Huffman编码相信学过数据结构这么课的都知道,概念也比较好理解,但是一般好理解的算法,在实际实现的过程中总是会遇到各种问题,一方面个人认为是对算法的实现过程不熟,另一方面在实际实现的过程中可以提升自己实现算法的能力,将自己的想法实现后还是比较满足的。下面是本人亲自实现的Huffman编码与解码的C语言实现,主要是记录一下自己当时的想法,供以后备忘吧。
数据结构定义如下:
typedef struct{ unsigned int weight; unsigned int parent,lchild,rchild; }HTNode, * HuffmanTree; typedef char * * HuffmanCode;
建Huffman树的过程是使用顺序结构数组存储树,由于没有度为一的节点,因此总数为2*n - 1个节点,n为叶子节点个数,也是待编码的字符个数。
建树的关键代码如下:
//建立Huffman树,初始化1到n号元素的parent都为0,每次从parent为0的元素中 //挑选最小的两个建树之后,将它们的parent都置为对应号码 for(i = n + 1; i <= m; i++) { int min1, min2; int j; for(j = 1; j <= i - 1; j++) if(HT[j].parent == 0) {min1 = j; break;} for(j = 1; j <= i - 1; j++) { if(HT[j].parent != 0) continue; if(HT[j].weight < HT[min1].weight) min1 = j; } HT[min1].parent = i; for(j = 1; j <= i - 1; j++) if(HT[j].parent == 0) {min2 = j; break;} for(j = 1; j <= i - 1; j++) { if(HT[j].parent != 0) continue; if(HT[j].weight < HT[min2].weight) min2 = j; } HT[min2].parent = i; HT[i].lchild = min1; HT[i].rchild = min2; HT[i].weight = HT[min1].weight + HT[min2].weight; }
编码过程是更加树的结构,对每个非叶子节点的左子树为‘0’,右子树为‘1’。实现如下:
//编码 HuffmanCode HC = (HuffmanCode)malloc(n*sizeof(char *)); char * cd = (char *)malloc(n*sizeof(char)); cd[n-1] = ‘\0‘; for(i = 0; i < n; i++) { int end = n - 1; int cur = i + 1; for (int a = HT[cur].parent; a != 0; cur = a, a = HT[a].parent) { if (HT[a].lchild == cur) cd[--end] = ‘0‘; else if (HT[a].rchild == cur) cd[--end] = ‘1‘; } HC[i] = (char*)malloc((n-end)*sizeof(char)); strcpy(HC[i], &cd[end]); } free(cd);
全部实现,封装在一个HuffmanEncode函数中。
HuffmanCode HuffmanEncode(HuffmanTree & HT, unsigned int * w, int n) { int m = 2 * n - 1; HT = (HuffmanTree)malloc((m+1)*sizeof(HTNode)); //第一个不用 int i; for(i = 1; i <= n; i++) { HT[i].weight = w[i-1]; HT[i].parent = 0; HT[i].lchild = 0; HT[i].rchild = 0; } for(i = n + 1;i <= m; i++) { HT[i].weight = 0; HT[i].parent = 0; HT[i].lchild = 0; HT[i].rchild = 0; } //建立Huffman树,初始化1到n号元素的parent都为0,每次从parent为0的元素中 //挑选最小的两个建树之后,将它们的parent都置为对应号码 for(i = n + 1; i <= m; i++) { int min1, min2; int j; for(j = 1; j <= i - 1; j++) if(HT[j].parent == 0) {min1 = j; break;} for(j = 1; j <= i - 1; j++) { if(HT[j].parent != 0) continue; if(HT[j].weight < HT[min1].weight) min1 = j; } HT[min1].parent = i; for(j = 1; j <= i - 1; j++) if(HT[j].parent == 0) {min2 = j; break;} for(j = 1; j <= i - 1; j++) { if(HT[j].parent != 0) continue; if(HT[j].weight < HT[min2].weight) min2 = j; } HT[min2].parent = i; HT[i].lchild = min1; HT[i].rchild = min2; HT[i].weight = HT[min1].weight + HT[min2].weight; } //编码 HuffmanCode HC = (HuffmanCode)malloc(n*sizeof(char *)); char * cd = (char *)malloc(n*sizeof(char)); cd[n-1] = ‘\0‘; for(i = 0; i < n; i++) { int end = n - 1; int cur = i + 1; for (int a = HT[cur].parent; a != 0; cur = a, a = HT[a].parent) { if (HT[a].lchild == cur) cd[--end] = ‘0‘; else if (HT[a].rchild == cur) cd[--end] = ‘1‘; } HC[i] = (char*)malloc((n-end)*sizeof(char)); strcpy(HC[i], &cd[end]); } free(cd); return HC; }
对于编码后得到的编码字符序列HC,结果过程只需找到对应的下标即可。如下:
int HuffmanDecode(HuffmanTree HT,char* code, int n) { int i = 0, r; for(int j = 1; j <= 2*n-1; j++) if(HT[j].parent == 0) {r = j;break;} while(code[i] == ‘0‘ || code[i] == ‘1‘) { if(code[i] == ‘0‘) r = HT[r].lchild; else if(code[i] == ‘1‘) r = HT[r].rchild; i++; } return r; }
最后,主函数中的调用如下,整个实现起来后比较方便。
int main() { printf("请输入待编码的字符个数:\n"); int n = 0; scanf("%d", &n); char codechar[n]; unsigned int weight[n]; printf("请输入编码字符:\n"); scanf("%s", codechar); for(int i = 0; i < n; i++) { printf("\n请输入第%d个字符对应的权重:\n", i+1); scanf("%d", &weight[i]); } HuffmanTree ht; HuffmanCode hc = HuffmanEncode(ht, weight, n); printf("\n构建的赫夫曼树如下(权重-左孩子-右孩子-父亲):\n"); for(int i = 1; i <= 2*n-1; i++) { printf("%d\t%d\t%d\t%d\n", ht[i].weight, ht[i].lchild, ht[i].rchild, ht[i].parent); } printf("\n每个字符对应的编码如下:\n"); for(int i = 0; i < n; i++) { printf("%c\t:\t%s\n", codechar[i], hc[i]); } printf("\n请输入待解码的编码字符串:"); char str[n]; scanf("%s", str); int hcindex = HuffmanDecode(ht, str, n); printf("%s编码的字符是:%c\n", str, codechar[hcindex-1]); return 0; }
结果如下:
时间: 2024-10-12 09:57:12