从中心以顺时针方向扩展螺旋打印二维数组 [英] Print 2-D Array in clockwise expanding spiral from center

问题描述

我保证是一个完美的方阵.我想从矩阵的中心开始，在这种情况下它是 matrix[2][2]，我知道如何计算中心 (int)(dimensions/2).我需要在以下向外螺旋图案中输出数组的内容.当然，该算法应该适用于任何完美的方阵.我不确定这个算法是否已经存在，我不想重新发明轮子.

I have an guaranteed to be a perfect square matrix. I want to start at the center of the matrix in this case it would be matrix[2][2], I know how to figure the center (int)(dimensions / 2). I need to output the contents of the array in this following outward spiral pattern. Of course the algorithm should work with any perfect square matrix. I wasn't sure if this algorithm already existed and I didn't want to re-invent the wheel.

int dimensions / 2;

21 22 23 24 25
20 7  8  9  10
19 6  1  2  11
18 5  4  3  12 
17 16 15 14 13

这个例子的输出应该是

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

推荐答案

让我们先确定模式..

偶数方阵，示例:4x4

Even-Size Square Matrix, Example: 4x4

  07 > 08 > 09 > 10
  ^               v
  06  (01)> 02   11
  ^          v    v
  05 < 04 < 03   12
                  v
 [16]< 15 < 14 < 13

Starting Point: [2, 2] ~ [SIZE/2, SIZE/2]

Ending Point: [4, 1] ~ [SIZE, 1]

Chains: Count(K-chain)=2 for K = 1..(SIZE-2)
        + 3 for Count = SIZE-1

奇数方阵，示例:5x5

Odd-Size Square Matrix, Example: 5x5

  21 > 22 > 23 > 24 >[25]
  ^
  20   07 > 08 > 09 > 10
  ^    ^               v
  19   06  (01)> 02   11
  ^    ^          v    v
  18   05 < 04 < 03   12 
  ^                    v
  17 < 16 < 15 < 14 < 13

Starting Point: [2, 2] ~ [⌊SIZE/2⌋, ⌊SIZE/2⌋]

Ending Point: [1, 5] ~ [1, SIZE]

Chains: Count(K-chain)=2 for K = 1..(SIZE-2)
        + 3 for Count = SIZE-1

实时代码

void print_spiral (int ** matrix, int size)
{
    int x = 0; // current position; x
    int y = 0; // current position; y
    int d = 0; // current direction; 0=RIGHT, 1=DOWN, 2=LEFT, 3=UP
    int c = 0; // counter
    int s = 1; // chain size

    // starting point
    x = ((int)floor(size/2.0))-1;
    y = ((int)floor(size/2.0))-1;

    for (int k=1; k<=(size-1); k++)
    {
        for (int j=0; j<(k<(size-1)?2:3); j++)
        {
            for (int i=0; i<s; i++)
            {
                std::cout << matrix[x][y] << " ";
                c++;

                switch (d)
                {
                    case 0: y = y + 1; break;
                    case 1: x = x + 1; break;
                    case 2: y = y - 1; break;
                    case 3: x = x - 1; break;
                }
            }
            d = (d+1)%4;
        }
        s = s + 1;
    }
}

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