Description
Input
Output
Sample Input
carbohydrate cart carburetor caramel caribou carbonic cartilage carbon carriage carton car carbonate
Sample Output
carbohydrate carboh cart cart carburetor carbu caramel cara caribou cari carbonic carboni cartilage carti carbon carbon carriage carr carton carto car car carbonate carbona
解题思路:使用 trie 树结构。在 trie 树节点中加入两个域count[] 和 next 。count[i] 表示有多少个单词经过这个节点。先将所有单词保存在 trie 树中,然后一个一个地查找,当到达某个节点使用 count[i] ==1 ,那么从根到该节点组成的字符串便是该单词的最短前缀。
#include <stdio.h> #include <malloc.h> #include <string.h> int t = 0; typedef struct Trie { int count[26]; //统计该字符出现的次数 struct Trie *next[26]; //26个字母,开辟26个空间 } Trie; void initTire(Trie *root, char *string) { Trie *trie = root; //根节点不含有数据,只有26个指针域 int i; int j; while(*string != '\0') { j = *string-'a'; if(trie->next[j] == NULL) { trie->next[j] = (Trie *)malloc(sizeof(Trie)); (trie->next[j])->count[j] = 1; trie = trie->next[j]; for(i = 0; i < 26; i++) { trie->next[i] = NULL; } } else { (trie->next[j])->count[j]++; trie = trie->next[j]; } string++; } } void findPrefix(Trie *root, char *string) { //寻找前缀 Trie *trie = root; int i = 0, j; while(*string != '\0') { j = *string-'a'; if((trie->next[j])->count[j] == 1) { //如果当前字母只出现一次,证明前缀字母到此结束 printf("%c", *string); break; } printf("%c", *string); trie = trie->next[j]; string++; } } int main() { int i = 0; int n; char str[3000][21]; char *pre; Trie *root; root = (Trie *)malloc(sizeof(Trie)); for(i = 0; i < 26; i++) { root->next[i] = NULL; } i = 0; while(scanf("%s", str[i]) != EOF) { initTire(root, str[i]); getchar(); i++; } n = i; for(i = 0; i < n; i++) { printf("%s ", str[i]); findPrefix(root, str[i]); printf("\n"); } return 0; }