这是一道笔试题来的，主要是用字典树来解决：
/****************************************** * * 描述：这是一道笔试题，要求按单词出现的次序打印出单词及其出现的次数; * * 思路：既然是笔试题，肯定要考虑时间复杂度，字典树要比哈斯的要好， * 因此我用了字典树. * * author: huangkq1989@scut2008 * blog:http://blog.csdn.net/kangquan2008 * * ****************************************/ #include <stdio.h> #include <stdlib.h> #include <string.h> #define SIZE 26 // 字符的种类数 #define DIFFERENCE_WORD_SIZE 10000 //不同的单词的个数 #define WORD_MAX_LENGTH 10 // 单词的最大长度 #define CHECK_RET(ret,info) \ if((ret) < 0) printf("error:%s\n",(info)),exit(EXIT_FAILURE) typedef struct Trie{ int cnt;// 该字符在该位置出现的次数 int times;// 标识为词尾 int sequence;// 出现的次序,0为初始值，从1开始 struct Trie * next[SIZE]; }*Trie, Node; typedef struct Result{ char data[WORD_MAX_LENGTH]; int times; }Result; int seq = 0; // 记录出现的次序 Result result[DIFFERENCE_WORD_SIZE]; int initial(Node ** node) { //Node * node = *node_arg; if(((*node) = malloc(sizeof(Node))) == NULL) return -1; (*node)->cnt = 1; (*node)->sequence = 0; (*node)->times = 0; for(int i=0; i<SIZE; i++) (*node)->next[i] = NULL; return 0; } int insert(Trie * tree2,char src[]) { Trie tree = *tree2; // 为再次插入做准备 if(tree == NULL)// 空树，注意字典树的根不存放字符 { CHECK_RET(initial(&tree),"not more memory"); *tree2 = tree; // 这步很重要，要注意，目的是把tree2指向正确分配了空间的地方 } int i=0; Node * node; while(src[i] != '\0') { int flag = 0; int ch_to_index = src[i] - 'a'; if(!(tree)->next[ch_to_index])// 未存在 { CHECK_RET(initial(&node),"not more memory"); (tree)->next[ch_to_index] = node; flag = 1; }else{// 存在 (tree)->next[ch_to_index]->cnt += 1; } if(src[i+1] == '\0') { if(flag) (tree)->next[ch_to_index]->sequence = seq++; (tree)->next[ch_to_index]->times += 1; break; } (tree) = (tree)->next[ch_to_index]; i++; } } void traverse_tree(Trie tree, char word[],int index) { if(tree == NULL) return ; for(int i=0; i<SIZE; i++) { if(tree->next[i]) { word[index] = i+'a'; if(tree->next[i]->times > 0) // 表示是一个单词 { word[index+1] = '\0'; //printf("in traverse: %s ",word); //printf("%d\n",tree->next[i]->sequence); strcpy(result[tree->next[i]->sequence].data,word); result[tree->next[i]->sequence].times = tree->next[i]->times; } traverse_tree(tree->next[i],word,index+1); } } } void display() { for(int i=0; i<seq; i++) printf("%s appears %d times\n", result[i].data, result[i].times); } int main() { char set[][10] = {"efgd","abc","abc","abcd","dbc","fdd"}; Trie tree = NULL; printf("trie tree has:\n"); for(int i=0; i<sizeof(set)/sizeof(set[0]); i++) printf("%s\n",set[i]),insert(&tree,set[i]); printf("-------------------\n"); printf("按单词出现的次序打印出单词及其出现的次数\n"); char tmp_word[WORD_MAX_LENGTH]; traverse_tree(tree,tmp_word,0); display(); return 0; } 转载请标明源博客：)
