Dada una array arr[][] de tamaño M*N , donde M es el número de filas y N es el número de columnas . La tarea es invertir las filas y columnas de la array alternativamente , es decir, comenzar invirtiendo la primera fila, luego la segunda columna, y así sucesivamente.
Ejemplos :
Entrada : arr[][] = {
{3, 4, 1, 8},
{11, 23, 43, 21},
{12, 17, 65, 91},
{71, 56, 34, 24}
}
Salida : {
{8, 56, 4, 24},
{11, 17, 43, 12},
{91, 65, 23, 21},
{71, 1, 34, 3}
}
Explicación : Operaciones a seguir:
- Invertir la primera fila
- Invertir la segunda columna
- Invertir la tercera fila
- Invertir la cuarta fila
Entrada : { {11, 23, 43, 21}, {12, 17, 65, 91}, {71, 56, 34, 24} }
Salida : { {21, 56, 23, 71}, {12, 17 , 65, 91}, {24, 34, 43, 11} }
Enfoque: la tarea se puede resolver simplemente ejecutando dos bucles while para recorrer filas y columnas alternativamente. Al final, imprima la array resultante.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program to find Reverse the
// rows and columns of a matrix alternatively.
#include <bits/stdc++.h>
using namespace std;
const int N = 4;
const int M = 4;
// Print matrix elements
void showArray(int arr[][N])
{
for (int i = 0; i < M; i++) {
for (int j = 0; j < N; j++)
cout << arr[i][j] << " ";
cout << endl;
}
}
// Function to Reverse the rows and columns
// of a matrix alternatively.
void reverseAlternate(int arr[][N])
{
int turn = 0;
while (turn < M && turn < N) {
if (turn % 2 == 0) {
int start = 0, end = N - 1, temp;
while (start < end) {
temp = arr[turn][start];
arr[turn][start] = arr[turn][end];
arr[turn][end] = temp;
start += 1;
end -= 1;
}
turn += 1;
}
if (turn % 2 == 1) {
int start = 0, end = M - 1, temp;
while (start < end) {
temp = arr[start][turn];
arr[start][turn] = arr[end][turn];
arr[end][turn] = temp;
start += 1;
end -= 1;
}
turn += 1;
}
}
}
// Driver code
int main()
{
int matrix[][N] = { { 3, 4, 1, 8 },
{ 11, 23, 43, 21 },
{ 12, 17, 65, 91 },
{ 71, 56, 34, 24 } };
reverseAlternate(matrix);
showArray(matrix);
}
Java
// Java program for the above approach
import java.util.*;
public class GFG
{
static int N = 4;
static int M = 4;
// Print matrix elements
static void showArray(int arr[][])
{
for (int i = 0; i < M; i++) {
for (int j = 0; j < N; j++)
System.out.print(arr[i][j] + " ");
System.out.println();
}
}
// Function to Reverse the rows and columns
// of a matrix alternatively.
static void reverseAlternate(int arr[][])
{
int turn = 0;
while (turn < M && turn < N) {
if (turn % 2 == 0) {
int start = 0, end = N - 1, temp;
while (start < end) {
temp = arr[turn][start];
arr[turn][start] = arr[turn][end];
arr[turn][end] = temp;
start += 1;
end -= 1;
}
turn += 1;
}
if (turn % 2 == 1) {
int start = 0, end = M - 1, temp;
while (start < end) {
temp = arr[start][turn];
arr[start][turn] = arr[end][turn];
arr[end][turn] = temp;
start += 1;
end -= 1;
}
turn += 1;
}
}
}
// Driver Code
public static void main(String args[])
{
int matrix[][] = { { 3, 4, 1, 8 },
{ 11, 23, 43, 21 },
{ 12, 17, 65, 91 },
{ 71, 56, 34, 24 } };
reverseAlternate(matrix);
showArray(matrix);
}
}
// This code is contributed by Samim Hossain Mondal.
Python3
# Python3 program to find Reverse the # rows and columns of a matrix alternatively. N = 4 M = 4 # Print matrix elements def showArray(arr): for i in range(M): for j in range(N): print(arr[i][j], end = " ") print() # Function to Reverse the rows and columns # of a matrix alternatively. def reverseAlternate(arr): turn = 0 while turn < M and turn < N: if (turn % 2 == 0): start = 0 end = N - 1 while (start < end): temp = arr[turn][start] arr[turn][start] = arr[turn][end] arr[turn][end] = temp start += 1 end -= 1 turn += 1 if (turn % 2 == 1): start = 0 end = M - 1 while (start < end): temp = arr[start][turn] arr[start][turn] = arr[end][turn] arr[end][turn] = temp start += 1 end -= 1 turn += 1 # Driver code matrix = [ [ 3, 4, 1, 8 ], [ 11, 23, 43, 21 ], [ 12, 17, 65, 91 ], [ 71, 56, 34, 24 ] ] reverseAlternate(matrix) showArray(matrix) # This code is contributed by Potta Lokesh
C#
// C# program to find Reverse the
// rows and columns of a matrix alternatively.
using System;
class GFG {
const int N = 4;
const int M = 4;
// Print matrix elements
static void showArray(int[, ] arr)
{
for (int i = 0; i < M; i++) {
for (int j = 0; j < N; j++)
Console.Write(arr[i, j] + " ");
Console.WriteLine();
}
}
// Function to Reverse the rows and columns
// of a matrix alternatively.
static void reverseAlternate(int[, ] arr)
{
int turn = 0;
while (turn < M && turn < N) {
if (turn % 2 == 0) {
int start = 0, end = N - 1, temp;
while (start < end) {
temp = arr[turn, start];
arr[turn, start] = arr[turn, end];
arr[turn, end] = temp;
start += 1;
end -= 1;
}
turn += 1;
}
if (turn % 2 == 1) {
int start = 0, end = M - 1, temp;
while (start < end) {
temp = arr[start, turn];
arr[start, turn] = arr[end, turn];
arr[end, turn] = temp;
start += 1;
end -= 1;
}
turn += 1;
}
}
}
// Driver code
public static void Main()
{
int[, ] matrix = { { 3, 4, 1, 8 },
{ 11, 23, 43, 21 },
{ 12, 17, 65, 91 },
{ 71, 56, 34, 24 } };
reverseAlternate(matrix);
showArray(matrix);
}
}
// This code is contributed by ukasp.
Javascript
<script>
// JavaScript program to find Reverse the
// rows and columns of a matrix alternatively.
const N = 4;
const M = 4;
// Print matrix elements
const showArray = (arr) => {
for (let i = 0; i < M; i++) {
for (let j = 0; j < N; j++)
document.write(`${arr[i][j]} `);
document.write("<br/>");
}
}
// Function to Reverse the rows and columns
// of a matrix alternatively.
const reverseAlternate = (arr) => {
let turn = 0;
while (turn < M && turn < N) {
if (turn % 2 == 0) {
let start = 0, end = N - 1, temp;
while (start < end) {
temp = arr[turn][start];
arr[turn][start] = arr[turn][end];
arr[turn][end] = temp;
start += 1;
end -= 1;
}
turn += 1;
}
if (turn % 2 == 1) {
let start = 0, end = M - 1, temp;
while (start < end) {
temp = arr[start][turn];
arr[start][turn] = arr[end][turn];
arr[end][turn] = temp;
start += 1;
end -= 1;
}
turn += 1;
}
}
}
// Driver code
let matrix = [[3, 4, 1, 8],
[11, 23, 43, 21],
[12, 17, 65, 91],
[71, 56, 34, 24]];
reverseAlternate(matrix);
showArray(matrix);
// This code is contributed by rakeshsahni
</script>
Producción
8 56 4 24 11 17 43 12 91 65 23 21 71 1 34 3
Complejidad de tiempo : O(M*N)
Complejidad de espacio : O(1), no se utiliza espacio extra adicional.
Publicación traducida automáticamente
Artículo escrito por pintusaini y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA