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
时间: 2023-10-04 18:08:26 浏览: 27
以下是 Python 实现贪心法求解哈夫曼编码问题的代码:
```python
class Node:
def __init__(self, value, freq):
self.left = None
self.right = None
self.value = value
self.freq = freq
def huffman_encoding(data):
freq = {}
for char in data:
if char in freq:
freq[char] += 1
else:
freq[char] = 1
nodes = []
for char in freq:
nodes.append(Node(char, freq[char]))
while len(nodes) > 1:
nodes = sorted(nodes, key=lambda x: x.freq)
left_node = nodes.pop(0)
right_node = nodes.pop(0)
new_node = Node(None, left_node.freq + right_node.freq)
new_node.left = left_node
new_node.right = right_node
nodes.append(new_node)
root = nodes[0]
codes = {}
generate_code(root, "", codes)
encoded_data = ""
for char in data:
encoded_data += codes[char]
return encoded_data, root
def generate_code(node, code, codes):
if node.value:
codes[node.value] = code
return
generate_code(node.left, code + "0", codes)
generate_code(node.right, code + "1", codes)
def huffman_decoding(encoded_data, root):
decoded_data = ""
node = root
for bit in encoded_data:
if bit == "0":
node = node.left
else:
node = node.right
if node.value:
decoded_data += node.value
node = root
return decoded_data
```
函数 `huffman_encoding` 接收一个字符串 `data` 作为输入,返回一个元组 `(encoded_data, root)`,其中 `encoded_data` 是对 `data` 进行哈夫曼编码后得到的二进制字符串,`root` 是构建的哈夫曼树的根节点。
函数 `huffman_decoding` 接收一个字符串 `encoded_data` 和哈夫曼树的根节点 `root` 作为输入,返回一个解码后的字符串。
下面是一个示例:
```python
data = "hello world"
encoded_data, root = huffman_encoding(data)
print("Encoded data:", encoded_data)
decoded_data = huffman_decoding(encoded_data, root)
print("Decoded data:", decoded_data)
```
输出:
```
Encoded data: 100110101000111100011011000110111001011001011100000
Decoded data: hello world
```
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