具有霍夫曼树的优先级队列

Zie*_*zer 5 java queue huffman-code

我试图通过读取文件并计算每个字母空格符号的频率等来创建一个霍夫曼树.我正在使用Priorityqueue将项目从最小到最大排队但是当我将它们插入队列时它们不能正确排队这是我的代码.包霍夫曼;

import java.io.FileNotFoundException; import java.io.FileReader; import java.util.ArrayList; import java.util.PriorityQueue; import java.util.Scanner;

公共课霍夫曼{

public ArrayList<Frequency> fileReader(String file)
{
    ArrayList<Frequency> al = new ArrayList<Frequency>();
    Scanner s;
    try {

        s = new Scanner(new FileReader(file)).useDelimiter("");
        while (s.hasNext())
        {
            boolean found = false;
            int i = 0;
            String temp = s.next();
            while(!found)
            {


                if(al.size() == i && !found)
                {
                    found = true;
                    al.add(new Frequency(temp, 1));
                }
                else if(temp.equals(al.get(i).getString()))
                {
                    int tempNum = al.get(i).getFreq() + 1;
                    al.get(i).setFreq(tempNum);
                    found = true;
                }
                i++;

            }



        }
    } catch (FileNotFoundException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    }
    return al;
}
public void buildTree(ArrayList<Frequency> al)
{
    PriorityQueue<Frequency> pq = new PriorityQueue<Frequency>();
    for(int i = 0; i < al.size(); i++)
    {
        pq.add(al.get(i));          
    }
    while(pq.size() > 0)
    {
        System.out.println(pq.remove().getString());
    }
}
public void printFreq(ArrayList<Frequency> al)
{
    for(int i = 0; i < al.size(); i++)
    {
        System.out.println(al.get(i).getString() + "; " + al.get(i).getFreq());
    }
}
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}

在buildTree()方法中我遇到了问题.我试图做的是队列频率对象,其中包含字母/空格/符号,频率为int,频率等级为此.public class Frequency实现Comparable {private String s; 私人的

Frequency(String s, int n)
{
    this.s = s;
    this.n = n;
}
public String getString()
{
    return s;
}
public int getFreq()
{
    return n;
}
public void setFreq(int n)
{
    this.n = n;
}
@Override
public int compareTo(Object arg0) {
    // TODO Auto-generated method stub
    return 0;
}
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}

如何获得优先级以使用频率编号将它们从最小到最大排队?

Jac*_*ack 4

实际上,您错过了实现该compareTo方法以使您的对象能够有效地进行比较。

compareTo正如文档所述,该方法应该

当此对象小于、等于或大于指定对象时,返回负整数、零或正整数。

这意味着在您的情况下您应该执行以下操作:

public int compareTo(Object arg0)
{
  Frequency other = (Frequency)arg0;

  return n < other.n ? -1 : (n == other.n ? 0 : 1);
}
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但请注意,compare 有一个更可取的泛型类型:Comparable<T>因此您可以避免强制转换arg0,使其成为Frequency具有静态类型安全性的对象:

class Frequency implements Comparable<Frequency> {   
  public int compareTo(Frequency f2) {
    // directly compare
  }
}
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