如何在 java 中编写计算 Anagrams 出现次数的算法？

Question

嗨，我想在 java 中编写字谜算法。我的要求是如果有人给出这样的字符串

Input: "aa aa odg dog gdo" 它的字谜计数应该是 2。谁能帮我解决这个问题？我尝试了一个解决方案，但它无法正常工作。

public static void main(String[] args) {
        String text = "a c b c run urn urn";
        String word = "urn";
        System.out.print(countAnagrams(text, word));

    }

    static boolean araAnagram(String s1,
                              String s2)
    {
        char[] ch1 = s1.toCharArray();
        char[] ch2 = s2.toCharArray();
        Arrays.sort(ch1);
        Arrays.sort(ch2);
        if (Arrays.equals(ch1, ch2))
            return true;
        else
            return false;
    }

    static int countAnagrams(String text, String word)
    {
        int N = text.length();
        int n = word.length();
        int res = 0;
        for (int i = 0; i <= N - n; i++) {

            String s = text.substring(i, i + n);
            if (araAnagram(word, s))
                res++;
        }
        return res;
    }

这个程序不符合我的要求请帮忙

Answer 1

假设有一种方法可以将一个单词转换成一个有序的字符序列，那么使用Stream可以计算出字谜的数量API:

static String anagram(String s) {
    char[] arr = s.toCharArray();
    Arrays.sort(arr);
    return new String(arr);
}

static long countAnagrams(String input) {
    if (null == input || input.isEmpty()) {
        return 0;
    }
    return Arrays.stream(input.split("\s+"))  // Stream of words
            .distinct()  // get unique words and
            // move them into Map<String, List<String>>, where List contains all anagrams contained in the input string
            .collect(Collectors.groupingBy(MyClass::anagram)) 
            .entrySet() // 
            .stream()   // stream of entries 
            .filter(e -> e.getValue().size() > 1) // filter values with at least two anagrams
            .peek(e -> System.out.println(e.getValue())) // debug print of the anagram list
            .count(); // and count them
}

测试：

String[] tests = {
    "aa aa odg dog gdo",
    "cars are very cool so are arcs and my os"
};

Arrays.stream(tests)
      .forEach(s -> System.out.printf("'%s' -> anagram count=%d%n", s, countAnagrams(s)));

输出

[odg, dog, gdo]
'aa aa odg dog gdo' -> anagram count=1
[so, os]
[cars, arcs]
'cars are very cool so are arcs and my os' -> anagram count=2

Answer 2

这个可以解决

public static boolean anagrams(String s1, String s2){
    if(!(s1.equalsIgnoreCase(s2))){
        char[] ch1 = s1.toCharArray();
        char[] ch2 = s2.toCharArray();
        Arrays.sort(ch1);
        Arrays.sort(ch2);
        return Arrays.equals(ch1,ch2);
    }
    return false;
}

public static String CountingAnagrams(String str) {
    int res = 0;
    String[] splitStr = str.split("\s+");
    for(int i = 0 ; i< splitStr.length; i++)
        for (int j = i + 1; j < splitStr.length; j++)
            if (anagrams(splitStr[i], splitStr[j]))
                res++;
    return String.valueOf(res-1);
}

如何在 java 中编写计算 Anagrams 出现次数的算法？

How to write algorithm for Count Occurrences of Anagrams in java?

java

algorithm

anagram