8

i wanted ask if there some algorithm ready, that allowed me to do this: i have a matrix m (col) x n (row) with m x n elements. I want give position to this element starting from center and rotating as a spiral, for example, for a matrix 3x3 i have 9 elements so defined:

 5  6  7
 4  9  8
 3  2  1

or for una matrix 4 x 3 i have 12 elements, do defined:

  8   9  10  1
  7  12  11  2
  6   5   4  3

or again, a matrix 5x2 i have 10 elements so defined:

    3  4
    7  8
   10  9
    6  5 
    2  1

etc. I have solved basically defining a array of integer of m x n elements and loading manually the value, but in generel to me like that matrix maked from algorithm automatically. Thanks to who can help me to find something so, thanks very much.

UPDATE

This code, do exactely about i want have, but not is in delphi; just only i need that start from 1 and not from 0. Important for me is that it is valid for any matrics m x n. Who help me to translate it in delphi?

(defun spiral (rows columns)  
  (do ((N (* rows columns))       
    (spiral (make-array (list rows columns) :initial-element nil))      
    (dx 1) (dy 0) (x 0) (y 0)    
   (i 0 (1+ i)))     
 ((= i N) spiral)   
 (setf (aref spiral y x) i)
    (let ((nx (+ x dx)) (ny (+ y dy)))  
    (cond       
       ((and (< -1 nx columns)       
       (< -1 ny rows)           
       (null (aref spiral ny nx)))    
    (setf x nx 
          y ny))  
     (t (psetf dx (- dy)                 
       dy dx)       
    (setf x (+ x dx)             
     y (+ y dy))))))) 


> (pprint (spiral 6 6))

#2A ((0  1  2  3  4  5)
    (19 20 21 22 23  6)
    (18 31 32 33 24  7)
    (17 30 35 34 25  8)
    (16 29 28 27 26  9)
    (15 14 13 12 11 10))


> (pprint (spiral 5 3))

#2A  ((0  1 2)
     (11 12 3)
     (10 13 4)
      (9 14 5)
      (8 7 6))

Thanks again very much.

4

4 回答 4

4

基于经典的螺旋算法。支持非方阵：

program SpiralMatrix;
{$APPTYPE CONSOLE}
uses
  SysUtils;

type
  TMatrix = array of array of Integer;

procedure PrintMatrix(const a: TMatrix);
var
  i, j: Integer;
begin
  for i := 0 to Length(a) - 1 do
  begin
    for j := 0 to Length(a[0]) - 1 do
      Write(Format('%3d', [a[i, j]]));
    Writeln;
  end;
end;

var
  spiral: TMatrix;
  i, m, n: Integer;
  row, col, dx, dy,
  dirChanges, visits, temp: Integer;
begin
  m := 3; // columns
  n := 3; // rows
  SetLength(spiral, n, m);
  row := 0;
  col := 0;
  dx := 1;
  dy := 0;
  dirChanges := 0;
  visits := m;
  for i := 0 to n * m - 1 do
  begin
    spiral[row, col] := i + 1;
    Dec(visits);
    if visits = 0 then
    begin
      visits := m * (dirChanges mod 2) + n * ((dirChanges + 1) mod 2) - (dirChanges div 2) - 1;
      temp := dx;
      dx := -dy;
      dy := temp;
      Inc(dirChanges);
    end;
    Inc(col, dx);
    Inc(row, dy);
  end;
  PrintMatrix(spiral);
  Readln;
end.

3×3：

1  2  3
8  9  4
7  6  5

4×3：

 1  2  3  4
10 11 12  5
 9  8  7  6

2×5：

 1  2
10  3
 9  4
 8  5
 7  6
于 2012-01-07T14:52:24.233 回答
1

给你！！！在 30 次语法错误之后...

在ideone.com 上，我进行了一些测试，它似乎运行良好。我认为您仍然可以在那里看到输出并自己运行它...

我在代码中添加了一些注释。足以理解大部分内容。主导航系统有点难以解释。简而言之，做一个螺旋是在第一个方向 1 次，第二个 1 次，第三个 2 次，第四个 2 次，第五个 3 次，3、4、4、5、5，等等。我使用我所谓的 aseed来step获得这种行为。

program test;

var
    w, h, m, n, v, d : integer; // Matrix size, then position, then value and direction.
    spiral : array of array of integer; // Matrix/spiral itself.
    seed, step : integer; // Used to travel the spiral.

begin
    readln(h);
    readln(w);
    setlength(spiral, h, w);
    v := w * h; // Value to put in spiral.
    m := trunc((h - 1) / 2);  // Finding center.
    n := trunc((w - 1) / 2);
    d := 0; // First direction is right.

    seed := 2;
    step := 1;

    // Travel the spiral.
    repeat
        // If in the sub-spiral, store value.
        if ((m >= 0) and (n >= 0) and (m < h) and (n < w)) then
        begin
            spiral[m, n] := v;
            v := v - 1;
        end;

        // Move!
        case d of
            0: n := n + 1;
            1: m := m - 1;
            2: n := n - 1;
            3: m := m + 1;
        end;

        // Plan trajectory.
        step := step - 1;
        if step = 0 then
        begin
            d := (d + 1) mod 4;
            seed := seed + 1;
            step := trunc(seed / 2);
        end;
    until v = 0;

    // Print the spiral.
    for m := 0 to (h - 1) do
    begin
        for n := 0 to (w - 1) do
        begin
            write(spiral[m, n], ' ');
        end;
        writeln();
    end;

end.

如果你真的需要它来打印文本螺旋，我会让你对齐数字。只需用空格填充它们。

编辑：

忘记了......为了让它在ideone上工作，我把参数放在2行作为输入。m，然后 n。

例如：

5
2

产量

3 4 
7 8 
10 9 
6 5 
2 1
于 2012-01-07T00:55:10.597 回答
-1

即使问题已经得到解答，这也是一种替代解决方案（可以说更简单）。解决方案在 python 中（使用 numpy 进行双向数组），但可以轻松移植。

基本思想是使用已知步数 (m*n) 作为结束条件，并在每次迭代时正确计算循环的下一个元素：

import numpy as np

def spiral(m, n):
    """Return a spiral numpy array of int with shape (m, n)."""
    a = np.empty((m, n), int)
    i, i0, i1 = 0, 0, m - 1
    j, j0, j1 = 0, 0, n - 1
    for k in range(m * n):
        a[i, j] = k
        if   i == i0 and     j0 <= j <  j1: j += 1
        elif j == j1 and     i0 <= i <  i1: i += 1
        elif i == i1 and     j0 <  j <= j1: j -= 1
        elif j == j0 and 1 + i0 <  i <= i1: i -= 1
        else:
            i0 += 1
            i1 -= 1
            j0 += 1
            j1 -= 1
            i, j = i0, j0
    return a

这里有一些输出：

>>> spiral(3,3)
array([[0, 1, 2],
       [7, 8, 3],
       [6, 5, 4]])
>>> spiral(4,4)
array([[ 0,  1,  2,  3],
       [11, 12, 13,  4],
       [10, 15, 14,  5],
       [ 9,  8,  7,  6]])
>>> spiral(5,4)
array([[ 0,  1,  2,  3],
       [13, 14, 15,  4],
       [12, 19, 16,  5],
       [11, 18, 17,  6],
       [10,  9,  8,  7]])
>>> spiral(2,5)
array([[0, 1, 2, 3, 4],
       [9, 8, 7, 6, 5]])
于 2014-12-11T12:59:40.933 回答
-1

这是您要完成的注释的 JavaScript 实现。

// return an array representing a matrix of size MxN COLxROW
function spiralMatrix(M, N) {
var result = new Array(M * N);
var counter = M * N;
// start position
var curCol = Math.floor((M - 1) / 2);
var curRow = Math.floor(N / 2);
// set the center
result[(curRow * M) + curCol] = counter--;
// your possible moves RIGHT, UP, LEFT, DOWN * y axis is flipped
var allMoves = [[1,0], [0,-1], [-1,0], [0,1]];
var curMove = 0;
var moves = 1; // how many times to make current Move, 1,1,2,2,3,3,4,4 etc
// spiral
while(true) {
 for(var i = 0; i < moves; i++) {
  // move in a spiral outward counter clock-wise direction
  curCol += allMoves[curMove][0];
  curRow += allMoves[curMove][1];
  // naively skips locations that are outside of the matrix bounds
  if(curCol >= 0 && curCol < M && curRow >= 0 && curRow < N) {
   // set the value and decrement the counter
   result[(curRow * M) + curCol] = counter--;
   // if we reached the end return the result
   if(counter == 0) return result;
  }
 }
 // increment the number of times to move if necessary UP->LEFT and DOWN->RIGHT
 if(curMove == 1 || curMove == 3) moves++;
 // go to the next move in a circular array fashion
 curMove = (curMove + 1) % allMoves.length;
}
}

该代码不是最有效的，因为它在没有首先检查它所走的位置是否有效的情况下天真地走着螺旋形。它仅在尝试对其设置值之前检查当前位置的有效性。

于 2012-01-06T18:36:02.027 回答