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题目： 给定一段产品的英文描述，包含M个英文单词，每个单词以空格分隔，无其他标点，再给定N个英文单词关键字。请说明思路并编程实现方法  String extractSummary(String description,String [ ] Keywords):目标是找出此产品描述中包含N个关键词（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，编程语言不限。

实现方法：在确保所有关键字都包含的情况下，每次从content尾向前挪动一个位置，都从content的头部到尾遍历一遍，碰上小的就付给result，直到完全遍历完

代码：

package test;import java.util.ArrayList;
import java.util.List;/*** @author hy*  2011/6/13*/
public class FindAbstract {static String content[] = { "a", "c", "d", "a", "c", "b", "d", "e", "a","a","b"};static String keyword[] = { "b", "c", "d" };static List<String> contentList = new ArrayList<String>();public static void main(String args[]) {List<String> result = new ArrayList<String>();int begin = 0;int end = content.length;// 将content内容从数组形式变换成List型for (int i = 0; i < end; i++)contentList.add(i, content[i]);// 输出给定的content和keywordSystem.out.print("[content]:  ");for (int i = 0; i < content.length; i++)System.out.print(content[i] + " ");System.out.println();System.out.print("[keyword]:  ");for (int i = 0; i < keyword.length; i++)System.out.print(keyword[i] + " ");System.out.println();// 输出最短摘要result = contentList;System.out.println("[AllMatch]：");for (end = content.length; end - begin >= keyword.length; end--) {for (begin = 0; end - begin >= keyword.length; begin++) {if (isAllHave(contentList.subList(begin, end), keyword)&& result.size() > contentList.subList(begin, end).size()){result = contentList.subList(begin, end);System.out.println("     "+result);}}begin = 0;}System.out.println("[ShortestMatch]: "+result);}// 是否都包含所有关键字static boolean isAllHave(List<String> arr, String key[]) {boolean is = false;int temp = 0;for (int i = 0; i < key.length; i++)if (isKeywordIn(arr, key[i]))temp++;if (temp == key.length)is = true;return is;}// 是否包含单个关键字static boolean isKeywordIn(List<String> arr, String key) {int i;for (i = 0; i < arr.size(); i++)if (arr.get(i) == key)return true;return false;}}

结果：

[content]:  a c d a c b d e a a b
[keyword]:  b c d
[AllMatch]：
     [c, d, a, c, b, d, e, a, a, b]
     [d, a, c, b, d, e, a, a, b]
     [a, c, b, d, e, a, a, b]
     [c, b, d, e, a, a, b]
     [c, b, d, e, a, a]
     [c, b, d, e, a]
     [c, b, d, e]
     [c, b, d]
[ShortestMatch]: [c, b, d]