hash sum mismatch哈夫曼编码码长怎么算？

hash sum mismatch 时间:2022-02-24 阅读:()

哈夫曼编码码长怎么算？

介绍一个不用构造哈夫曼树的方法来计算哈夫曼编码的长度的算法，本算法的时间复杂度为O(nlogn)，空间复杂度为O(N)。下面是具体代码。 #include? #include? #include? #include? #include? #include? #include? #include? using?namespace?std; #define?BUFF_SIZE?4096 #define?HASH_SIZE?256 char?buff[BUFF_SIZE];????????//缓冲区? int??hash[HASH_SIZE];????????//统计每个字符出现的次数 int?heap[(HASH_SIZE<<1)+2]; int?pos[HASH_SIZE+1][3];//内部节点?,?0和1记录子结点位置,3节录当前的深度 int?tlen[HASH_SIZE+1];????????//记录每个叶子结点的深度 int??fd;??????????????????????//文件描述符 int?sym_num?;???????????//文件中出现的符号数量 int?SUM(0);??????????????//文件中字符总数 //初始化程序 void?init(const?char?*?pathname) { ???????memset(buff?,?0?,?sizeof(buff)); ???????memset(hash?,?0?,?sizeof(hash)); ???????memset(heap?,?0?,?sizeof(heap)); ???????memset(pos??,?0?,?sizeof(pos)); ???????memset(tlen?,?0?,?sizeof(tlen)); ???????//打开文件 ???????fd?=?open(pathname?,?O_RDONLY); ???????if(fd?SUM?+=?strlen(buff); ??????????????for(int?i=strlen(buff)?-?1;i>=0;i--) ?????????????????????hash[(unsigned?int)(buff[i]?&?0xFF)]++; ???????} ???????//记录出现的符号数量; ???????for(int?i?=?HASH_SIZE?-?1;?i?>=?0;?i--) ??????????????if(hash[i])sym_num++; } //建立一个最小堆 void?build_min_heap() { ???????for(int?i=sym_num;i>0;i--){ ??????????????int?p?=?i?>>?1?,?j?=?i; ??????????????while(p?>=?1){ ?????????????????????if(heap[heap[p]]?>?heap[heap[j]]) ????????????????????????????std::swap(heap[j]?,?heap[p]); ?????????????????????j?=?p;?p?>>=?1; ??????????????} ???????} } //每次取出最小数之后重新调整堆, //h?指推中元素的个数 void?heap_adjust(int?h) { ???????int?t?=?1?,?p?,?q?,?l; ???????while(t1){ ??????????????//取出最小数 ??????????????m1?=?heap[heap[1]]; ??????????????pos[h][0]?=?heap[1]; ??????????????heap[1]?=?heap[h]; ??????????????h--; ??????????????heap_adjust(h); ??????????????//取出次小数 ??????????????m2?=?heap[heap[1]]; ??????????????pos[h+1][1]?=?heap[1]; ??????????????//最后数和次小数之和放在堆的最后一个位置 ??????????????heap[h+1]?=?m1?+?m2; ??????????????//重新指向最新合并的结点 ??????????????heap[1]?=?h+1; ??????????????heap_adjust(h); ???????} ???????//统计编码长度?,?线性时间统计 ???????int?ts?=?sym_num?<sum?=?0; ???????for(int?i=1;i<=sym_num;i++) ??????????????sum?+=?tlen[i]?*?heap[sym_num?+?i]; ???????cout<

展开全文