例如，假设有宽度和高度，6和4分别来自用户输入，并且存储在2D阵列中的输入（也来自用户输入）是：
0 1 2 2 1 0
1 0 0 0 0 1
1 0 0 0 0 1
0 1 1 1 1 0
有没有办法翻转x轴和y轴？ 我想要做的是改变
0 1 2 2 1 0
1 0 0 0 0 1
1 0 0 0 0 1
0 1 1 1 1 0
到
0 1 1 0
1 0 0 1
2 0 0 1
2 0 0 1
1 0 0 1
0 1 1 0
以下代码，
scanf("%d %d", &width, &height);
int board[height][width];
for(i = 0; i < height; i++)
{
for(j = 0; j < width; j++)
{
scanf("%d", &input);
board[i][j] = input;
}
}
并通过
for(i = 0; i < width; i++)
{
for(j = 0; j < height; j++)
{
printf("%d", board[j][i]);
}
printf("\n");
}
，这打印出我期望的输出，但它实际上并没有改变它的位置......我不能改变第一个编码部分，因为我已经用它来做其他的工作。无论如何通过添加其他方法或新板来解决问题？
有人可以帮帮我吗？如果有人帮助我，我将非常感激！ 感谢
答案 0 :(得分：0)
只需声明第二个数组，其指定的行数等于width，列数等于height，并将值从souirce数组复制到第二个数组。
例如
#include <stdio.h>
int main(void)
{
size_t height, width;
printf( "Enter the height and the width of the array: " );
scanf( "%zu %zu", &height, &width );
int a[height][width];
puts( "Enter values for elements of the array" );
for ( size_t i = 0; i < height; i++ )
{
printf( "%zu row: ", i + 1 );
for ( size_t j = 0; j < width; j++ ) scanf( "%d", &a[i][j] );
}
int b[width][height];
for ( size_t i = 0; i < height; i++ )
{
for ( size_t j = 0; j < width; j++ ) b[j][i] = a[i][j];
}
puts( "\nSource array is" );
for ( size_t i = 0; i < height; i++ )
{
for ( size_t j = 0; j < width; j++ ) printf( "%d ", a[i][j] );
putchar( '\n' );
}
puts( "\nReversed array is" );
for ( size_t i = 0; i < width; i++ )
{
for ( size_t j = 0; j < height; j++ ) printf( "%d ", b[i][j] );
putchar( '\n' );
}
return 0;
}
程序输出可能看起来像
Enter the height and the width of the array: 4 6
Enter values for elements of the array
1 row: 0 1 2 2 1 0
2 row: 1 0 0 0 0 1
3 row: 1 0 0 0 0 1
4 row: 0 1 1 1 1 0
Source array is
0 1 2 2 1 0
1 0 0 0 0 1
1 0 0 0 0 1
0 1 1 1 1 0
Reversed array is
0 1 1 0
1 0 0 1
2 0 0 1
2 0 0 1
1 0 0 1
0 1 1 0
另一种方法是动态分配第一个和辅助数组。 例如
#include <stdio.h>
#include <stdlib.h>
int main(void)
{
size_t height, width;
printf( "Enter the height and the width of the array: " );
scanf( "%zu %zu", &height, &width );
int **a = malloc( height * sizeof( int * ) );
for ( size_t i = 0; i < height; i++ ) a[i] = malloc( width * sizeof( int ) );
puts( "Enter values for elements of the array" );
for ( size_t i = 0; i < height; i++ )
{
printf( "%zu row: ", i + 1 );
for ( size_t j = 0; j < width; j++ ) scanf( "%d", &a[i][j] );
}
int **b = malloc( width * sizeof( int * ) );
for ( size_t i = 0; i < width; i++ ) b[i] = malloc( height * sizeof( int ) );
for ( size_t i = 0; i < height; i++ )
{
for ( size_t j = 0; j < width; j++ ) b[j][i] = a[i][j];
}
puts( "\nSource array is" );
for ( size_t i = 0; i < height; i++ )
{
for ( size_t j = 0; j < width; j++ ) printf( "%d ", a[i][j] );
putchar( '\n' );
}
for ( size_t i = 0; i < height; i++ ) free( a[i] );
free( a );
a = b;
puts( "\nReversed array is" );
for ( size_t i = 0; i < width; i++ )
{
for ( size_t j = 0; j < height; j++ ) printf( "%d ", a[i][j] );
putchar( '\n' );
}
for ( size_t i = 0; i < width; i++ ) free( a[i] );
free( a );
return 0;
}
答案 1 :(得分：0)
这可能是最简单的方法。避免在输入过程中复制和存储元素。但是，如果您无法更改输入过程，则必须使用以下方法复制元素：
void reverse_matrix(int c, int r, int board[][r], int board2[][c])
代码：
#include <stdio.h>
void print(int c, int r, int board[][r] )
{
for(int i = 0; i < c; i++)
{
for(int j = 0; j < r; j++)
{
printf("%d ", board[i][j]);
}
printf("\n");
}
}
void reverse_matrix(int c, int r, int board[][r], int board2[][c] )
{
for(int i = 0; i < c; i++)
{
for(int j = 0; j < r; j++)
{
board2[j][i] = board[i][j];
}
}
}
int main()
{
int width, height, input;
scanf("%d %d", &width, &height);
int board[height][width];
int board2[width][height];
for(int i = 0; i < height; i++)
{
for(int j = 0; j < width; j++)
{
scanf("%d", &input);
board[i][j] = input;
// board2[j][i] = input; // if you can add it
}
}
printf("First matrix:\n");
print(height, width, board);
reverse_matrix(height, width, board, board2); // alternative
printf("Second matrix:\n");
print(width, height, board2);
return 0;
}
输出：
2
3
1
2
3
4
5
6
First matrix:
1 2
3 4
5 6
Second matrix:
1 3 5
2 4 6