通过按顺序交替颠倒行和列来将给定的矩阵洗牌 K 次
给定一个大小为M * N的矩阵arr[][] ，其中M是行数， N是列数，以及一个整数K。任务是在K步中对该矩阵进行混洗，以便在每一步中，矩阵的行和列交替反转，即从第一步反转第一行开始，然后在第二步中反转第一列，然后在第二步中反转第二行第三步以此类推。
注意：如果K超过M或N ，则反转再次从第一行或第一列开始。
例子：
Input: arr[][] = {{3, 4, 1, 8},
{11, 23, 43, 21},
{12, 17, 65, 91},
{71, 56, 34, 24}}
K = 4
Output: {{71, 56, 4, 3},
{21, 17, 23, 12},
{11, 43, 65, 91},
{8, 1, 34, 24}}
Explanation: Steps to be followed:
Reverse the first row
Reverse the first column
Reverse the second row
Reverse the second column
Input: arr[][] = {{11, 23, 43, 21},
{12, 17, 65, 91},
{71, 56, 34, 24}}
K = 8
Output: {{11, 43, 56, 21},
{91, 65, 17, 12},
{24, 34, 23, 71}}
方法：这个问题可以通过简单地运行两个循环来交替遍历行和列来解决，直到K变为0。最后，打印结果矩阵。请按照以下步骤操作：
- 运行一个循环遍历矩阵。
- 在每次迭代中反转正确的行和列。
- 执行此操作 K 次。
- 打印最后一个循环。
下面是上述方法的实现
C++
// C++ code to implement the above approach
#include
using namespace std;
const int M = 3;
const int N = 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 shuffle matrix
void reverseAlternate(int arr[][N], int K)
{
int turn = 0;
int row = 0, col = 0;
while (turn < K) {
// Reverse the row
if (turn % 2 == 0) {
int start = 0, end = N - 1, temp;
while (start < end) {
temp = arr[row % M][start];
arr[row % M][start] =
arr[row % M][end];
arr[row % M][end] = temp;
start += 1;
end -= 1;
}
row++;
turn++;
}
// Reverse the column
else {
int start = 0, end = M - 1, temp;
while (start < end) {
temp = arr[start][col % N];
arr[start][col % N] =
arr[end][col % N];
arr[end][col % N] = temp;
start += 1;
end -= 1;
}
col++;
turn++;
}
}
}
// Driver code
int main()
{
int matrix[M][N] = { { 11, 23, 43, 21 },
{ 12, 17, 65, 91 },
{ 71, 56, 34, 24 } };
int K = 8;
reverseAlternate(matrix, K);
showArray(matrix);
return 0;
} Java
// Java code to implement the above approach
import java.io.*;
class GFG {
public static int M = 3;
public static int N = 4;
// Print matrix elements
public 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 shuffle matrix
public static void reverseAlternate(int arr[][], int K)
{
int turn = 0;
int row = 0, col = 0;
while (turn < K) {
// Reverse the row
if (turn % 2 == 0) {
int start = 0, end = N - 1, temp;
while (start < end) {
temp = arr[row % M][start];
arr[row % M][start] =
arr[row % M][end];
arr[row % M][end] = temp;
start += 1;
end -= 1;
}
row++;
turn++;
}
// Reverse the column
else {
int start = 0, end = M - 1, temp;
while (start < end) {
temp = arr[start][col % N];
arr[start][col % N] =
arr[end][col % N];
arr[end][col % N] = temp;
start += 1;
end -= 1;
}
col++;
turn++;
}
}
}
// Driver code
public static void main (String[] args)
{
int matrix[][] = { { 11, 23, 43, 21 },
{ 12, 17, 65, 91 },
{ 71, 56, 34, 24 } };
int K = 8;
reverseAlternate(matrix, K);
showArray(matrix);
}
}
// This code is contributed by Shubham SinghPython3
# Python code to implement the above approach
M = 3
N = 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 shuffle matrix
def reverseAlternate(arr, K):
turn = 0
row = 0
col = 0
while (turn < K):
# Reverse the row
if (turn % 2 == 0):
start = 0
end = N - 1
temp = 0
while (start < end):
temp = arr[row % M][start]
arr[row % M][start] = arr[row % M][end]
arr[row % M][end] = temp
start += 1
end -= 1
row += 1
turn += 1
#Reverse the column
else:
start = 0
end = M - 1
temp = 0
while (start < end):
temp = arr[start][col % N]
arr[start][col % N] = arr[end][col % N]
arr[end][col % N] = temp
start += 1
end -= 1
col += 1
turn += 1
# Driver code
matrix = [[11, 23, 43, 21] ,
[12, 17, 65, 91] ,
[71, 56, 34, 24]]
K = 8
reverseAlternate(matrix, K)
showArray(matrix)
# This code is contributed by Shubham SinghC#
// C# code to implement the above approach
using System;
class GFG {
public static int M = 3;
public static int N = 4;
// Print matrix elements
public 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 shuffle matrix
public static void reverseAlternate(int[, ] arr, int K)
{
int turn = 0;
int row = 0, col = 0;
while (turn < K) {
// Reverse the row
if (turn % 2 == 0) {
int start = 0, end = N - 1, temp;
while (start < end) {
temp = arr[row % M, start];
arr[row % M, start] = arr[row % M, end];
arr[row % M, end] = temp;
start += 1;
end -= 1;
}
row++;
turn++;
}
// Reverse the column
else {
int start = 0, end = M - 1, temp;
while (start < end) {
temp = arr[start, col % N];
arr[start, col % N] = arr[end, col % N];
arr[end, col % N] = temp;
start += 1;
end -= 1;
}
col++;
turn++;
}
}
}
// Driver code
public static void Main(string[] args)
{
int[, ] matrix = { { 11, 23, 43, 21 },
{ 12, 17, 65, 91 },
{ 71, 56, 34, 24 } };
int K = 8;
reverseAlternate(matrix, K);
showArray(matrix);
}
}
// This code is contributed by ukasp.Javascript
输出
11 43 56 21
91 65 17 12
24 34 23 71 时间复杂度： O(K * max(M, N))
辅助空间： O(1)