huffman编码——原理与实现
哈夫曼算法原理
Wikipedia上面说的非常清楚了，这里我就不再赘述，直接贴过来了。
1952年, David A. Huffman提出了一个不同的算法，这个算法能够为不论什么的可能性提供出一个理想的树。香农-范诺编码（Shanno-Fano）是从树的根节点到叶子节点所进行的的编码，哈夫曼编码算法却是从相反的方向，暨从叶子节点到根节点的方向编码的。
- 为每一个符号建立一个叶子节点，并加上其对应的发生频率
- 当有一个以上的节点存在时，进行下列循环:
- 把这些节点作为带权值的二叉树的根节点，左右子树为空
- 选择两棵根结点权值最小的树作为左右子树构造一棵新的二叉树，且至新的二叉树的根结点的权值为其左右子树上根结点的权值之和。
- 把权值最小的两个根节点移除
- 将新的二叉树增加队列中.
- 最后剩下的节点暨为根节点，此时二叉树已经完毕。
演示样例
-
符号 A B C D E 计数 15 7 6 6 5 概率 0.38461538 0.17948718 0.15384615 0.15384615 0.12820513
在这样的情况下，D，E的最低频率和分配分别为0和1，分组结合概率的0.28205128。如今最低的一双是B和C，所以他们就分配0和1组合结合概率的0.33333333在一起。这使得BC和DE所以0和1的前面加上他们的代码和它们结合的概率最低。然后离开仅仅是一个和BCDE，当中有前缀分别为0和1，然后结合。这使我们与一个单一的节点，我们的算法是完整的。
可得A代码的代码长度是1比特，其余字符是3比特。
-
字符 A B C D E 代码 0 100 101 110 111
Entropy：
Pseudo-code
1: begin
2: count frequencies of single characters (source units)
3: output(frequencies using Fibonacci Codes of degree 2)
4: sort them to non-decreasing sequence
5: create a leaf node (character, frequency c, left son = NULL, right son = NULL)
6: of the tree for each character and put nodes into queue F
7: while (|F|>=2) do
8: begin
9: pop the first two nodes (u1, u2) with the lowest
10: frequencies from sorted queue
11: create a node evaluated with sum of the chosen units,
12: successors are chosen units (eps, c(u1)+c(u2), u1, u2)
13: insert new node into queue
14: end
15: node evaluate with way from root to leaf node (left son 1, right son 0)
16: create output from coded intput characters
17: end哈夫曼算法实现
实现的时候我们用vector<bool>记录每一个char的编码；用map<char,vector<bool>>表示整个字典；
就得到了以下的代码（以下有两个，一对一错）：
先放出来这个错的，以示警戒：
/************************************************************************/
/* File Name: Huffman.cpp
* @Function: Lossless Compression
@Author: Sophia Zhang
@Create Time: 2012-9-26 10:40
@Last Modify: 2012-9-26 11:10
*/
/************************************************************************/
#include"iostream"
#include "queue"
#include "map"
#include "string"
#include "iterator"
#include "vector"
#include "algorithm"
using namespace std;
#define NChar 8 //suppose use at most 8 bits to describe all symbols
#define Nsymbols 1<<NChar //can describe 256 symbols totally (include a-z, A-Z)
typedef vector<bool> Huff_code;//8 bit code of one char
map<char,Huff_code> Huff_Dic; //huffman coding dictionary
class HTree
{
public :
HTree* left;
HTree* right;
char ch;
int weight;
HTree(){left = right = NULL; weight=0;}
HTree(HTree* l,HTree* r,int w,char c){left = l; right = r; weight=w; ch=c;}
~HTree(){delete left; delete right;}
int Getweight(){return weight?weight:left->weight+right->weight;}
bool Isleaf(){return !left && !right; }
bool operator < (const HTree tr) const
{
return tr.weight < weight;
}
};
HTree* BuildTree(int *frequency)
{
priority_queue<HTree*> QTree;
//1st level add characters
for (int i=0;i<Nsymbols;i++)
{
if(frequency[i])
QTree.push(new HTree(NULL,NULL,frequency[i],(char)i));
}
//build
while (QTree.size()>1)
{
HTree* lc = QTree.top();
QTree.pop();
HTree* rc = QTree.top();
QTree.pop();
HTree* parent = new HTree(lc,rc,parent->Getweight(),(char)256);
QTree.push(parent);
}
//return tree root
return QTree.top();
}
void Huffman_Coding(HTree* root, Huff_code& curcode)
{
if(root->Isleaf())
{
Huff_Dic[root->ch] = curcode;
return;
}
Huff_code& lcode = curcode;
Huff_code& rcode = curcode;
lcode.push_back(false);
rcode.push_back(true);
Huffman_Coding(root->left,lcode);
Huffman_Coding(root->right,rcode);
}
int main()
{
int freq[Nsymbols] = {0};
char *str = "this is the string need to be compressed";
//statistic character frequency
while (*str!='\0')
freq[*str++]++;
//build tree
HTree* r = BuildTree(freq);
Huff_code nullcode;
nullcode.clear();
Huffman_Coding(r,nullcode);
for(map<char,Huff_code>::iterator it = Huff_Dic.begin(); it != Huff_Dic.end(); it++)
{
cout<<(*it).first<<'\t';
Huff_code vec_code = (*it).second;
for (vector<bool>::iterator vit = vec_code.begin(); vit!=vec_code.end();vit++)
{
cout<<(*vit)<<endl;
}
}
}
上面这段代码，我执行出来不正确。在调试的时候发现了一个问题，就是QTree优先队列中的排序出了问题，说来也是，上面的代码中，我重载小于号是对HTree object做的；而实际上我建树时用的是指针，那么优先级队列中元素为指针时该怎么办呢？
那我们将friend bool operator >(Node node1,Node node2)改动为friend bool operator >(Node* node1,Node* node2)，也就是传递的是Node的指针行不行呢?
答案是不能够，由于依据c++primer中重载操作符中讲的“程序猿仅仅能为类类型或枚举类型的操作数定义重载操作符，在把操作符声明为类的成员时，至少有一个类或枚举类型的參数依照值或者引用的方式传递”，也就是说friend bool operator >(Node* node1,Node* node2)形參中都是指针类型的是不能够的。我们仅仅能再建一个类，用当中的重载（）操作符作为优先队列的比較函数。
就得到了以下正确的代码：
/************************************************************************/
/* File Name: Huffman.cpp
* @Function: Lossless Compression
@Author: Sophia Zhang
@Create Time: 2012-9-26 10:40
@Last Modify: 2012-9-26 12:10
*/
/************************************************************************/
#include"iostream"
#include "queue"
#include "map"
#include "string"
#include "iterator"
#include "vector"
#include "algorithm"
using namespace std;
#define NChar 8 //suppose use 8 bits to describe all symbols
#define Nsymbols 1<<NChar //can describe 256 symbols totally (include a-z, A-Z)
typedef vector<bool> Huff_code;//8 bit code of one char
map<char,Huff_code> Huff_Dic; //huffman coding dictionary
/************************************************************************/
/* Tree Class elements:
*2 child trees
*character and frequency of current node
*/
/************************************************************************/
class HTree
{
public :
HTree* left;
HTree* right;
char ch;
int weight;
HTree(){left = right = NULL; weight=0;ch ='\0';}
HTree(HTree* l,HTree* r,int w,char c){left = l; right = r; weight=w; ch=c;}
~HTree(){delete left; delete right;}
bool Isleaf(){return !left && !right; }
};
/************************************************************************/
/* prepare for pointer sorting*/
/*because we cannot use overloading in class HTree directly*/
/************************************************************************/
class Compare_tree
{
public:
bool operator () (HTree* t1, HTree* t2)
{
return t1->weight> t2->weight;
}
};
/************************************************************************/
/* use priority queue to build huffman tree*/
/************************************************************************/
HTree* BuildTree(int *frequency)
{
priority_queue<HTree*,vector<HTree*>,Compare_tree> QTree;
//1st level add characters
for (int i=0;i<Nsymbols;i++)
{
if(frequency[i])
QTree.push(new HTree(NULL,NULL,frequency[i],(char)i));
}
//build
while (QTree.size()>1)
{
HTree* lc = QTree.top();
QTree.pop();
HTree* rc = QTree.top();
QTree.pop();
HTree* parent = new HTree(lc,rc,lc->weight+rc->weight,(char)256);
QTree.push(parent);
}
//return tree root
return QTree.top();
}
/************************************************************************/
/* Give Huffman Coding to the Huffman Tree*/
/************************************************************************/
void Huffman_Coding(HTree* root, Huff_code& curcode)
{
if(root->Isleaf())
{
Huff_Dic[root->ch] = curcode;
return;
}
Huff_code lcode = curcode;
Huff_code rcode = curcode;
lcode.push_back(false);
rcode.push_back(true);
Huffman_Coding(root->left,lcode);
Huffman_Coding(root->right,rcode);
}
int main()
{
int freq[Nsymbols] = {0};
char *str = "this is the string need to be compressed";
//statistic character frequency
while (*str!='\0')
freq[*str++]++;
//build tree
HTree* r = BuildTree(freq);
Huff_code nullcode;
nullcode.clear();
Huffman_Coding(r,nullcode);
for(map<char,Huff_code>::iterator it = Huff_Dic.begin(); it != Huff_Dic.end(); it++)
{
cout<<(*it).first<<'\t';
std::copy(it->second.begin(),it->second.end(),std::ostream_iterator<bool>(cout));
cout<<endl;
}
}
Reference：