【问题标题】:Infinite Recursion, StackOverError in huffman tree无限递归,霍夫曼树中的 StackOverError
【发布时间】:2012-11-25 09:17:53
【问题描述】:

我正在开发一个霍夫曼编码程序,我几乎完成了,但我陷入了无限递归循环。有谁知道这是哪里出错了?

这是我得到的错误:

Exception in thread "main" java.lang.StackOverflowError
at sun.nio.cs.SingleByteEncoder.encodeLoop(SingleByteEncoder.java:130)
at java.nio.charset.CharsetEncoder.encode(CharsetEncoder.java:544)
at sun.nio.cs.StreamEncoder.implWrite(StreamEncoder.java:252)
at sun.nio.cs.StreamEncoder.write(StreamEncoder.java:106)
at java.io.OutputStreamWriter.write(OutputStreamWriter.java:190)
at java.io.BufferedWriter.flushBuffer(BufferedWriter.java:111)
at java.io.PrintStream.write(PrintStream.java:476)
at java.io.PrintStream.print(PrintStream.java:619)
at java.io.PrintStream.println(PrintStream.java:756)
at HuffmanNode.buildTree(hw4.java:63)
at HuffmanNode.buildTree(hw4.java:64)
at HuffmanNode.buildTree(hw4.java:64)
at HuffmanNode.buildTree(hw4.java:64)
at HuffmanNode.buildTree(hw4.java:64)
at HuffmanNode.buildTree(hw4.java:64)
at HuffmanNode.buildTree(hw4.java:64)
at HuffmanNode.buildTree(hw4.java:64)

并且输出是不断的 5:1, 5:4, 5:2, 重复

我的数据文件如下所示:

a
a
a
a
d
d
d
d
d
d
d
d
k
k
k
k
k
k
f
f
f
f
f
f
h
h
h
h
h
h
b
b
b
b
b
b
b
b
n
n
n
n
n
n
n
e
e
e
e
e
i
i
i
i
i
i
i
i
l
k
j
a
n
s
g
l
k
j
a
s
v
o
i
j
a
s
d
l
k
g
k
n
m
a
s
d
k
l
o
v
h
a
s
d
g
z

我的代码是

    import java.util.*;
import java.io.*;

class HuffmanNode implements Comparable<HuffmanNode>{
HuffmanNode right;
HuffmanNode left;
HuffmanNode parent;
int count;          
String letter;

public HuffmanNode(){}

public HuffmanNode (String letter, int count){
this.letter = letter;
this.count = count;
}
public HuffmanNode (String letter, int count, HuffmanNode parent, HuffmanNode left, HuffmanNode right){
    this.letter = letter;
    this.count = count;
    this.left = left;
    this.right = right;
    this.parent = parent;
}

public void setCount(int count){
this.count = count;
}

public int getCount(){
return count;
}

public void setRight(HuffmanNode right){
this.right = right;
}

public HuffmanNode getRight(HuffmanNode right){
return right;
}

public void setLeft(HuffmanNode left){
this.left = left;
}

public HuffmanNode getLeft(HuffmanNode left){
return left;
}       
public void setParent(HuffmanNode right){
this.left = left;
}   
public HuffmanNode getParent(HuffmanNode parent){
return parent;
}

public void buildTree(HuffmanNode node){
    if (node.compareTo(this) <= 0 && left != null){
    System.out.println(node.getCount() + ":" + this.count);
    left.buildTree(node);
    }
    else if (node.compareTo(this) <= 0 && left == null){
    this.left = node;
    node.parent = this;
    }
    else if (node.compareTo(this) > 0 && right != null){
    System.out.println(node.getCount() + ":" +this.count);
    right.buildTree(node);
    }
    else if (node.compareTo(this) > 0 && right == null){
    this.right = node;
    node.parent = this;
    }
}


public int compareTo(HuffmanNode x){
return this.count - x.count;
}
public void genCode(String s){
    if(left != null){
    left.genCode(s + "0");
    }
    if(right != null){
    right.genCode(s + "1"); 
    }
    if (left == null && right == null){
    System.out.println(s);
    }
}
}

public class hw4{
public static void main (String []args)throws IOException{

//ask user to enter file name
System.out.printf("Enter a file location and name to encode [press Enter]: ");
Scanner input = new Scanner(System.in);
String filename = input.next();

//Gets file name from Scanner and checks to see if valid
File file = new File(filename);
//if (!file.isFile()){
//System.out.printf("Enter a file location and name to encode [press Enter]: ");
//}
Scanner text = new Scanner(file);

String[] letters = {"a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"};
int[] freq = {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0};

String letter;
String tempStr;
int tempInt;

    while(text.hasNext()){
        letter = text.next();
        //System.out.printf("%s\n", letter);

                    char c = letter.charAt(0);
        int index = c - 97;
        freq[index]++;     
    }

    for(int i=0; i <25; i++){
    System.out.printf("%s:%d\n", letters[i], freq[i]);
    }
    System.out.printf("\n");

    for (int n=0; n <25; n++) {
        for (int i=0; i <25; i++) {
            if (freq[i] > freq[i+1]) {
                // exchange elements
                tempInt = freq[i];  
                tempStr = letters[i]; 
                freq[i] = freq[i+1];
                letters[i] = letters[i+1];  
                freq[i+1] = tempInt;
                letters[i+1] = tempStr;
            }
        }   
       } 

    PriorityQueue<HuffmanNode> huffmanList = new PriorityQueue<HuffmanNode>();

    for(int i=0; i <26; i++){
    System.out.printf("%s:%d\n", letters[i], freq[i]);
        if(freq[i] > 0){
        huffmanList.add(new HuffmanNode(letters[i],freq[i]));
        }
    }

    HuffmanNode root = new HuffmanNode();

    while(huffmanList.size() > 1){
    HuffmanNode x = huffmanList.poll();
    HuffmanNode y = huffmanList.poll();
    HuffmanNode result = new HuffmanNode("-", x.getCount() + y.getCount(), null, x, y);
        if(root == null){
        root = result;
        }
        else{
        root.buildTree(result); 
        }
    huffmanList.add(result);                    
    }
     root.genCode(" ");
}
}

【问题讨论】:

    标签: java encoding recursion huffman-code


    【解决方案1】:

    你的造树有问题。

    while(huffmanList.size() > 1){
        HuffmanNode x = huffmanList.poll();
        HuffmanNode y = huffmanList.poll();
    

    你从队列中取出最轻的两棵树,

        HuffmanNode result = new HuffmanNode("-", x.getCount() + y.getCount(), null, x, y);
    

    并合并它们,形成一棵以权重总和为权重的树 - 到目前为止,一切都很好。

        if(root == null){     // never happens, but doesn't matter
            root = result;
        }
        else{
            root.buildTree(result);
    

    然后将新形成的树插入root

        }
        huffmanList.add(result);                    
    }
    

    并将其重新添加到队列中。

    现在,让我们考虑一个以

    开头的队列
    (a,1), (b,2), (c,3), (d,3), (e,3), ...
    

    root = new HuffmanNode();root 设置为(null, 0)

    首先,将ab 节点合并,得到&lt;(a,1) | (-,3) | (b,2)&gt;。插入root 产生

             (null,0)
              /    \
            null  (-,3)
                  /   \
                (a,1) (b,2)
    

    因为3 &gt; 0。队列是

    <(a,1) | (-,3) | (b,2)>, (c,3), (d,3), (e,3) ...
    

    插入合并树后[合并树也可以插入到几个权重为3的节点之后,时间会长一些]。

    现在两个最轻的树被弹出并合并,得到

    <AB | (-,6) | (c,3)>
    

    (缩写为AB = &lt;(a,1) | (-,3) | (b,2)&gt;)。然后将该树插入到root 树中。 6 &gt; 0,所以它被插入到root6 &gt; 3的右孩子,所以它被插入到(-,3)6 &gt; 2的右孩子,所以它成为(b,2)节点的右孩子。但是,新合并的树的左孩子和root 的右孩子指的是同一个对象,所以在插入之后,你就有了

                       __________
                      |          |
                      v          |
        (null,0)   (-,6)         |
        /      \   /   \         |
      null     (-,3)   (c,3)     |
               /   \             |
           (a,1)   (b,2)         |
                       \_________|
    

    本应是一棵树的循环。接下来,(d,3)(e,3) 两个节点被弹出并合并,给出一个权重为 6 的树,当该树被插入到 root 图中时,它会循环。

    不同的插入行为和/或不同的字母权重会改变细节,但事实上,在root.buildTree(result);huffmanList.add(result); 之后,队列包含对root 顶部的图形的引用,只要你有最初有足够的节点。而且一旦有足够的循环,buildTree() 调用不会陷入无限循环的可能性很小。

    你根本不应该打电话给root.buildTree(result)。通过简单地合并队列中最轻的两个并重新插入结果来构造树,直到队列只包含一棵树。

    while(huffmanList.size() > 1){
        HuffmanNode x = huffmanList.poll();
        HuffmanNode y = huffmanList.poll();
        HuffmanNode result = new HuffmanNode("-", x.getCount() + y.getCount(), null, x, y);
        huffmanList.add(result);                    
    }
    root = huffmanList.poll();
    

    【讨论】:

    • 丹。一个真正壮观和直观的答案。我非常感谢您花时间以如此勤奋的方式诊断代码。非常感谢。
    【解决方案2】:

    尝试改变

    if(root.equals("null")){
    

    if(root == null){
    

    还有

    尝试像这样缩短您众多的 if 代码

     char c = letter.charAt(0);
     int index = c - 97;
     freq[index]++;
    

    【讨论】:

    • @user1093111,好的,你必须改变它。将引用与 String 进行比较完全没有意义。
    • 我一会儿检查一下。我试着把它变成一个字符串,因为我收到一个错误,说它不具有可比性。撤消直到我下课时可以到达那一点,并修复频率 if 语句
    • 很奇怪,我粘贴了,效果很好?你有什么代码?
    • @Dimitri。排队是什么意思
    • @NikolayKuznetsov k 我做了编辑,怎么还有错误
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