如何迭代四维链接的二维数据网格，就像它是一个二维数组？ [英] How can I iterate over a quadruple linked 2-dimensional grid of data as if it were a 2-dimensional array?

问题描述

我的网格结构是：

如何迭代四维链接的二维网格数据？ >

  typedef bool tile; 
 
 struct BLOCK; 
 typedef struct BLOCK block; 
 
 struct BLOCK {
 const block * to_the_left; 
 const block * above; 
 const block * to_the_right; 
 const block * below; 
 
瓦数据; 
}; 
 
 typedef struct {
 const block * start; 
}地图; 
  

我需要能够遍历这个网格，就像它是一个二维数组，所以我可以以围绕开始块的屏幕显示地图的瓦片。

PSS我最喜欢在C中看到一个解决方案（这是我为这个项目编写的代码），C ++，Haskell或者Java代码，因为这些都是我所熟悉的语言，但任何语言都很好。我只需要算法。

P.S.S.S。为了清楚起见，通过像二维数组一样迭代，我的意思是我需要获得一个索引到x和y位置作为变量。例如，我需要调用mvaddch（y，x，'＃'）。

解决方案

订单好像是二维数组

  BLOCK * process = upper_left_block; 
 BLOCK * leftmost = process; 
 while（true）
 {
 //  process = process-> to_the_right; 
 if（process == null）
 {
 process = leftmost-> below; 
 leftmost = process 
} 
 if（process == null）
 break; 
} 
  

How can I iterate over a quadruple linked 2-dimensional grid of data as if it were a 2-dimensional array?

My grid structure is:

typedef bool tile;

struct BLOCK;
typedef struct BLOCK block;

struct BLOCK {
 const block * to_the_left;
 const block * above;
 const block * to_the_right;
 const block * below;

 tile data;
};

typedef struct {
 const block * start;
} map;

I need to be able to iterate over this grid like it is a 2-dimensional array so I can display tiles of the map on the screen centering around the start block.

P.S.S. I would most preferable like to see a solution in C, (this is what I'm coding in for this project), C++, Haskell, or Java code as those are languages I know well, but any language is fine. I just need the algorithm.

P.S.S.S. For clarity, by iterate over like a 2-dimensional array I mean I need to get an index into the x and y position as variables. For example, I need to do call mvaddch(y,x,'#').

解决方案

this will iterate in the same order as if it were a 2 dimensional array

    BLOCK * process = upper_left_block;
    BLOCK * leftmost = process;
    while(true)
    {
        //<do stuff here>
        process = process->to_the_right;
        if(process == null)
        {
            process = leftmost->below;
            leftmost = process;
        }
        if(process == null)
            break;
    }

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