Huffman编码算法之Java实现

Huffman编码介绍

Huffman编码处理的是字符以及字符对应的二进制的编码配对问题,分为编码和解码,目的是压缩字符对应的二进制数据长度。我们知道字符存贮和传输的时候都是二进制的(计算机只认识0/1),那么就有字符与二进制之间的mapping关系。字符属于字符集(Charset), 字符需要通过编码(encode)为二进制进行存贮和传输,显示的时候需要解码(decode)回字符,字符集与编码方法是一对多关系(Unicode可以用UTF-8,UTF-16等编码)。理解了字符集,编码以及解码,满天飞的乱码问题也就游刃而解了。以英文字母小写a为例, ASCII编码中,十进制为97,二进制为01100001。ASCII的每一个字符都用8个Bit(1Byte)编码,假如有1000个字符要传输,那么就要传输8000个Bit。问题来了,英文中字母e的使用频率为12.702%,而z为0.074%,前者是后者的100多倍,但是确使用相同位数的二进制。可以做得更好,方法就是可变长度编码,指导原则就是频率高的用较短的位数编码,频率低的用较长位数编码。Huffman编码算法就是处理这样的问题。

Huffman编码Java实现

Huffman编码算法主要用到的数据结构是完全二叉树(full binary tree)和优先级队列。后者用的是java.util.PriorityQueue,前者自己实现(都为内部类),代码如下:

static class Tree {
		private Node root;

		public Node getRoot() {
			return root;
		}

		public void setRoot(Node root) {
			this.root = root;
		}
	}

	static class Node implements Comparable<Node> {
		private String chars = "";
		private int frequence = 0;
		private Node parent;
		private Node leftNode;
		private Node rightNode;

		@Override
		public int compareTo(Node n) {
			return frequence - n.frequence;
		}

		public boolean isLeaf() {
			return chars.length() == 1;
		}

		public boolean isRoot() {
			return parent == null;
		}

		public boolean isLeftChild() {
			return parent != null && this == parent.leftNode;
		}

		public int getFrequence() {
			return frequence;
		}

		public void setFrequence(int frequence) {
			this.frequence = frequence;
		}

		public String getChars() {
			return chars;
		}

		public void setChars(String chars) {
			this.chars = chars;
		}

		public Node getParent() {
			return parent;
		}

		public void setParent(Node parent) {
			this.parent = parent;
		}

		public Node getLeftNode() {
			return leftNode;
		}

		public void setLeftNode(Node leftNode) {
			this.leftNode = leftNode;
		}

		public Node getRightNode() {
			return rightNode;
		}

		public void setRightNode(Node rightNode) {
			this.rightNode = rightNode;
		}
	}

统计数据

既然要按频率来安排编码表,那么首先当然得获得频率的统计信息。我实现了一个方法处理这样的问题。如果已经有统计信息,那么转为Map<Character,Integer>即可。如果你得到的信息是百分比,乘以100或1000,或10000。总是可以转为整数。比如12.702%乘以1000为12702,Huffman编码只关心大小问题。统计方法实现如下:

public static Map<Character, Integer> statistics(char[] charArray) {
		Map<Character, Integer> map = new HashMap<Character, Integer>();
		for (char c : charArray) {
			Character character = new Character(c);
			if (map.containsKey(character)) {
				map.put(character, map.get(character) + 1);
			} else {
				map.put(character, 1);
			}
		}

		return map;
	}

构建树

构建树是Huffman编码算法的核心步骤。思想是把所有的字符挂到一颗完全二叉树的叶子节点,任何一个非页子节点的左节点出现频率不大于右节点。算法为把统计信息转为Node存放到一个优先级队列里面,每一次从队列里面弹出两个最小频率的节点,构建一个新的父Node(非叶子节点), 字符内容刚弹出来的两个节点字符内容之和,频率也是它们的和,最开始的弹出来的作为左子节点,后面一个作为右子节点,并且把刚构建的父节点放到队列里面。重复以上的动作N-1次,N为不同字符的个数(每一次队列里面个数减1)。结束以上步骤,队列里面剩一个节点,弹出作为树的根节点。代码如下:

private static Tree buildTree(Map<Character, Integer> statistics,
			List<Node> leafs) {
		Character[] keys = statistics.keySet().toArray(new Character[0]);

		PriorityQueue<Node> priorityQueue = new PriorityQueue<Node>();
		for (Character character : keys) {
			Node node = new Node();
			node.chars = character.toString();
			node.frequence = statistics.get(character);
			priorityQueue.add(node);
			leafs.add(node);
		}

		int size = priorityQueue.size();
		for (int i = 1; i <= size - 1; i++) {
			Node node1 = priorityQueue.poll();
			Node node2 = priorityQueue.poll();

			Node sumNode = new Node();
			sumNode.chars = node1.chars + node2.chars;
			sumNode.frequence = node1.frequence + node2.frequence;

			sumNode.leftNode = node1;
			sumNode.rightNode = node2;

			node1.parent = sumNode;
			node2.parent = sumNode;

			priorityQueue.add(sumNode);
		}

		Tree tree = new Tree();
		tree.root = priorityQueue.poll();
		return tree;
	}

编码

某个字符对应的编码为,从该字符所在的叶子节点向上搜索,如果该字符节点是父节点的左节点,编码字符之前加0,反之如果是右节点,加1,直到根节点。只要获取了字符和二进制码之间的mapping关系,编码就非常简单。代码如下:

public static String encode(String originalStr,
			Map<Character, Integer> statistics) {
		if (originalStr == null || originalStr.equals("")) {
			return "";
		}

		char[] charArray = originalStr.toCharArray();
		List<Node> leafNodes = new ArrayList<Node>();
		buildTree(statistics, leafNodes);
		Map<Character, String> encodInfo = buildEncodingInfo(leafNodes);

		StringBuffer buffer = new StringBuffer();
		for (char c : charArray) {
			Character character = new Character(c);
			buffer.append(encodInfo.get(character));
		}

		return buffer.toString();
	}

private static Map<Character, String> buildEncodingInfo(List<Node> leafNodes) {
		Map<Character, String> codewords = new HashMap<Character, String>();
		for (Node leafNode : leafNodes) {
			Character character = new Character(leafNode.getChars().charAt(0));
			String codeword = "";
			Node currentNode = leafNode;

			do {
				if (currentNode.isLeftChild()) {
					codeword = "0" + codeword;
				} else {
					codeword = "1" + codeword;
				}

				currentNode = currentNode.parent;
			} while (currentNode.parent != null);

			codewords.put(character, codeword);
		}

		return codewords;
	}

解码

因为Huffman编码算法能够保证任何的二进制码都不会是另外一个码的前缀,解码非常简单,依次取出二进制的每一位,从树根向下搜索,1向右,0向左,到了叶子节点(命中),退回根节点继续重复以上动作。代码如下:

public static String decode(String binaryStr,
			Map<Character, Integer> statistics) {
		if (binaryStr == null || binaryStr.equals("")) {
			return "";
		}

		char[] binaryCharArray = binaryStr.toCharArray();
		LinkedList<Character> binaryList = new LinkedList<Character>();
		int size = binaryCharArray.length;
		for (int i = 0; i < size; i++) {
			binaryList.addLast(new Character(binaryCharArray[i]));
		}

		List<Node> leafNodes = new ArrayList<Node>();
		Tree tree = buildTree(statistics, leafNodes);

		StringBuffer buffer = new StringBuffer();

		while (binaryList.size() > 0) {
			Node node = tree.root;

			do {
				Character c = binaryList.removeFirst();
				if (c.charValue() == '0') {
					node = node.leftNode;
				} else {
					node = node.rightNode;
				}
			} while (!node.isLeaf());

			buffer.append(node.chars);
		}

		return buffer.toString();
	}

测试以及比较

以下测试Huffman编码的正确性(先编码,后解码,包括中文),以及Huffman编码与常见的字符编码的二进制字符串比较。代码如下:

public static void main(String[] args) {
		String oriStr = "Huffman codes compress data very effectively: savings of 20% to 90% are typical, "
				+ "depending on the characteristics of the data being compressed. 中华崛起";
		Map<Character, Integer> statistics = statistics(oriStr.toCharArray());
		String encodedBinariStr = encode(oriStr, statistics);
		String decodedStr = decode(encodedBinariStr, statistics);

		System.out.println("Original sstring: " + oriStr);
		System.out.println("Huffman encoed binary string: " + encodedBinariStr);
		System.out.println("decoded string from binariy string: " + decodedStr);

		System.out.println("binary string of UTF-8: "
				+ getStringOfByte(oriStr, Charset.forName("UTF-8")));
		System.out.println("binary string of UTF-16: "
				+ getStringOfByte(oriStr, Charset.forName("UTF-16")));
		System.out.println("binary string of US-ASCII: "
				+ getStringOfByte(oriStr, Charset.forName("US-ASCII")));
		System.out.println("binary string of GB2312: "
				+ getStringOfByte(oriStr, Charset.forName("GB2312")));
	}

	public static String getStringOfByte(String str, Charset charset) {
		if (str == null || str.equals("")) {
			return "";
		}

		byte[] byteArray = str.getBytes(charset);
		int size = byteArray.length;
		StringBuffer buffer = new StringBuffer();
		for (int i = 0; i < size; i++) {
			byte temp = byteArray[i];
			buffer.append(getStringOfByte(temp));
		}

		return buffer.toString();
	}

	public static String getStringOfByte(byte b) {
		StringBuffer buffer = new StringBuffer();
		for (int i = 7; i >= 0; i--) {
			byte temp = (byte) ((b >> i) & 0x1);
			buffer.append(String.valueOf(temp));
		}

		return buffer.toString();
	}

参考链接

    原文作者:游程编码问题
    原文地址: https://blog.csdn.net/kimylrong/article/details/17022319
    本文转自网络文章,转载此文章仅为分享知识,如有侵权,请联系博主进行删除。
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