(数据结构)Huffman树和编码

最新推荐文章于 2025-11-30 16:20:41 发布

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本文详细阐述了如何使用Huffman编码算法构建字符编码过程，包括构造Huffman树、求解编码以及释放工作空间等关键步骤。通过实例演示，帮助读者理解并掌握Huffman编码的基本原理和应用。

#include <STDIO.H>
#include <stdlib.h>
#include <string.h>

typedef unsigned int unint;
#define MAX 1000;

typedef struct  
{
	unint weight;
	unint parent, lchild, rchild;
}HTNode, *HuffmanTree;

typedef char ** HuffmanCode;


void Select(HuffmanTree &HT, unint n, unint &s1num, unint &s2num)
{
	unint tempMin, tempNum;
	unint i;
	unint s1, s2;
	
	s1 = s2 = MAX;

	for(i = 1; i <= n; i++)
		if( (0 == HT[i].parent) && (HT[i].weight < s2) )
		{
			s2 = HT[i].weight;
			s2num = i;
			if(s2 < s1)
			{
				tempMin = s2;
				s2 = s1;
				s1 = tempMin;

				tempNum = s2num;
				s2num = s1num;
				s1num = tempNum;
			}
		}
	if (0 == HT[s2num].lchild)
	{
		tempNum = s2num;
		s2num = s1num;
		s1num = tempNum;
	}
}

void HuffmanCoding( HuffmanTree &HT, HuffmanCode &HC, unint *w, unint n )
{	// w存放n个字符的权值(均>0)，构造Huffman树HT，并求出n个字符的Huffman编码HC。
	unint m, i, start, c, s2, s1, f;

	HuffmanTree p;
	char *cd;

	if (n <= 1)		return;
	m = 2 * n - 1;
	HT = (HuffmanTree) malloc ( (m + 1) * sizeof(HTNode) );// 0号单元未用

	for (p = HT + 1, i = 1; i <= n; ++i, ++p, ++w)
	{
	//	*p = { *w, 0, 0, 0 };
		p->weight = *w;
		p->parent = 0;
		p->lchild = 0;
		p->rchild = 0;
	}
	for (; i <= m; ++i, ++p)
	{
		//*p = { 0, 0, 0, 0};
		p->weight = 0;
		p->parent = 0;
		p->lchild = 0;
		p->rchild = 0;
	}
	for (i = n+1; i <= m; ++i)// 建Huffman树
	{
		Select(HT, i - 1, s1, s2);
		// 在HT[1..i-1]选择parent为0且weight最小的两个结点，其序号分别为s1和s2。
		HT[s1].parent = i;
		HT[s2].parent = i;
		HT[i].lchild = s1;
		HT[i].rchild = s2;
		HT[i].weight = HT[s1].weight + HT[s2].weight;
	}

	//----从叶子到根逆向求每个字符的Huffman编码------------
	HC = (HuffmanCode) malloc ( (n + 1) * sizeof(char *) );// 分配n个字符编码的头指针向量
	cd = (char *) malloc (n * sizeof(char));// 分配求编码的工作空间
	cd[n - 1] = '\0';						// 编码结束符
	for (i = 1; i <= n; ++i)				// 逐个字符求Huffman编码
	{
		start = n - 1;
		for (c = i, f = HT[i].parent; f != 0; c = f, f = HT[f].parent)// 从叶子到根逆向求编码
			if (c == HT[f].lchild)
				cd[-- start] = '0';
			else
				cd[-- start] = '1';
		HC[i] = (char *) malloc ( (n - start) * sizeof(char) );// 为第i个字符编码分配空间
		strcpy(HC[i], &cd[start]);								// 从cd复制编码(串)到HC
		printf("%s\n",HC[i]);
		free(HC[i]);
	}
	free(HT);// 释放工作空间
	free(cd);
	free(HC);
}

int main()
{
	HuffmanTree HT;
	HuffmanCode HC;
	unint w[8]={5, 29, 7, 8, 14, 23, 3, 11};
	unint n = 8;

	HuffmanCoding(HT, HC, w, n);

	return 0;
}

不清楚有木有错~ 求大牛带