2014-04-02 抽取最短英文描述（阿里巴巴面试题 有bug不是最优）

题目：

Alibaba笔试题：给定一段产品的英文描述，包含M个英文单词，每个英文单词以空格分隔，无其他标点符号；再给定N个英文单词关键字，请说明思路并编程实现方法

char* extractSummary(char[] description,char[] *key words)

目标是找出此产品描述中包含N个关键字（每个关键词至少出现一次）的长度最短的子串，作为产品简介输出。（不限编程语言）20分。

思路：

首先找出这N个单词在该M个单词出现的最开始出现的单词X，和最后出现的单词Y，然后分别去掉单词X,再次判别单词是否在X,Y之间，若在，后移找到新的单词X,一次操作；对于Y，可以进行同样的操作，依次前移，得到最后的最短字串单词为止。

程序实现：

#include<stdio.h>
#include<stdlib.h>
#include<string.h>

char *left, *right;
char character[1024];
char **keywords;
int number;

char * str_reverse_str(char *word)
{
	char *p = NULL;
	char *temp = NULL;
	char *count = word;
	char *characters = character;

	while((*characters != '\0') && (p = strstr(characters, word))){
		temp = p;
		while(count && *count != '\0'){
			p++;
			count++;
		}
		characters = p;
		count = word;
	}
	
	return temp;
}

int is_key_word(char *words)
{
	int i;
	for(i = 0; i < number; i++){		//obtain the first and last
		if(strcmp(words, keywords[i]) == 0)
			return 1;
	}
	return 0;
}
	
void obtain_first_word(char *first)
{
	left = first;

	char p[30];	//obtain the word
	char *s = p;
	*s++ = ' ';
	first++;
	while(first  && *first != ' '){
		*s++ = *first++;
	}
	*s++ = ' ';
	*s = '\0';
	
	if(strstr(first, p)){
		obtain_first_word(first);
	}else if(!is_key_word(p)){
		obtain_first_word(first);
	} 
}

void obtain_last_word(char *last)
{
	right = last;
	char p[30];
	char *s = p;
	*s++ = ' '; 			
	last--;
	last--;
	char *p2 = last;
	while(p2  && *p2 != ' '){
		s++;
		p2--;
	}
	*s++ =' ';
	*s = '\0';
	s--;
	s--;
	while(last  && *last != ' '){
		*s-- = *last--;
	}
	char *temp = last;
	temp++;
	char c;
	if(temp){
		c = *temp;
		*temp = '\0';
	}
	if(strstr(left, p)){
		obtain_last_word(temp);
	} else if(!is_key_word(p)){
		obtain_last_word(temp);
	} else {
		*temp = c;
	}
}

char * extract_summary()
{
	int i;
	char *first = NULL, *last = NULL, *p ;

	for(i = 0; i < number; i++){		//obtain the first and last
		if((p = strstr(character, keywords[i]))){
			if(!first ||first > p)
				first = p;
		}

		if((p = str_reverse_str(keywords[i]))){
			if(!last ||last < p)
				last = p;
			//printf("%d\n", 101);
		}
	}	

	last++;
	while(last && *last != ' '){
		last++;
	}
	*last++ = ' ';	
	*last = '\0';
	obtain_first_word(first);
	obtain_last_word(last);
	printf("\nthe minest character as follows:\n%s", ++left);
}

int main(void)
{
	char *s = character;			//set the the side of character to ' '
	*s++ = ' ';
	printf("Please input the character sequence as follows:\n");
	gets(s);
	while(s && *s != '\0'){
		s++;
	}
	*s++ = ' ';
	*s = '\0';
	
	printf("Please input  the number of keywords:\n");
	scanf("%d", &number);
	printf("Please input  the  keyword as follows:\n");
	keywords = (char **)malloc(sizeof(char *) * number);	
	int i;
	for(i = 0; i < number; i++){
		keywords[i] = (char *)malloc(sizeof(char) * 30);
		s = keywords[i];		//set the the side of character to ' '
		*s++ = ' '; 
		scanf("%s", s);
		while(s && *s != '\0'){
			s++;
		}
		*s++ = ' ';
		*s ='\0';
	}

	extract_summary();
	
	for(i = 0; i < number; i++){
		free(keywords[i]);
	}
	free(keywords);
	
	return 0;
}