns */
for (i = k; i <= m; ++i)
stk.push(a[i][n]);
n--;
/* Print the last row from the remaining
rows */
if ( k <= m)
{
for (i = n; i >= l; --i)
stk.push(a[m][i]);
m--;
}
/* Print the first column from the remaining
columns */
if (l <= n)
{
for (i = m; i >= k; --i)
stk.push(a[i][l]);
l++;
}
}
while (stk.length!=0)
{
document.write(stk[stk.length-1] + " ");
stk.pop();
}
}
/* Driver program to test above function */
let mat = [[1, 2, 3, 4, 5],
[6, 7, 8, 9, 10],
[11, 12, 13, 14, 15],
[16, 17, 18, 19, 20]];
antiSpiralTraversal(mat.length - 1, mat[0].length - 1,
mat);
// This code is contributed by avanitrachhadiya2155
</script>
Producción
12 13 14 9 8 7 6 11 16 17 18 19 20 15 10 5 4 3 2 1
Complejidad temporal: O(R + C).
Espacio Auxiliar: O(R + C).
- Recorrido diagonal de Matrix
- Array de impresión en forma de espiral
- Imprimir una array dada en forma de zigzag
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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