首頁猿問檢查字符串中是否存在字符集 - 改進

檢查字符串中是否存在字符集 - 改進

Java

慕碼人2483693 2023-02-16 15:41:37

如果兩個英語單詞只包含相同的字母，則它們是相似的。例如，food 和 good 不相似，但 dog 和 good 相似。（如果 A 與 B 相似，則 A 中的所有字母都包含在 B 中，B 中的所有字母都包含在 A 中。）給定一個單詞 W 和一個單詞列表 L，找到 L 中與 W 相似的所有單詞。將單詞計數打印到標準輸出。例子：輸入（標準輸入）：love velo low vole lovee volvell lowly lower lover levo loved love lovee lowe lowes lovey lowan lowa evolve loves volvelle lowed love輸出（標準輸出）：14解釋：L中與love相近的詞是 velo vole lovee volvell lover levo loved love lovee lovey evolve loves volvelle love最多14.所以我目前的解決方案如下：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;}我想知道我的簡單代碼是否有任何改進。Aho-Corasick 是否適用解決方案？

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5 回答

鳳凰求蠱

TA貢獻1825條經驗獲得超4個贊

由于只有 26 個字母，而 an 中有 32 位int，因此 anint足以容納有關單詞中出現的字母的所有信息：

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);

}

反對回復 2023-02-16

楊魅力

TA貢獻1811條經驗獲得超6個贊

我建議簡化正則表達式，不需要前瞻，簡單的“^[love]*$”就可以了。

private static String buildPattern(String s) {

String pattern = "^[";

for (int i = 0; i < s.length(); i++) {

pattern += s.charAt(i);

}

pattern += "]*$";

return pattern;

}

反對回復 2023-02-16

HUH函數

TA貢獻1836條經驗獲得超4個贊

數 10 應該成功！

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";

Predicate<Character> p = x -> s.indexOf(x) > -1 ? true : false;

List<String> asList = Arrays.asList(arr);

asList.stream().forEach(x -> {

List<Character> chars = new ArrayList<>();

for (int i = 0; i < x.length(); i++) {

chars.add(x.charAt(i));

}

boolean anyMatch = chars.stream().allMatch(p);

if (anyMatch)

count++;

});

System.out.println(count);

反對回復 2023-02-16

慕桂英546537

TA貢獻1848條經驗獲得超10個贊

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());

}

結果：10

反對回復 2023-02-16

揚帆大魚

TA貢獻1799條經驗獲得超9個贊

我只計算了 10 個應該成功的，無論是在我的實施中還是在我手動檢查時。

很簡單，就是比較每個單詞中的字母集合是否相等

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());

}

反對回復 2023-02-16

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檢查字符串中是否存在字符集 - 改進

檢查字符串中是否存在字符集 - 改進

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