Question

在C＃或任何已知的库中是否有任何方法可以打印尺寸大小的所有可能性，例如在数组声明中？

例如：

[2, 2]

给出：

[0,0] [0,1] [0,2] [1,0] [1,1] [1,2] [2,0] [2,1] [2,2]

我希望它适用于所有尺寸（1D，2D，3D，4D ......）

Answer 1

我建议生成所有可能的项目

private static IEnumerable<String> Ranges(params int[] ranges) {
  int[] current = new int[ranges.Length];

  bool hasItems = true;

  yield return String.Format("[{0}]", String.Join(",", current));

  while (hasItems) {
    hasItems = false;

    for (int i = 0; i < current.Length; ++i) {
      if (current[i] < ranges[i]) {
        hasItems = true;

        current[i] = current[i] + 1;

        for (int j = i - 1; j >= 0; --j)
          current[j] = 0;

        yield return String.Format("[{0}]", String.Join(",", current));

        break;
      }
    }
  }
}

您的样本：

 int[] source = new int[] { 2, 2 };

 // [0,0] [1,0] [2,0] [0,1] [1,1] [2,1] [0,2] [1,2] [2,2]
 Console.Write(String.Join(" ", Ranges(source)));

另一项测试（3个维度）

  int[] source = new int[] { 2, 1, 3};

  // [0,0,0] [1,0,0] [2,0,0] [0,1,0] [1,1,0] [2,1,0] [0,0,1] [1,0,1] [2,0,1] [0,1,1] 
  // [1,1,1] [2,1,1] [0,0,2] [1,0,2] [2,0,2] [0,1,2] [1,1,2] [2,1,2] [0,0,3] [1,0,3] 
  // [2,0,3] [0,1,3] [1,1,3] [2,1,3]
  Console.Write(String.Join(" ", Ranges(source)));

编辑：如果您想使用实际的多维数组，可以使用 Linq 获取维度数组：

  // d is 3D array
  // please, notice, that type of the array (string) doesn't play any role
  string[,,] d = new string[2, 3, 1];

  // array of dimensions: int[] {2, 3, 1}
  int[] dims = Enumerable
    .Range(0, d.Rank)
    .Select(dim => d.GetLength(dim))
    .ToArray();

  Console.Write(String.Join(" ", Ranges(dims)));

Answer 2

一点点IEnumerable魔法！

    public static IEnumerable<IEnumerable<int>> GetCombinations(IEnumerable<int> dimensions)
    {
        if (!dimensions.Any())
        {
            yield return Enumerable.Empty<int>();
            yield break;
        }

        var first = dimensions.First();

        foreach (var subSolution in GetCombinations(dimensions.Skip(1)))
        {
            for (var i = 0; i < first + 1; i++)
            {
                yield return new[] { i }.Concat(subSolution);
            }
        }
    }

Answer 3

一个简单的解决方案，可以从任何多维数组动态生成索引输出：

static void Main(string[] args)
{
    int[,] arr = new int[2, 2];
    // int[,,] arr = new int[3, 2, 3];
    printArrayIndexes(arr);
}

private static void printArrayIndexes(object arr)
{
    var dimensArr = arr as Array;
    List<int> indexList = new List<int>();
    for (int dimension = 0; dimension < dimensArr.Rank; dimension++)
    {
        indexList.Add(0);
    }
    bool hasItems = true;
    while (hasItems)
    {
        hasItems = false;
        Console.WriteLine(String.Format("[{0}]", String.Join(",", indexList)));
        for (int i = 0; i < indexList.Count; i++)
        {
            if (indexList[i] < dimensArr.GetLength(i)) {
                hasItems = true;
                indexList[i]++;
                break;
            } else {
                indexList[i] = 0;
            }
        }
    }
}

使用您的示例int[,] arr = new int[2,2]：

[0,0] [1,0] [2,0] [0,1] [1,1] [2,1] [0,2] [1,2] [2,2]

使用任何3D数组int[,,] arr = new int[3,2,3]：

[0,0,0] [1,0,0] [2,0,0] [3,0,0] [0,1,0] [1,1,0] [2,1,0] [3,1,0]
[0,2,0] [1,2,0] [2,2,0] [3,2,0] [0,0,1] [1,0,1] [2,0,1] [3,0,1]
[0,1,1] [1,1,1] [2,1,1] [3,1,1] [0,2,1] [1,2,1] [2,2,1] [3,2,1]
[0,0,2] [1,0,2] [2,0,2] [3,0,2] [0,1,2] [1,1,2] [2,1,2] [3,1,2]
[0,2,2] [1,2,2] [2,2,2] [3,2,2] [0,0,3] [1,0,3] [2,0,3] [3,0,3]
[0,1,3] [1,1,3] [2,1,3] [3,1,3] [0,2,3] [1,2,3] [2,2,3] [3,2,3]

所以你可以使用1D，2D，3D，4D阵列......

Answer 4

你可以自己做这样的事情：

void IndexPermutations(int range1, int range2)
{
  for(int i=0; i<range1; i++)
  {
   for(int j=0; j<range2; j++)
   {
    Console.WriteLine("[{0},{1}]",i,j);
   }
  }
}

Answer 5

您可以使用递归解决方案。

public void IndexPermutations(int dimensionSize, int minValue, int maxValue, Action<List<int>> action)
{
    IndexPermutationsInternal(dimensionSize, minValue, maxValue, action, new List<int>());
}

private void IndexPermutationsInternal(
    int dimensionSize,
    int minValue,
    int maxValue,
    Action<List<int>> action,
    List<int> current)
{
    if (dimensionSize == current.Count)
    {
        action(current);
    }
    else
    {
        for (int i = minValue; i <= maxValue; i++)
        {
            current.Add(i);
            IndexPermutationsInternal(dimensionSize, minValue, maxValue, action, current);
            current.RemoveAt(current.Count - 1);
        }
    }
}

打印所有尺寸的可能性

5 个答案: