德尔福阵列的电源组

Question

我正在尝试编写一个函数,它将在输入和返回数组的数组上采用数组,包含所有可能的输入数组子集(没有空元素的幂集).例如输入:[1, 2, 3]结果将是[[1], [2], [3], [1, 2], [1, 3], [2, 3], [1, 2, 3]].

这个函数在python中完成了这个工作:

def list_powerset(lst):
    result = [[]]
    for x in lst:
        result += [subset + [x] for subset in result]
    result.pop(0)
    return result

但我正在寻找在Delphi中实现它.这有可能以这种方式完成,还是应该寻找其他东西？

Answer 1

type
  TIdArray = array of Integer;
  TPowerSet = array of TIdArray;

function PowerSet(Ids: TIdArray): TPowerSet;
// Implementation loosely based on the explanation on
// http://www.mathsisfun.com/sets/power-set.html
var
  TotalCombinations: Integer;
  TotalItems: Integer;
  Combination: Integer;
  SourceItem: Integer;
  ResultItem: Integer;
  Bit, Bits: Integer;
begin
  TotalItems := Length(Ids);

  // Total number of combination for array of n items = 2 ^ n.
  TotalCombinations := 1 shl TotalItems;

  SetLength(Result, TotalCombinations);

  for Combination := 0 to TotalCombinations - 1 do
  begin
    // The Combination variable contains a bitmask that tells us which items
    // to take from the array to construct the current combination.
    // Disadvantage is that because of this method, the input array may contain
    // at most 32 items.

    // Count the number of bits set in Combination. This is the number of items
    // we need to allocate for this combination.
    Bits := 0;
    for Bit := 0 to TotalItems - 1 do
      if Combination and (1 shl Bit) <> 0 then
        Inc(Bits);

    // Allocate the items.
    SetLength(Result[Combination], Bits);

    // Copy the right items to the current result item.
    ResultItem := 0;

    for SourceItem := 0 to TotalItems - 1 do
      if Combination and (1 shl SourceItem) <> 0 then
      begin
        Result[Combination][ResultItem] := Ids[SourceItem];
        Inc(ResultItem);
      end;
  end;

end;