我有一个4集。

  {A1，A2，A3，...}
{B1，B1，B3，...}
{C1，C2，C3，...}
{D1，D2，D3 ...}
 

我需要用下面的规则来查找所有可能的组合：

示例组合：

  {A1，B1，B2，C1，C2，C3，D1}
{A2，B1，B2，C1，C2，C3，D1}
 

解决方案

Visual Studio的时间太长加载，所以我这样做是在JavaScript中，因为我可以测试在我的控制台。这将打印出所有的选择。 （此外，它听起来更像是一个什么是算法呢？而不是有什么算法，这种的在C＃的？）

 函数makeGroup（ID，计数）{
  VAR的结果= [];
  对于（VAR I = 1; I＆LT; =计数;我++）{
    result.push（ID + I）;
  }
  返回结果;
}

功能选择（组数）{
  如果（计数=== 1）{
    VAR的结果= [];
    对于（VAR I = 0; I＆LT; group.length;我++）{
      result.push（[组[I]）;
    }
    返回结果;
  }
  如果（计数=== 2）{
    VAR的结果= [];
    对于（VAR I = 0; I＆LT; group.length;我++）{
      对于（VAR J = 0; J＆LT; group.length; J ++）{
        如果（我== J和！＆安培;
            I＆LT; j）条{
          result.push（[组[I]，组[J]）;
        }
      }
    }
    返回结果;
  }
  如果（计数=== 3）{
    VAR的结果= [];
    对于（VAR I = 0; I＆LT; group.length;我++）{
      对于（VAR J = 0; J＆LT; group.length; J ++）{
        为（变种K = 0; K＆LT; group.length; k ++）{
          如果（我== J和！＆安培;我== K＆放大器;！＆安培;Ĵ== K＆放大器;！＆安培;
              I＆LT; J＆安培;＆安培; J＆LT; K）{
            result.push（[组[I]，组[J]，组[K]）;
          }
        }
      }
    }
    返回结果;
  }
}

VAR A组= makeGroup（'A'，2）;
变种B组= makeGroup（'B'，2）;
变种C组= makeGroup（C，3）;
变种D组= makeGroup（D，1）;

选择（A组，1）.forEach（函数（）{
  选择（B组，2）.forEach（函数（二）{
    选择（C组，3）.forEach（函数（C）{
      选（D组，1）.forEach（功能（D）{
        的console.log（一个++ B [0] ++ B [1] ++ C [0] ++ C [1] ++ C [2] ++ D [0]）;
      }）;
    }）;
  }）;
}）;
 

输出示例：

  A1 B1，B2，C1 C2 C3 D1
A2 B1，B2，C1 C2 C3 D1
 

要看到它的工作的一个例子，你可以这样做：

  VAR testGroup = makeGroup（'T'，4）;

VAR选中1 =选择（testGroup，1）;
VAR choose2 =选择（testGroup，2）;
VAR choose3 =选择（testGroup，3）;

的console.log（JSON.stringify（选中1））;
的console.log（JSON.stringify（choose2））;
的console.log（JSON.stringify（choose3））;
 

和看到，它选择正确的：

  [T1]，[T2]，[T3]，[T4]
[[T1，T2]，[T1，T3]，[T1，T4]，[T2，T3]，[T2，T4] [T3，T4]
[[T1，T2，T3]，[T1，T2，T4]，[T1，T3，T4]，[T2，T3 ，T4]]
 

I have a 4 collections.

{ A1, A2, A3, ...}
{ B1, B1, B3, ...}
{ C1, C2, C3, ...}
{ D1, D2, D3, ...}

I need to find all of the possible combinations using the following rules:

Example combinations:

{A1, B1, B2, C1, C2, C3, D1}
{A2, B1, B2, C1, C2, C3, D1}

Visual Studio was taking too long to load, so I did it in JavaScript since I could test in my console. This will print out all the choices. (Also it sounds more like a "what's the algorithm for this?" not "what's the algorithm for this in C#?")

function makeGroup(id, count) {
  var result = [];
  for (var i = 1; i <= count; i++) {
    result.push(id + i);
  }
  return result;
}

function choose(group, count) {
  if (count === 1) {
    var result = [];
    for (var i = 0; i < group.length; i++) {
      result.push([group[i]]);
    }
    return result;
  }
  if (count === 2) {
    var result = [];
    for (var i = 0; i < group.length; i++) {
      for (var j = 0; j < group.length; j++) {
        if (i !== j && 
            i < j) {
          result.push([group[i], group[j]]);
        }
      }
    }
    return result;
  }
  if (count === 3) {
    var result = [];
    for (var i = 0; i < group.length; i++) {
      for (var j = 0; j < group.length; j++) {
        for (var k = 0; k < group.length; k++) {
          if (i !== j && i !== k && j !== k &&
              i < j && j < k) {
            result.push([group[i], group[j], group[k]]);
          }
        }
      }
    }
    return result;
  }
}

var groupA = makeGroup('A', 2);
var groupB = makeGroup('B', 2);
var groupC = makeGroup('C', 3);
var groupD = makeGroup('D', 1);

choose(groupA, 1).forEach(function (a) { 
  choose(groupB, 2).forEach(function (b) {
    choose(groupC, 3).forEach(function (c) {
      choose(groupD, 1).forEach(function (d) {
        console.log(a + " " + b[0] + " " + b[1] + " " + c[0] + " " + c[1] + " " + c[2] + " " + d[0]);
      });
    });
  });
});

Example output:

A1 B1 B2 C1 C2 C3 D1
A2 B1 B2 C1 C2 C3 D1

To see an example of it working, you can do:

var testGroup = makeGroup('T', 4);

var choose1 = choose(testGroup, 1);
var choose2 = choose(testGroup, 2);
var choose3 = choose(testGroup, 3);

console.log(JSON.stringify(choose1));
console.log(JSON.stringify(choose2));
console.log(JSON.stringify(choose3));

And see that it chooses correctly:

[["T1"],["T2"],["T3"],["T4"]]
[["T1","T2"],["T1","T3"],["T1","T4"],["T2","T3"],["T2","T4"],["T3","T4"]]
[["T1","T2","T3"],["T1","T2","T4"],["T1","T3","T4"],["T2","T3","T4"]]