数据结构—哈夫曼树

原理：参考趣学数据结构

代码：

#include<stdio.h>
#include<stdlib.h>
#define N 100
#define INF 2^31-1
typedef struct fNode {//哈夫曼树中每个节点的信息int c;//字符int parent;//父节点，左右孩子，权重int lchild, rchild;int weight;
}fNode;
typedef struct rNode {//存储单个的编码字符的编码序列int r[N];int start;//有效编码起始的位置int length;
}rNode;
void huffMan(fNode fnode[],int n) {//构造哈夫曼树  选取二个最小的没有父节点的结点合并，以此类推for (int i = 0; i < n - 1; i++) {//n个字符n-1次的构造即可构造哈夫曼树int min1=INF, min2 = INF;int u =-1,v = -1;for (int j = 0; j < n + i; j++) {//找二个最小的没有父节点的if (fnode[j].weight < min1 && fnode[j].parent==-1) {min2 = min1;//最值同时往前推v = u;min1 = fnode[j].weight;u = j;}else if(fnode[j].weight<min2 && fnode[j].parent == -1) {min2 = fnode[j].weight;v = j;}}fnode[n + i].weight = min1 + min2;fnode[n + i].lchild = u;fnode[n + i].rchild = v;fnode[u].parent = fnode[v].parent = n + i;//更新父节点}
}
void findHuffManCodePath(fNode fnode[],rNode rnode[],int n) {//寻找每个字符编码表示rNode temp;int start=n-1;//最坏的哈夫曼树为一条链表for (int i = 0; i < n; i++) {start = n - 1;//每个字符的编码从叶子节点向根节点遍历int p = fnode[i].parent;int tempv = i;while (p != -1) {//p不等于-1表示有父节点if (tempv == fnode[p].lchild) {temp.r[start] = 0;}else {temp.r[start] = 1;}start--;tempv = p;p = fnode[p].parent;}for (int j = start + 1; j <= n - 1; j++) {//更新每个字符的编码数组rnode[i].r[j] = temp.r[j];rnode[i].start = start + 1;}rnode[i].length = n - start - 1;}int sum = 0;//遍历每个字符的编码数组for (int j = 0; j < n; j++) {//n个字符的编码遍历printf("%d的哈夫曼编码为:", fnode[j].c);for (int k = rnode[j].start; k <= n - 1; k++) {printf("%d", rnode[j].r[k]);}sum += (fnode[j].weight*rnode[j].length);printf("  ");}printf("\n");printf("哈夫曼编码长度为:%d\n",sum);printf("\n");
}
int main() {fNode fnode[N];rNode rnode[N];int u;printf("请输入编码字符的个数:");scanf_s("%d", &u);int length = u;int c;int weight=0;//http://c.biancheng.net/view/1796.html 输入字符参考for (int k = 0; k < u;k++) {//初始化字符编码的信息 //c = getchar();//输入字符scanf_s("%d %d", &c,&weight);fnode[k].c = c;fnode[k].weight = weight;}u = 2 * length - 1;//哈夫曼总共有2n-1个结点for(int i=0;i<u;i++){//初始化结点的左右孩子和父节点信息fnode[i].lchild = -1;fnode[i].rchild = -1;fnode[i].parent = -1;}huffMan(fnode, length);findHuffManCodePath(fnode, rnode, length);system("pause");return 0;
}

测试截图：

时间复杂度O(n x n),空间复杂度O(n)

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