内存不足

Question

我正在努力解决古老的字谜问题。多亏了许多教程，我能够迭代一组字符串，递归地找到所有的排列，然后将它们与英语单词列表进行比较。我发现的问题是，在大约三个单词之后（通常是关于“变形”之类的东西），我会得到一个OutOfMemory错误。我试着把我的批分成小的集合，因为它似乎是消耗我所有内存的递归部分。但即使只是“变形”也把它锁起来了...

Scanner scanner = new Scanner(resource.getInputStream());
   while (scanner.hasNext()) {
       String s = scanner.nextLine();
        uniqueWords.add(s.toLowerCase());
   }

List<List<String>> subSets = Lists.partition(new ArrayList(uniqueWords), SET_SIZE);

for (List<String> set: subSets) {
      // tried created as class attribute & injection, no difference 
      AnagramGenerator anagramGenerator = new AnagramGenerator();
      List<Word> anagrams = anagramGenerator.createWordList(set);
      wordsRepository.save(anagrams);
      LOGGER.info("Inserted {} records into the database", anagrams.size());
 }

public class AnagramGenerator {

private Map<String, List<String>> map = new Hashtable<>();
public List<Word> createWordList(List<String> dictionary) {

   buildAnagrams(dictionary);

   List<Word> words = new ArrayList<>();
   for (Map.Entry<String, List<String>> entry : map.entrySet()) {
       words.add(new Word(entry.getKey(), entry.getValue()));
   }
    return words;
   }

private Map<String, List<String>> buildAnagrams(List<String> dictionary) {

        for (String str : dictionary) {
            String key = sortString(str);
            if (map.get(key) != null) {
                map.get(key).add(str.toLowerCase());
            } else {
                if (str.length() < 2) {
                    map.put(key, new ArrayList<>());
                } else {
                    Set<String> permutations = permutations(str);
                    Set<String> anagramList = new HashSet<>();

                    for (String temp : permutations) {
                        if (dictionary.contains(temp) && !temp.equalsIgnoreCase(str)) {
                            anagramList.add(temp);
                        }
                    }
                    map.put(key, new ArrayList<>(anagramList));
                }
            }
        }
        return map;
    }

   private Set<String> permutations(String str) {    
        if (str.isEmpty()) {
            return Collections.singleton(str);
        } else {
            Set<String> set = new HashSet<>();
            for (int i = 0; i < str.length(); i++)
                for (String s : permutations(str.substring(0, i) + str.substring(i + 1)))
                    set.add(str.charAt(i) + s);
            return set;
        }
    }

编辑：根据出色的反馈，我已经将生成器从排列更改为工作查找：

public class AnagramGenerator {
private Map<String, Set<String>> groupedByAnagram = new HashMap<String, Set<String>>();

    private Set<String> dictionary;

    public AnagramGenerator(Set<String> dictionary) {

        this.dictionary = dictionary;
    }

 public List<Word> searchAlphabetically() {

        List<Word> words = new ArrayList<>();
        for (String word : dictionary) {
            String key = sortString(word);
            if (!groupedByAnagram.containsKey(key)) {
                groupedByAnagram.put(key, new HashSet<>());
            }
            if (!word.equalsIgnoreCase(key)) {
                groupedByAnagram.get(key).add(word);
            }
        }

        for (Map.Entry<String, Set<String>> entry : groupedByAnagram.entrySet()) {
            words.add(new Word(entry.getKey(), new ArrayList(entry.getValue())));
        }

        return words;
    }
 private String sortString(String goodString) {

        char[] letters = goodString.toLowerCase().toCharArray();
        Arrays.sort(letters);
        return new String(letters);
    }

它有一点更多的调整，所以我不添加一个词，因为它是自己的字形，但除此之外，这似乎是炽热的快速。而且，代码要干净得多。谢谢大家！

Answer 1

正如较长的单词所指出的那样，排列的数量很快就变得巨大。

Debian上的/usr/share/dict/British-English有99,156行。有更长的单词列表，但让我们用它作为一个例子。

一个九个字母的单词的排列数是9！=362,880

10! milliseconds = ~1 hour
12! milliseconds = ~5.54 days
15! milliseconds = ~41.44 years

 sorted_input = sort_alphabetically(input_word)
 for each dictionary_word // probably a file readline()
     sorted_dictionary_word = sort_alphabetically(dictionary_word)
     if(sorted_dictionary_word = sorted_input)
         it's an anagram! Handle it
     end 
 end

  multimap = new MultiMap<String, String> // or whatever

  def build_dict:
      for each dictionary_word // probably a file readline()
          multimap.add(
               sort_alphabetically(dictionary_word), 
               dictionary_word)
      end
  end

  def lookup_anagrams(word):
      return multimap.get(sort_alphabetically(word))
  end 

这占用了适度的内存（整个字典，加上一点键和映射开销），但意味着一旦创建了结构，您就可以一遍又一遍地查询，成本确实非常低。

如果你想找到两个字的字谜，你将需要一个更复杂和有趣的算法。但即便如此，避免暴力强迫整个排列搜索空间对你的成功至关重要。