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).例如对于输入：[1,2,3],结果将是[[1],[2],[3],[1,2],[2,3]].
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实现,还是应该寻找其他东西？
解决方法
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;