哈夫曼案例

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成绩⽐比重: 在70~89分之间占⽤用了了70% 但是都是需要经过3次判断才能得到正确的结果. 那么如果数量量集⾮非常⼤大时,这样的⽐比较就会出现效率问题.

这颗树最后:WPL = 1 * 5 + 2 * 15 + 3 *40 +4 * 30 + 4 * 10 = 315

如果用哈夫曼树最后:

屏幕快照 2020-05-18 下午6.36.04.png

这颗树最后WPL = 5 * 3 + 15 * 3 + 40 * 2 + 30 * 2 +10 * 2 = 220
相比哈弗曼树就比第一个颗树要快的多

哈夫曼树构建

初始化哈夫曼⼆二叉树
循环不不断找到结点中,最⼩小的2个结点值. 加⼊入到哈夫曼树中.

typedef struct HaffNode{
    int weight;
    int flag;
    int parent;
    int leftChild;
    int rightChild;
}HaffNode;

typedef struct Code//存放哈夫曼编码的数据元素结构
{
    int bit[MaxBit];//数组
    int start;  //编码的起始下标
    int weight;//字符的权值
}Code;

初始化哈夫曼树

//1.
//根据权重值,构建哈夫曼树;
//{2,4,5,7}
//n = 4;
void Haffman(int weight[],int n,HaffNode *haffTree){
    
    int j,m1,m2,x1,x2;
    
    //1.哈夫曼树初始化
    //n个叶子结点. 2n-1
    for(int i = 0; i < 2*n-1;i++){
        
        if(i<n)
            haffTree[i].weight = weight[i];
        else
            haffTree[i].weight = 0;
        
        haffTree[i].parent = 0;
        haffTree[i].flag = 0;
        haffTree[i].leftChild = -1;
        haffTree[i].rightChild = -1;
    }
    
    
    //2.构造哈夫曼树haffTree的n-1个非叶结点
    for (int i = 0; i< n - 1; i++){
         m1 = m2 = MaxValue;
         x1 = x2 = 0;
        //2,4,5,7
        for (j = 0; j< n + i; j++)//循环找出所有权重中，最小的二个值--morgan
        {
            if (haffTree[j].weight < m1 && haffTree[j].flag == 0)
            {
                m2 = m1;
                x2 = x1;
                m1 = haffTree[j].weight;
                x1 = j;
            } else if(haffTree[j].weight<m2 && haffTree[j].flag == 0)
            {
                m2 = haffTree[j].weight;
                x2 = j;
            }
        }
        
        //3.将找出的两棵权值最小的子树合并为一棵子树
        haffTree[x1].parent = n + i;
        haffTree[x2].parent = n + i;
        //将2个结点的flag 标记为1,表示已经加入到哈夫曼树中
        haffTree[x1].flag = 1;
        haffTree[x2].flag = 1;
        //修改n+i结点的权值
        haffTree[n + i].weight = haffTree[x1].weight + haffTree[x2].weight;
        //修改n+i的左右孩子的值
        haffTree[n + i].leftChild = x1;
        haffTree[n + i].rightChild = x2;
    }
    
}

哈夫曼编码

/*
 9.2 哈夫曼编码
 由n个结点的哈夫曼树haffTree构造哈夫曼编码haffCode
 //{2,4,5,7}
 */
void HaffmanCode(HaffNode haffTree[], int n, Code haffCode[])
{
    //1.创建一个结点cd
    Code *cd = (Code * )malloc(sizeof(Code));
    int child, parent;
    //2.求n个叶结点的哈夫曼编码
    for (int i = 0; i<n; i++)
    {
        //从0开始计数
        cd->start = 0;
        //取得编码对应权值的字符
        cd->weight = haffTree[i].weight;
        //当叶子结点i 为孩子结点.
        child = i;
        //找到child 的双亲结点;
        parent = haffTree[child].parent;
        //由叶结点向上直到根结点
        while (parent != 0)
        {
            if (haffTree[parent].leftChild == child)
                cd->bit[cd->start] = 0;//左孩子结点编码0
            else
                cd->bit[cd->start] = 1;//右孩子结点编码1
            //编码自增
            cd->start++;
            //当前双亲结点成为孩子结点
            child = parent;
            //找到双亲结点
            parent = haffTree[child].parent;
        }
        
         int temp = 0;

        for (int j = cd->start - 1; j >= 0; j--){
            temp = cd->start-j-1;
            haffCode[i].bit[temp] = cd->bit[j];
        }
      
        //把cd中的数据赋值到haffCode[i]中.
        //保存好haffCode 的起始位以及权值;
        haffCode[i].start = cd->start;
        //保存编码对应的权值
        haffCode[i].weight = cd->weight;
    }
}

结果

int main(int argc, const char * argv[]) {
    // insert code here...
    printf("Hello, 哈夫曼编码!\n");
    int i, j, n = 4, m = 0;
    
    //权值
    int weight[] = {2,4,5,7};
    
    //初始化哈夫曼树, 哈夫曼编码
    HaffNode *myHaffTree = malloc(sizeof(HaffNode)*2*n-1);
    Code *myHaffCode = malloc(sizeof(Code)*n);
    
    //当前n > MaxN,表示超界. 无法处理.
    if (n>MaxN)
    {
        printf("定义的n越界，修改MaxN!");
        exit(0);
    }
    
    //1. 构建哈夫曼树
    Haffman(weight, n, myHaffTree);
    //2.根据哈夫曼树得到哈夫曼编码
    HaffmanCode(myHaffTree, n, myHaffCode);
    //3.
    for (i = 0; i<n; i++)
    {
        printf("Weight = %d\n",myHaffCode[i].weight);
        for (j = 0; j<myHaffCode[i].start; j++)
            printf("%d",myHaffCode[i].bit[j]);
        m = m + myHaffCode[i].weight*myHaffCode[i].start;
         printf("\n");
    }
    printf("Huffman's WPS is:%d\n",m);
/*
 Hello, 哈夫曼编码!
 Weight = 2
 110
 Weight = 4
 111
 Weight = 5
 10
 Weight = 7
 0
 Huffman's WPS is:35
 Program ended with exit code: 0
 */
    return 0;
}