用Java实现哈夫曼编码
我有一个程序 但是运行结果不对 而且 也没有界面设计
请各位高手帮忙修改下 用JAVA实现 要有界面设计 谢谢
import java.io.*;
public class Huffumancode {


static float sum=0;
static String s;
public static void main(String[] args) {
// TODO Auto-generated method stub
new Huffumancode().run();

}
public void run()
{
try
{

FileReader fr=new FileReader("WordAndFreq.txt");
BufferedReader br=new BufferedReader(fr);
String s=null;
Node []A=new Node[24686];
A[0]=new Node();
A[0].freq=24685;
int i=1;
String []st=new String [2];
while((s=br.readLine())!=null)
{
st=s.split("\\s");
A[i]=new Node();
A[i].data=st[0];
A[i].freq=Integer.parseInt(st[1]);
i++;
}
System.out.println(A[24685].data +A[24685].freq);
heap_sort(A);
A[0].freq=24685;
build_huffmantree(A);

System.out.println("A[1]ede pin ldu"+A[1].freq);

FileWriter fw=new FileWriter("output.txt");
BufferedWriter bw=new BufferedWriter(fw);
preorder(A[1],bw);
bw.flush();

float average=sum/3330999;
String s1="the average length of the code is :".concat(Float.toString(average));
bw.write(s1);
bw.close();
System.out.println(average);



}
catch(Exception e)
{
System.out.println(e);
}
}
public void min_Heapify( Node []A, int i)
{  

int l=2*i;
int r=2*i+1;
int small;
if(l<=A[0].freq&&A[l].freq<A[i].freq)
small=l;
else small=i;
if (r<=A[0].freq&&A[r].freq<A[small].freq)
small=r;
//System.out.println("haha ");
if (small!=i)
{
Node node1=new Node();
node1=A[small];
A[small]=A[i];
A[i]=node1;
min_Heapify( A , small);

//depth_visited(A[1]);
}
}
public void build_min_heap(Node [] B)
{
int len=(int)B[0].freq;
for(int j=len/2;j>0;j--)
min_Heapify(B,j);
}
public void heap_sort(Node [] b)
{
build_min_heap(b);
int k=(int)b[0].freq;
for(int i=k;i>=2;i--)
{
Node node1=new Node();
node1=b[1];b[1]=b[i];
b[i]=node1;
b[0].freq=b[0].freq-1;
min_Heapify(b,1);

}

}
public void build_huffmantree(Node []B)
{
int len=(int)B[0].freq;
Node node;
while(len>1)
{  
//System.out.println("len 长度是:"+len);
//System.out.println("B[len] 的频率"+B[len].freq);
//System.out.println("B[len-1] 的频率"+B[len-1].freq);
node=new Node();
node.freq=B[len].freq+B[len-1].freq;



node.left=B[len];
node.right=B[len-1];
B[len]=null;
B[len-1]=null;
 
//System.out.println("node 的频率："+node.freq);


int k=len-2;
while(k>=1)
{
if(B[k].freq<node.freq)
B[k+1]=B[k];
else 
break;
k--;

}
 
B[k+1]=node;
//System.out.println(B[k+1].freq);
//System.out.println(B[9].freq) ;
len--;
node=null;
//sort(B,len);

//System.out.println(index--);
 
}
}
public void preorder(Node node , BufferedWriter bt)
{
if(node!=null)
{  

if(node.left!=null&&node.right!=null)
{
if(node.code!=null)
node.left.code="0".concat(node.code);
else
node.left.code="0";