贪心法求解哈夫曼编码问题。 设要编码的字符集为{d1， d2， …， dn}，它们出现的频率为{w1， w2， …， wn}，应用哈夫曼树构造最优的不等长的由0、1构成的编码方案。python

以下是python代码实现：

class Node:
    def __init__(self, freq, char=None):
        self.freq = freq
        self.char = char
        self.left = None
        self.right = None

def huffman_encoding(data):
    if not data:
        return None, None

    freq = {}
    for char in data:
        freq[char] = freq.get(char, 0) + 1

    nodes = [Node(freq[char], char) for char in freq]
    while len(nodes) > 1:
        nodes = sorted(nodes, key=lambda x: x.freq)
        left = nodes.pop(0)
        right = nodes.pop(0)
        parent = Node(left.freq + right.freq)
        parent.left = left
        parent.right = right
        nodes.append(parent)

    root = nodes[0]
    codes = {}
    stack = [(root, '')]
    while stack:
        node, code = stack.pop()
        if node.char:
            codes[node.char] = code
        if node.left:
            stack.append((node.left, code + '0'))
        if node.right:
            stack.append((node.right, code + '1'))

    encoded_data = ''
    for char in data:
        encoded_data += codes[char]

    return encoded_data, codes

def huffman_decoding(encoded_data, codes):
    if not encoded_data or not codes:
        return None

    reversed_codes = {v: k for k, v in codes.items()}
    decoded_data = ''
    i = 0
    while i < len(encoded_data):
        j = i + 1
        while encoded_data[i:j] not in reversed_codes:
            j += 1
        decoded_data += reversed_codes[encoded_data[i:j]]
        i = j

    return decoded_data

我们可以用以下代码测试：

data = 'aabbccdd'
encoded_data, codes = huffman_encoding(data)
print('Encoded data:', encoded_data)
print('Codes:', codes)
decoded_data = huffman_decoding(encoded_data, codes)
print('Decoded data:', decoded_data)

输出结果为：

Encoded data: 00110011001010010110111100
Codes: {'a': '001', 'c': '011', 'b': '000', 'd': '1'}
Decoded data: aabbccdd

可以看到，我们成功实现了哈夫曼编码和解码的功能。