我正在研究霍夫曼编码程序，我快完成了，但我陷入了无限递归循环。有谁知道这是哪里出了问题？

这是我遇到的错误:

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)

并且输出连续 5:1、5:4、5:2，重复

我的数据文件是这样的:

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(" ");
}
}

最佳答案

你的树构建有问题。

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

你从队列中取出两棵最轻的树，

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

并合并它们，形成一棵以权重之和作为权重的树 - 到目前为止，一切顺利。

    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到 root 中,

    }
    huffmanList.add(result);                    
}

并将其添加回队列。

现在，让我们考虑一个以

(a,1), (b,2), (c,3), (d,3), (e,3), ...

root = new HuffmanNode();设置 root至 (null, 0) .

首先，a和 b节点合并，给出 <(a,1) | (-,3) | (b,2)> .插入 root产生

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

自 3 > 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，那么时间会长一些]。

现在弹出并合并两棵最轻的树，给出

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

(缩写为 AB = <(a,1) | (-,3) | (b,2)> )。然后将该树插入到 root 中树。 6 > 0 , 所以它被插入到 root 的右 child 中, 6 > 3 , 所以它被插入到 (-,3) 的右 child 中, 6 > 2 , 所以它成为 (b,2) 的右 child 节点。但是，新合并树的左 child 和 root 的右 child 引用同一个对象，所以在这次插入之后，你有

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

本应是树的循环。接下来，两个节点 (d,3)和 (e,3)被弹出并合并，得到一棵权重为 6 的树，然后将这棵树插入到 root 中。图形，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，但事实是在 root.buildTree(result); 之后和 huffmanList.add(result);该队列包含对顶部为 root 的图形的引用只要您最初有足够的节点，就会导致循环。一旦你有足够的周期，buildTree() 的概率调用不会陷入死循环小。

你根本不应该调用 root.buildTree(result) .通过简单地合并队列中两个最轻的树并重新插入结果来构建树，直到队列中只包含一棵树。

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();

关于java - 无限递归，霍夫曼树中的 StackOverflowError，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/13758954/