输出双螺旋矩阵

实验室师兄找工作时在有道遇到的一道笔试题：打印双螺旋矩阵。
双螺旋矩阵的定义如下，矩阵的最中心是1，往上是2，右拐3，向下4，然后依次5、6，7...构成一条顺序增大的螺旋线，此外，如果从中心往下走的话，也是一条对称的螺旋线。题目是给定一个矩阵维度N，将其打印出来，示例如下。要求在纸上把代码写完整，时间半小时左右。

25 14 15 16 17 18 19

24 13 6 7 8 9 20

23 12 5 2 3 10 21

22 11 4 1 4 11 22

21 10 3 2 5 12 23

20 9 8 7 6 13 24

19 18 17 16 15 14 25

看上去似乎挺简单，但是做的时候还是遇到了一些麻烦，最后参考师兄的思路，我给出了下面的实现。基本思路是，先打印中心，然后一圈一圈地将外围矩阵打印出来。打印每圈的时候，先分别计算出左上角和右下角的坐标（Pos）和值，然后沿着螺旋线的方向，从左上角出发打印这一圈矩阵的上边和右边；从右下角出发打印下边和左边。代码如下，能够正确运行，欢迎提出意见。

#include <stdio.h>

#include <stdlib.h>

#include <math.h>

typedef struct {

int row;

int column;

} Position;

void print_screws_matrix( int size )

{

int **matrix = (int **) malloc( size*sizeof(int*) );

int i,j; // just for loop

for( i = 0; i < size; ++i )

matrix[i] = (int *) malloc( size*sizeof(int) );

/* initialize matrix center */

Position centerPos;

centerPos.row = ( size - 1 )/2;

centerPos.column = ( size - 1 )/2;

matrix[ centerPos.row ][ centerPos.column ] = 1;

int screwNum;

/* screwNum of matrix center is 0 */

for( screwNum = 1; screwNum <= (size-1)/2; ++screwNum ) {

/* get the beginning positions for each screw */

Position screwUpRightPos, screwDownLeftPos;

screwUpRightPos.row = centerPos.row - screwNum;

screwUpRightPos.column = centerPos.column - screwNum + 1;

screwDownLeftPos.row = centerPos.row + screwNum;

screwDownLeftPos.column = centerPos.column + screwNum - 1;

/* initialize beginning positions */

matrix[ screwUpRightPos.row ][ screwUpRightPos.column ]

= matrix[ screwUpRightPos.row + 1 ][ screwUpRightPos.column ] + 1;

matrix[ screwDownLeftPos.row ][ screwDownLeftPos.column ]

= matrix[ screwDownLeftPos.row - 1 ][ screwDownLeftPos.column ] + 1;

/* calculate up and down sides of screw */

for( i = 0; i < 2*screwNum - 1; ++i ) {

screwUpRightPos.column++;

matrix[ screwUpRightPos.row ][ screwUpRightPos.column ]

= matrix[ screwUpRightPos.row][ screwUpRightPos.column - 1 ] + 1;

/* calculate up and down sides of screw */

for( i = 0; i < 2*screwNum - 1; ++i ) {

screwUpRightPos.column++;

matrix[ screwUpRightPos.row ][ screwUpRightPos.column ]

= matrix[ screwUpRightPos.row][ screwUpRightPos.column - 1 ] + 1;

screwDownLeftPos.column--;

matrix[ screwDownLeftPos.row ][ screwDownLeftPos.column ]

= matrix[ screwDownLeftPos.row ][ screwDownLeftPos.column + 1 ] + 1;

}

/* calculate right and left sides of screw */

for( i = 0; i < 2*screwNum; ++i ) {

screwUpRightPos.row++;

matrix[ screwUpRightPos.row ][ screwUpRightPos.column ]

= matrix[ screwUpRightPos.row - 1][ screwUpRightPos.column ] + 1;

screwDownLeftPos.row--;

matrix[ screwDownLeftPos.row ][ screwDownLeftPos.column ]

= matrix[ screwDownLeftPos.row + 1][ screwDownLeftPos.column ] + 1;

}

}

/* print matrix */

for( i = 0; i < size; ++i ) {

for( j = 0; j < size; ++j )

printf( "%4d ",matrix[ i ][ j ]);

printf( "/n" );

}

for( i = 0; i < size; ++i )

free( matrix[i] );

free( matrix );

}

int main()

{

print_screws_matrix( 21 );

return 0;

}

参考：http://blog.csdn.net/dennis101/article/details/3053739