2

我有一个字符串和一个单词数组，我必须编写代码来查找字符串的所有子字符串，这些子字符串以任意顺序包含数组中的所有单词。该字符串不包含任何特殊字符/数字，每个单词用空格分隔。

例如：

给定的字符串：

aaaa aaaa aaaa aaaa cccc bbbb bbbb bbbb bbbb aaaa bbbb cccc

数组中的单词：

aaaa
bbbb
cccc

输出样本：

aaaa aaaa aaaa aaaa cccc bbbb bbbb bbbb bbbb    

aaaa aaaa aaaa aaaa cccc bbbb    

aaaa cccc bbbb bbbb bbbb bbbb    

cccc bbbb bbbb bbbb bbbb aaaa  

aaaa cccc bbbb

我已经使用 for 循环实现了这一点，但这非常低效。

我怎样才能更有效地做到这一点？

我的代码：

    for(int i=0;i<str_arr.length;i++)
    {
        if( (str_arr.length - i) >= words.length)
        {
            String res = check(i);
            if(!res.equals(""))
            {
                System.out.println(res);
                System.out.println("");
            }
            reset_all();
        }
        else
        {
            break;
        }
    }

public static String check(int i)
{
    String res = "";
    num_words = 0;

    for(int j=i;j<str_arr.length;j++)
    {
        if(has_word(str_arr[j]))
        {
            t.put(str_arr[j].toLowerCase(), 1);
            h.put(str_arr[j].toLowerCase(), 1);

            res = res + str_arr[j]; //+ " ";

            if(all_complete())
            {
                return res;
            }

            res = res + " ";
        }
        else
        {
            res = res + str_arr[j] + " ";
        }

    }
    res = "";
    return res;
}
4

2 回答 2

1

我的第一种方法类似于以下伪代码

  for word:string {
    if word in array {
      for each stored potential substring {
        if word wasnt already found {
          remove word from notAlreadyFoundList
          if notAlreadyFoundList is empty {
            use starting pos and ending pos to save our substring
          }
        }
      store position and array-word as potential substring
  }

这应该具有不错的性能，因为您只遍历字符串一次。

[编辑]

这是我的伪代码的一个实现，试试看它的性能是好是坏。它的工作假设是一旦找到最后一个单词就找到了匹配的子字符串。如果您真的想要所有匹配项，请更改标记的行//ALLMATCHES：

class SubStringFinder {
    String textString = "aaaa aaaa aaaa aaaa cccc bbbb bbbb bbbb bbbb aaaa bbbb cccc";
    Set<String> words = new HashSet<String>(Arrays.asList("aaaa", "bbbb", "cccc"));

    public static void main(String[] args) {
        new SubStringFinder();
    }

    public SubStringFinder() {
        List<PotentialMatch> matches = new ArrayList<PotentialMatch>();
        for (String textPart : textString.split(" ")) {
            if (words.contains(textPart)) {
                for (Iterator<PotentialMatch> matchIterator = matches.iterator(); matchIterator.hasNext();) {
                    PotentialMatch match = matchIterator.next();
                    String result = match.tryMatch(textPart);
                    if (result != null) {
                        System.out.println("Match found: \"" + result + "\"");
                        matchIterator.remove(); //ALLMATCHES - remove this line
                    }
                }
                Set<String> unfound = new HashSet<String>(words);
                unfound.remove(textPart);
                matches.add(new PotentialMatch(unfound, textPart));
            }// ALLMATCHES add these lines 
             // else {
             // matches.add(new PotentialMatch(new HashSet<String>(words), textPart));
             // }
        }
    }

    class PotentialMatch {
        Set<String> unfoundWords;
        StringBuilder stringPart;
        public PotentialMatch(Set<String> unfoundWords, String part) {
            this.unfoundWords = unfoundWords;
            this.stringPart = new StringBuilder(part);
        }
        public String tryMatch(String part) {
            this.stringPart.append(' ').append(part);
            unfoundWords.remove(part);                
            if (unfoundWords.isEmpty()) {
                return this.stringPart.toString();
            }
            return null;
        }
    }
}
于 2012-06-27T10:39:09.893 回答
0

这是另一种方法：

public static void main(String[] args) throws FileNotFoundException {
    // init
    List<String> result = new ArrayList<String>();
    String string = "aaaa aaaa aaaa aaaa cccc bbbb bbbb bbbb bbbb aaaa bbbb cccc";
    String[] words = { "aaaa", "bbbb", "cccc" };
    // find all combs as regexps (e.g. "(aaaa )+(bbbb )+(cccc )*cccc", "(aaaa )+(cccc )+(bbbb )*bbbb")
    List<String> regexps = findCombs(Arrays.asList(words));
    // compile and add
    for (String regexp : regexps) {
        Pattern p = Pattern.compile(regexp);
        Matcher m = p.matcher(string);
        while (m.find()) {
            result.add(m.group());
        }
    }
    System.out.println(result);
}

private static List<String> findCombs(List<String> words) {
    if (words.size() == 1) {
        words.set(0, "(" + Pattern.quote(words.get(0)) + " )*" + Pattern.quote(words.get(0)));
        return words;
    }
    List<String> list = new ArrayList<String>();
    for (String word : words) {
        List<String> tail = new LinkedList<String>(words);
        tail.remove(word);
        for (String s : findCombs(tail)) {
            list.add("(" + Pattern.quote(word) + " ?)+" + s);
        }
    }
    return list;
}

这将输出：

[aaaa bbbb cccc, aaaa aaaa aaaa aaaa cccc bbbb bbbb bbbb bbbb, cccc bbbb bbbb bbbb bbbb aaaa]

我知道结果并不完整：你只得到了可用的组合，完全扩展，但你得到了所有的组合。

于 2012-06-27T13:11:06.297 回答