数据结构 树 Huffman编码

1． Huffman树不是唯一的，这是因为在选择具有最小权值的2个节点时会出现权值相等的情况。

编码效率是不是一样的？不一样，但相差不大。

如何选择最小的？

2． 一组数，找最小的两个，算法的复杂度是O（n + ceil(log(n)) - 2），使用堆排序。

3． Huffman编码算法用于压缩。

程序如下：

#include<stdio.h>
#include<stdlib.h>
#include<string.h>

typedefstruct...{
unsignedintweight;
unsignedintparent;
unsignedintleft;
unsignedintright;
}HNode;

typedefchar**HCode;

voidselect(HNode*htree,inti,int&s1,int&s2)...{
for(intj=1;j<=i;j++)...{
if(htree[j].parent!=0)...{
continue;
}else...{
s1=j;
break;
}
}

for(j++;j<=i;j++)...{
if(htree[j].parent!=0)...{
continue;
}else...{
s2=j;
break;
}
}

if(htree[s1].weight>htree[s2].weight)...{
inttmp=s2;
s2=s1;
s1=tmp;
}

for(j=1;j<=i;j++)...{
if(htree[j].parent==0)...{
if(htree[j].weight<=htree[s2].weight)...{
if(htree[j].weight<htree[s1].weight)...{
s2=s1;
s1=j;
}else...{
if(j!=s1)...{
s2=j;
}
}
}
}
}
}

voidHuffmanCoding(HNode*htree,HCode&hcode,int*w,intn)...{
if(n<=1)...{
return;
}

intm=2*n-1;
inti=0;

htree=(HNode*)malloc(sizeof(HNode)*(m+1));

HNode*p=htree;
for(i=1;i<=n;i++)...{
p++;

(*p).weight=w[i-1];

(*p).parent=(*p).left=(*p).right=0;
}

for(;i<=m;i++)...{
p++;
(*p).weight=(*p).parent=(*p).left=(*p).right=0;
}


ints1=0,s2=0;
for(i=n+1;i<=m;i++)...{
select(htree,i-1,s1,s2);
htree[s1].parent=i;
htree[s2].parent=i;
htree[i].left=s1;
htree[i].right=s2;
htree[i].weight=htree[s1].weight+htree[s2].weight;
}


hcode=(HCode)malloc(sizeof(char*)*(n+1));

char*ch=(char*)malloc(n*sizeof(char));
ch[n-1]='

执行结果为：

Huffman编码为：
a:11111
b:10
c:1110
d:000
e:110
f:01
g:11110
h:001
1111110000110
abde
Pressanykeytocontinue