Question

如果两个英语单词仅包含相同的字母，则它们是相似的。例如，食物和食物不相似，但是狗和食物相似。（如果A与B相似，则A中的所有字母都包含在B中，而B中的所有字母都包含在A中。）

给出单词W和单词L的列表，找到L中所有与W相似的单词。将单词计数打印到标准输出。

示例：

输入（stdin）：

love
velo low vole lovee volvell lowly lower lover levo loved love lovee lowe lowes lovey lowan lowa evolve loves volvelle lowed love

输出（标准输出）：

14

说明：

L中与爱相似的词是 velo vole lovee volvell lover levo loved love lovee lovey evolve loves volvelle love

最多14。

所以我当前的解决方案如下：

 public static void main(String[] args) {
    String[] arr = new String[]{"velo", "low", "vole", "lovee", "volvell", "lowly", "lower", "lover", "levo", "loved", "love",
            "lovee", "lowe", "lowes", "lovey", "lowan", "lowa", "evolve", "loves", "volvelle", "lowed", "love"};
    String s = "love";
    int result = 0;

    Pattern p = Pattern.compile(buildPattern(s));

    for (String val : arr) {
        if (p.matcher(val).find()) result++;
    }

    System.out.println(result);
}

private static String buildPattern(String s) {
    String pattern = "^";
    for (int i = 0; i < s.length(); i++) {
        pattern += "(?=.*" + s.charAt(i) + ")";
    }
    return pattern;
}

我想知道我的简单代码是否有任何改进。

Aho-Corasick是否适用解决方案？

Answer 1

由于只有26个字母，并且int中有32位，因此int足够大，可以容纳单词中出现哪些字母的所有信息：

static int getFingerprint(String s)
{
    int result=0;
    for (int i = s.length()-1; i>=0; --i) {
        char c = s.charAt(i);
        if (c>='a' && c<='z')
            result |= 1<<(int)(c-'a');
        else if (c>='A' && c<='Z')
            result |= 1<<(int)(c-'A');
    }
    return result;
}

public static void main(String[] args) {
    String[] arr = new String[]{"velo", "low", "vole", "lovee", "volvell", "lowly", "lower", "lover", "levo", "loved", "love",
        "lovee", "lowe", "lowes", "lovey", "lowan", "lowa", "evolve", "loves", "volvelle", "lowed", "love"};
    String s = "love";

    int fingerprint = getFingerprint(s);

    int matches = 0;
    for (String item : arr) {
        if (getFingerprint(item)==fingerprint)
            ++matches;
    }
    System.out.println(matches);
}

Answer 2

我建议简化正则表达式，不需要先行，简单的“ ^ [love] * $”应该可以解决问题。

<input type="number" class="form-control" matInput name="value" placeholder="xxx" (change)="xxx()" formControlName="value">

Answer 3

我会尽量避免使用正则表达式，而是自己检查字母。

public static void main(String[] args)
{
  String[] arr = new String[]{"velo", "low", "vole", "lovee", "volvell", "lowly", "lower", "lover", "levo", "loved", "love",
          "lovee", "lowe", "lowes", "lovey", "lowan", "lowa", "evolve", "loves", "volvelle", "lowed", "love"};
  String s = "love";
  int result = 0;

  for (String word : arr)
  {
    if (isSimilar(s, word))
    {
      result++;
    }
  }

  System.out.println(result);
}

private static boolean isSimilar(String word, String test)
{
  for (char c : test.toCharArray())
  {
    if (word.indexOf(c) == -1)
    {
      return false;
    }
  }
  return true;
}

虽然目前我上面的示例仅返回10？

Answer 4

在实现和手动检查时，我只算出应该成功的10个。

就像比较每个单词中的字母是否相等一样简单

public static void main(String... args)
{
    String word = "love";
    List<String> strs = Arrays.asList(
        "velo", "low", "vole", "lovee", "volvell", "lowly", "lower", "lover", "levo", "loved", "love",
        "lovee", "lowe", "lowes", "lovey", "lowan", "lowa", "evolve", "loves", "volvelle", "lowed", "love"
    );

    System.out.println(
        strs.stream()
           .filter(str -> chars(word).equals(chars(str)))
           .count()
    );
}

private static Set<Character> chars(String word)
{
    return word.chars()
        .mapToObj(ch -> (char) ch)
        .collect(Collectors.toSet());
}

Answer 5

public static void main(String[] args) {
    String[] arr = new String[]{"velo", "low", "vole", "lovee", "volvell", "lowly", "lower", "lover", "levo", "loved", "love",
            "lovee", "lowe", "lowes", "lovey", "lowan", "lowa", "evolve", "loves", "volvelle", "lowed", "love"};
    String s = "love";

    Set<Character> searchWordCharacters = getDistinctCharacters(s);
    long result = Stream.of(arr)
            .map(Scratch::getDistinctCharacters)
            .filter(wordCharacters -> wordCharacters.size() == searchWordCharacters.size())
            .filter(wordCharacters -> wordCharacters.containsAll(searchWordCharacters))
            .peek(System.out::println)
            .count();
    System.out.println(result);
}

private static Set<Character> getDistinctCharacters(String word) {
    return word.chars()
            .mapToObj(i -> (char) i)
            .collect(Collectors.toSet());
}

结果：10

Answer 6

应成功计数10个！

import seaborn as sns
sns.barplot(x,y)

Answer 7

import java.util.Arrays;

class SomeClass {
    public static void main(String[] args) {
        String[] arr = new String[]{"velo", "low", "vole", "lovee", "volvell", "lowly", "lower", "lover", "levo", "loved", "love",
                "lovee", "lowe", "lowes", "lovey", "lowan", "lowa", "evolve", "loves", "volvelle", "lowed", "love"};
        String s = "love";
        int count = 0;

        boolean[] characters_state = new boolean[26];
        Arrays.fill(characters_state, false);
        for(int i = 0; i < s.length(); i++) {
            characters_state[s.charAt(i) - 'a'] = true;
        }

        for(int i = 0; i < arr.length; i++) {
            if (check(arr[i], characters_state.clone())) {
                count++;
            }
        }
        System.out.println(count);
    }

    static boolean check(String s, boolean[] characters_state) {
        for(int i = 0; i < s.length(); i++) {
            if(!characters_state[s.charAt(i) - 'a']) {
                return false;
            }
        }
        return true;
    }
}

输出

10

real    0m0,210s
user    0m0,206s
sys 0m0,025s

检查字符串中是否存在字符集-改进

7 个答案: