打印尺寸的所有可能性

Question

在C＃或任何已知的库中，是否有任何方式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] 

我想它为所有的维（1D，2D，3D，4D ...）

Answer 1

我建议产生所有项目可能

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; 
     } 
    } 
    } 
} 

你的样品：

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))); 

另一个测试（3名维）

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：如果你要使用实际的多维数组，你可以使用Linq来获得尺寸列表：

// 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

你可以工作使它成为像这样的东西：

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 3

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); 
     } 
    } 
} 

Answer 4

有点IEnumerable的魔法！

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 5

一个简单的解决方案，它从任何多维数组动态地生成的索引输出：

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; 
      } 
     } 
    } 
} 

使用您的样品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阵列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] 

所以你可以使用一维，二维，三维，四维阵列...