递归生成(n-1)的Delphi代码！！从 n=2*k 个元素中设置 (1*3*5*7...n-1)

Delphi code for recursive generation of (n-1)!! sets (1*3*5*7...n-1) from n=2*k elements

var
  A: TArray<Integer>;

  procedure Swap(i, j: integer);
  var
    t : integer;
  begin
    t := A[i];
    A[i] := A[j];
    A[j] := t;
  end;

  procedure MakePairs(Start: Integer; Pairs: string);
  var
    i: Integer;
  begin
    if Start >= Length(A) then
      Writeln(Pairs)
    else
    for i := Start + 1 to High(A) do begin
      Swap(Start + 1, i); //store used element in the array beginning
      MakePairs(Start + 2, Pairs + Format('(%d,%d)', [A[Start], A[Start + 1]]));
      Swap(Start + 1, i); //get it back
    end;
  end;

begin
  A := TArray<Integer>.Create(1,2,3,4,5,6);
  //be sure that array length is even!!!
  MakePairs(0, '');
  Writeln(PairCount);

输出:

(1,2)(3,4)(5,6)
(1,2)(3,5)(4,6)
(1,2)(3,6)(5,4)
(1,3)(2,4)(5,6)
(1,3)(2,5)(4,6)
(1,3)(2,6)(5,4)
(1,4)(3,2)(5,6)
(1,4)(3,5)(2,6)
(1,4)(3,6)(5,2)
(1,5)(3,4)(2,6)
(1,5)(3,2)(4,6)
(1,5)(3,6)(2,4)
(1,6)(3,4)(5,2)
(1,6)(3,5)(4,2)
(1,6)(3,2)(5,4)
15

加法
也适用于奇数长度数组的变体(奇怪的排序)

Addition
Variant that works with odd-length array too (weird ordering)

  procedure MakePairs(Start: Integer; Pairs: string);
  var
    i: Integer;
    OddFlag: Integer;
  begin
    if Start >= Length(A) then
      Memo1.Lines.Add(Pairs)
    else begin
      Oddflag := (High(A) - Start) and 1;
      for i := Start + OddFlag to High(A) do begin
        Swap(Start + OddFlag, i);
        if OddFlag = 1 then
          MakePairs(Start + 2, Pairs + Format('(%d,%d)', [A[Start], A[Start + 1]]))
        else
          MakePairs(Start + 1, Pairs);
        Swap(Start + OddFlag, i);
      end;
    end;
  end;

对于 (1,2,3,4,5):

for (1,2,3,4,5):

(2,3)(4,5)
(2,4)(3,5)
(2,5)(4,3)
(1,3)(4,5)
(1,4)(3,5)
(1,5)(4,3)
(2,1)(4,5)
(2,4)(1,5)
(2,5)(4,1)
(2,3)(1,5)
(2,1)(3,5)
(2,5)(1,3)
(2,3)(4,1)
(2,4)(3,1)
(2,1)(4,3)
15

现在不相关:
如果每一对只出现一次(从你的例子中 n=4 不清楚)，那么你可以使用 循环赛算法

Not relevant now:
If every pair should occur just once (it is not clear from your example with n=4), then you can use round-robin tournament algorithm

此处为 n=4 个案例