请教各位专家高手，帮我把这个算法转成JAVA，急用，谢谢！

Question

请教各位专家高手，帮我把这个算法转成JAVA，或者给我ID3或C4.5的java主要代码，不过要有详细的注释，谢谢！分数我会继续加的，只要能解决，我不会吝惜积分的。。
//*********************************************
//决策树ID3算生成决策树的源程序
//*********************************************
procedure TFrmShibie.Generate_decision_tree(b: TwoArr; bn: Integer;
  attribute_test_list: array of integer; classCount, sn, ai, aj: Integer);
var
same_class,i,j,k,t,l,ll,lll,count_Positive,temp1:integer;
  p1,p2,Entropy_Es:double;
  temp:array[0..s_max-1] of integer;
  b1:TwoArr;
  temp_b:array[0..s_max-1,0..N-1,0..M+1] of integer;
  TypeCountArr:array of Integer;
  AttriArr:array of Integer;
  len:Integer;
  AttriArrCount:array of Integer;
  q:TArr;
begin
//定义数组a记录目前待分的训练样本集，定义数组b记录目前要分结点中所含的训练样本集
    //same_class用来记数，判别samples是否都属于同一个类
Trip:=Trip+1;
  path_a[leaves][Trip]:=ai;
  path_b[leaves][Trip]:=aj;
if(bn=0) then
  begin//待分结点的样本集为空时，加上一个树叶，标记为训练集中最普通的类
//记录路径与前一路径相同
for i:=1 to Trip-1 do
    begin
if(path_a[leaves][i]=0) then
      begin
        path_a[leaves][i]:=path_a[leaves-1][i];
        path_b[leaves][i]:=path_b[leaves-1][i];
end;
    end;
    for i:=1 to Trip-1 do
    begin
      path_a[leaves][0]:=most;
    end;
        //修改树的深度
if(path_a[leaves][Trip]=av[path_b[leaves][Trip]-1]) then
    begin
      for i:=Trip downto 2 do
      begin
        if(path_a[leaves][i]=av[path_b[leaves][i]-1]) then
        begin
          dec(Trip);
        end
        else
        begin
          break;
        end;
      end;
    end;
    dec(Trip);
inc(leaves);
end
  else
  begin
    same_class:=1;  //相同类别的测试集数目
    for i:=0 to bn-2 do
    begin
  if (b[i][0]=b[i+1][0]) then
        inc(same_class);
    end;
    if(same_class=bn) then
    begin  //假如所有测试集在某个属性上取相同值
    //记录路径与前一路径相同
      for i:=1 to Trip-1 do
      begin
        if (path_a[leaves][i]=0) then
        begin
           path_a[leaves][i]:=path_a[leaves-1][i];
           path_b[leaves][i]:=path_b[leaves-1][i];
        end;
      end;
      path_a[leaves][0]:=b[0][0]; //如果所有的测试集都是属于同一个类型，就以该类别标记叶子
  //修改树的深度
  if (path_a[leaves][Trip]=av[path_b[leaves][Trip]-1]) then
      begin
   for i:=Trip to 2 do
       begin
          if(path_a[leaves][i]=av[path_b[leaves][i]-1]) then
            dec(Trip)
          else
            break;
       end;
      end;
      dec(Trip);
      inc(leaves);
      //未分类的样本集减少
      l:=-1;
      lll:=0;
      for i:=0 to count do  //count+1为训练集样本数
      begin
        for j:=0 to bn-1 do
        begin
          if(s[i][M+1]=b[j][M+1]) then   //数组s[i][j]用来记录第i个训练样本的第j个属性值
                                         //M定义为候选属性的个数
            inc(lll);
        end;
        if(lll=0) then
        begin
          inc(l);
          for ll:=0 to M+1 do
            a[l][ll]:=s[i][ll];
        end;
      end;
      k:=-1;
      for i:=0 to l-1 do
      begin
        inc(k);
        for ll:=0 to M+1 do
          s[k][ll]:=a[i][ll];
      end;
      count:=count-bn;
    end
    else  //假如不是所有测试集在某个属性上取相同值
    begin
      if sn=1 then
      begin
        for i:=0 to bn-1 do
        begin
          path[i][0]:=s[i][0];
          path[i][1]:=s[i][1];
        end;
      end;
      if(sn=0) then  //属性个数为0时
      begin
        for i:=1 to Trip-1 do
        begin
    if(path_a[leaves][i]=0) then
          begin
      path_a[leaves][i]:=path_a[leaves-1][i];
            path_b[leaves][i]:=path_b[leaves-1][i];
    end;
        end;
        ll:=0;
        lll:=0;
        //判断类别
        for i:=0 to bn-1 do
        begin
          if(b[i][0]=0) then
            inc(ll)
          else
            inc(lll);
        end;
        if(ll>lll) then
        begin
          path_a[leaves][0]:=0;
        end
        else
        begin
          path_a[leaves][0]:=1;
        end;
    if(path_a[leaves][Trip]=av[path_b[leaves][Trip]-1]) then
        begin
      for i:=Trip downto 2 do
          begin
       if(path_a[leaves][i]=av[path_b[leaves][i]-1]) then
              dec(Trip)
           else
      break;
          end;
        end;
        dec(Trip);
        inc(leaves);
        //未分类的样本集减少
        l:=-1;
        lll:=0;
        for i:=0 to count do
        begin
          for j:=0 to bn-1 do
              if(s[i][M+1]=b[j][M+1]) then inc(lll);
          if(lll=0) then
          begin
             inc(l);
             for ll:=0 to M+1 do
                 a[l][ll]:=s[i][ll];
          end;
        end;
        k:=-1;
        for i:=0 to l-1 do
        begin
          inc(k);
          for ll:=0 to M+1 do
            s[k][ll]:=a[i][ll];
        end;
        count:=count-bn;
      end
      else
      begin
        count_Positive:=0;
        SetLength(TypeCountArr,classCount);
        for i:=0 to classCount-1 do
        begin
          for j:=0 to bn-1 do
          begin
            if s[j][0]=i then
              inc(TypeCountArr[i]);
          end;
        end;
        Entropy_Es:=Entropy(TypeCountArr,bn);
        //计算信息增益
        for i:=1 to sn-1 do
        begin
          len:=1;
          setlength(AttriArr,len);
          setlength(AttriArrCount,len);
          AttriArr[0]:=s[0][attribute_test_list[i]];
          AttriArrCount[0]:=1;
          for k:=0 to bn-1 do
          begin
            for j:=0 to high(AttriArr) do
            begin
              if AttriArr[j]=s[k][attribute_test_list[i]] then
              begin
                inc(AttriArrCount[j]);
              end
              else
              begin
                inc(len);
                setlength(AttriArr,len);
                setlength(AttriArrCount,len);
                AttriArr[len-1]:=s[k][attribute_test_list[i]];
                AttriArrCount[len-1]:=1;
              end;
            end;
          end;
          setlength(q,sn*len*classCount);
          for j:=0 to high(AttriArr) do
          begin
            for k:=0 to sn-1 do
            begin
              for t:=0 to bn-1 do
              begin
                if (AttriArr[j]=s[i][attribute_test_list[i]]) and
                    (attribute_test_list[k]=s[i][0]) then
                begin
                  inc(q[i][j][k]);
                end;
              end;
            end;
          end;
          Gain[attribute_test_list[i]-1]:=GainA(bn,TypeCountArr,AttriArrCount,q,i);
        end;
        //找出信息赢取最大值,用j记录哪个候选属性的信息赢取最大
        max_Gain:=Gain[0];
        j:=attribute_test_list[0]-1;
        for i:=0 to sn-1 do
          if(max_Gain<Gain[attribute_test_list[i]-1]) then
          begin
            max_Gain:=Gain[attribute_test_list[i]-1];
            j:=attribute_test_list[i]-1;
          end;
        temp1:=-1;
        for i:=1 to av[j] do
        begin
          temp[i]:=-1;
          for k:=0 to bn-1 do
            if(b[k][j+1]=i) then
            begin
             inc(temp[i]);
             for l:=0 to M+1 do
              temp_b[i][temp[i]][l]:=b[k][l];
            end;
        end;
        //对于每一个分支使用递归函数重复生成树
        for i:=1 to av[j] do
        begin
          for k:=0 to temp[i] do
            for l:=0 to M+1 do
              b1[k][l]:=temp_b[i][k][l];
            if(i=1) then
            begin
             l:=0;
             for ll:=0 to sn-1 do
             begin
               if(attribute_test_list[ll]-1<>j) then attribute_test_list[l]:=attribute_test_list[ll];
               inc(l);
             end;
             Generate_decision_tree(b1,k,attribute_test_list,classCount,l,i,j+1);
             dec(sn);
            end
            else
            begin
              Generate_decision_tree(b1,k,attribute_test_list,classCount,sn,i,j+1);
              if(i=av[j]) then
                attribute_test_list[sn]:=j+1;
            end;
          end;
        end;
    end;
  end;
end;

Answer 1

转个毛
直接用jni调用算了

Answer 2

这个是Delphi代码，大家帮忙看一下吧！

Answer 3

没做过Delphi，这个代码看得晕。

Answer 4

看到很多for晕了
回复内容太短了！

Answer 5

没时间仔细看，大概的意思是不是要实现树型复选框之类的？？
若是的话，你可以用html来做，也是很容易的事。

Answer 6

procedure TFrmShibie.Generate_decision_tree(b: TwoArr; bn: Integer;
  attribute_test_list: array of integer; classCount, sn, ai, aj: Integer);public void Generate_decision_tree(
   TwoArr b
   ,int bn
   ,int[] attribute_test_list
   ,int classCount
   ,int sn
   ,int ai
   ,int aj
)
{
  ....
}

Answer 7

程序不全，TwoArr类没有，补齐他先