Question

我正在努力了解如何解决我的家庭作业问题。我正在尝试创建一个将用Java编码和解码消息的霍夫曼树。我有字符串和频率。

[a=10, b=15, c=12, e=3, nl=4, sp=13, t=1].

我知道，对于Huffman Tree，您可以使用两个最低频率并将它们放入一棵树中，其频率为总和。我知道使用优先级队列我可以将所有频率插入其中，并使用remove()方法取出2个最低频率。然后将它们加在一起以获得它们的重量，然后将该重量插入队列并重复。

最终的树应该保持

的重量

[58=root, root.left = 33, root.right = 25]   
[33.left = 18,  18.left = 8,  8.left = 4]

我不确定如何开始实现一个能够用频率创建树并显示树的霍夫曼树代码。我看看其他代码，似乎他们都是从Streaming Input Code创建的。

任何帮助都会让我前进。提前谢谢！

我想用以下格式打印出来:(预订遍历）

58  
- 33  
- - 18  
- - - 8  
- - - - 4  
- - - - - 1:t  
- - - - - 3:e  
- - - -  4:nl  
- - - 10:a  
- - 15:b  
- 25  
- - 12:c  
- - 13:sp

Answer 1

import java.util.PriorityQueue;

public class Node implements Comparable<Node> {
    Node left;
    Node right;
    Node parent;
    String text;
    int frequency;

    public Node(String textIn, int frequencyIn) {
        text = textIn;
        frequency = frequencyIn;
    }

    public Node(int frequencyIn) {
        text = "";
        frequency = frequencyIn;
    }

    public int compareTo(Node n) {
        if (frequency < n.frequency) {
            return -1;
        }
        else if(frequency > n.frequency) {
            return 1;
        }
        return 0;
    }

    public static void printFromPreOrder(Node n, String dashes) {
        // print with colon if leaf node
        if (n.text != "") {
            System.out.println(dashes + n.frequency + ":" + n.text);
        }
        else {
            System.out.println(dashes + n.frequency);
        }

        // Start recursive on left child then right
        if (n.left != null) {
            printFromPreOrder(n.left, dashes + "-");
        }
        if (n.right != null) {
            printFromPreOrder(n.right, dashes + "-");
        }
    }

    // Returns root node to pass to printFromPreOrder
    public static Node makeHuffmanTree(int frequencies[], String text[]) {
        PriorityQueue<Node> queue = new PriorityQueue<Node>();
        for (int i = 0; i < text.length; i++) {
            Node n = new Node(text[i], frequencies[i]);
            queue.add(n);
        }
        Node root = null;
        while (queue.size() > 1)  {
            Node least1 = queue.poll();
            Node least2 = queue.poll();
            Node combined = new Node(least1.frequency + least2.frequency);
            combined.right = least1;
            combined.left = least2;
            least1.parent = combined;
            least2.parent = combined;
            queue.add(combined);
            // Keep track until we actually find the root
            root = combined;
        }
        return root;
    }

    public static void main(String[] args) {
        int frequencies[] = {10, 15, 12, 3, 4, 13, 1};
        String text[] = {"a", "b", "c", "e", "nl", "sp", "t"};
        Node root = Node.makeHuffmanTree(frequencies, text);
        Node.printFromPreOrder(root, "");
    }
}

这对你有用。我已经包含了您的示例，但它应该适用于任意数量的频率和文本。只需确保频率和文本大小相同，因为我没有进行错误检查。

具有给定频率的霍夫曼树如何混淆如何开始？ Java的

1 个答案: