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1 |
| -# |
| 1 | +#include<stdlib.h> |
| 2 | +#include<stdio.h> |
| 3 | +#include<string.h> |
| 4 | + |
| 5 | +//哈夫曼树结点结构 |
| 6 | +typedef struct { |
| 7 | + int weight;//结点权重 |
| 8 | + int parent, left, right;//父结点、左孩子、右孩子在数组中的位置下标 |
| 9 | +}HTNode, *HuffmanTree; |
| 10 | +//动态二维数组,存储哈夫曼编码 |
| 11 | +typedef char ** HuffmanCode; |
| 12 | + |
| 13 | +//HT数组中存放的哈夫曼树,end表示HT数组中存放结点的最终位置,s1和s2传递的是HT数组中权重值最小的两个结点在数组中的位置 |
| 14 | +void Select(HuffmanTree HT, int end, int *s1, int *s2) |
| 15 | +{ |
| 16 | + int min1, min2; |
| 17 | + //遍历数组初始下标为 1 |
| 18 | + int i = 1; |
| 19 | + //找到还没构建树的结点 |
| 20 | + while(HT[i].parent != 0 && i <= end){ |
| 21 | + i++; |
| 22 | + } |
| 23 | + min1 = HT[i].weight; |
| 24 | + *s1 = i; |
| 25 | + |
| 26 | + i++; |
| 27 | + while(HT[i].parent != 0 && i <= end){ |
| 28 | + i++; |
| 29 | + } |
| 30 | + //对找到的两个结点比较大小,min2为大的,min1为小的 |
| 31 | + if(HT[i].weight < min1){ |
| 32 | + min2 = min1; |
| 33 | + *s2 = *s1; |
| 34 | + min1 = HT[i].weight; |
| 35 | + *s1 = i; |
| 36 | + }else{ |
| 37 | + min2 = HT[i].weight; |
| 38 | + *s2 = i; |
| 39 | + } |
| 40 | + //两个结点和后续的所有未构建成树的结点做比较 |
| 41 | + for(int j=i+1; j <= end; j++) |
| 42 | + { |
| 43 | + //如果有父结点,直接跳过,进行下一个 |
| 44 | + if(HT[j].parent != 0){ |
| 45 | + continue; |
| 46 | + } |
| 47 | + //如果比最小的还小,将min2=min1,min1赋值新的结点的下标 |
| 48 | + if(HT[j].weight < min1){ |
| 49 | + min2 = min1; |
| 50 | + min1 = HT[j].weight; |
| 51 | + *s2 = *s1; |
| 52 | + *s1 = j; |
| 53 | + } |
| 54 | + //如果介于两者之间,min2赋值为新的结点的位置下标 |
| 55 | + else if(HT[j].weight >= min1 && HT[j].weight < min2){ |
| 56 | + min2 = HT[j].weight; |
| 57 | + *s2 = j; |
| 58 | + } |
| 59 | + } |
| 60 | +} |
| 61 | + |
| 62 | +//HT为地址传递的存储哈夫曼树的数组,w为存储结点权重值的数组,n为结点个数 |
| 63 | +void CreateHuffmanTree(HuffmanTree *HT, int *w, int n) |
| 64 | +{ |
| 65 | + if(n<=1) return; // 如果只有一个编码就相当于0 |
| 66 | + int m = 2*n-1; // 哈夫曼树总节点数,n就是叶子结点 |
| 67 | + *HT = (HuffmanTree) malloc((m+1) * sizeof(HTNode)); // 0号位置不用 |
| 68 | + HuffmanTree p = *HT; |
| 69 | + // 初始化哈夫曼树中的所有结点 |
| 70 | + for(int i = 1; i <= n; i++) |
| 71 | + { |
| 72 | + (p+i)->weight = *(w+i-1); |
| 73 | + (p+i)->parent = 0; |
| 74 | + (p+i)->left = 0; |
| 75 | + (p+i)->right = 0; |
| 76 | + } |
| 77 | + //从树组的下标 n+1 开始初始化哈夫曼树中除叶子结点外的结点 |
| 78 | + for(int i = n+1; i <= m; i++) |
| 79 | + { |
| 80 | + (p+i)->weight = 0; |
| 81 | + (p+i)->parent = 0; |
| 82 | + (p+i)->left = 0; |
| 83 | + (p+i)->right = 0; |
| 84 | + } |
| 85 | + //构建哈夫曼树 |
| 86 | + for(int i = n+1; i <= m; i++) |
| 87 | + { |
| 88 | + int s1, s2; |
| 89 | + Select(*HT, i-1, &s1, &s2); |
| 90 | + (*HT)[s1].parent = (*HT)[s2].parent = i; |
| 91 | + (*HT)[i].left = s1; |
| 92 | + (*HT)[i].right = s2; |
| 93 | + (*HT)[i].weight = (*HT)[s1].weight + (*HT)[s2].weight; |
| 94 | + } |
| 95 | +} |
| 96 | +//HT为哈夫曼树,HC为存储结点哈夫曼编码的二维动态数组,n为结点的个数 |
| 97 | +void HuffmanCoding(HuffmanTree HT, HuffmanCode *HC,int n){ |
| 98 | + *HC = (HuffmanCode) malloc((n+1) * sizeof(char *)); |
| 99 | + char *cd = (char *)malloc(n*sizeof(char)); //存放结点哈夫曼编码的字符串数组 |
| 100 | + cd[n-1] = '\0';//字符串结束符 |
| 101 | + |
| 102 | + for(int i=1; i<=n; i++){ |
| 103 | + //从叶子结点出发,得到的哈夫曼编码是逆序的,需要在字符串数组中逆序存放 |
| 104 | + int start = n-1; |
| 105 | + //当前结点在数组中的位置 |
| 106 | + int c = i; |
| 107 | + //当前结点的父结点在数组中的位置 |
| 108 | + int j = HT[i].parent; |
| 109 | + // 一直寻找到根结点 |
| 110 | + while(j != 0){ |
| 111 | + // 如果该结点是父结点的左孩子则对应路径编码为0,否则为右孩子编码为1 |
| 112 | + if(HT[j].left == c) |
| 113 | + cd[--start] = '0'; |
| 114 | + else |
| 115 | + cd[--start] = '1'; |
| 116 | + //以父结点为孩子结点,继续朝树根的方向遍历 |
| 117 | + c = j; |
| 118 | + j = HT[j].parent; |
| 119 | + } |
| 120 | + //跳出循环后,cd数组中从下标 start 开始,存放的就是该结点的哈夫曼编码 |
| 121 | + (*HC)[i] = (char *)malloc((n-start)*sizeof(char)); |
| 122 | + strcpy((*HC)[i], &cd[start]); |
| 123 | + } |
| 124 | + //使用malloc申请的cd动态数组需要手动释放 |
| 125 | + free(cd); |
| 126 | +} |
| 127 | +//打印哈夫曼编码的函数 |
| 128 | +void PrintHuffmanCode(HuffmanCode htable,int *w,int n) |
| 129 | +{ |
| 130 | + printf("Huffman code : \n"); |
| 131 | + for(int i = 1; i <= n; i++) |
| 132 | + printf("%d code = %s\n",w[i-1], htable[i]); |
| 133 | +} |
| 134 | +int main(void) |
| 135 | +{ |
| 136 | + int w[7] = {3,12,7,4,2,8,11}; |
| 137 | + int n = 7; |
| 138 | + HuffmanTree htree; |
| 139 | + HuffmanCode htable; |
| 140 | + CreateHuffmanTree(&htree, w, n); |
| 141 | + HuffmanCoding(htree, &htable, n); |
| 142 | + PrintHuffmanCode(htable,w, n); |
| 143 | + return 0; |
| 144 | +} |
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