我试图从2个数组中提取类似的字符串,我设法这样做,除了它们是重复的.即数组1 {“箭头”,“箭头”,“sycophant”}和数组2(“箭头”,“sycophant”,“bulbasaur”}将给我输出{“箭头”,“箭头”,“sycophant”虽然我只想尝试一次箭头.有什么建议吗？

public static void main(String[] args) {
    String[] words1 = { "sycophant", "rattle", "zinc", "alloy", "tunnel", "arrow" };
    String[] words2 = { "sycophant", "arrow", "arrow" };

    // String prefix = "a";
    // String substring = "at";
    // char[] letters = { 'a', 'b' };

    // String[] output = wordsStartingWith(words1, prefix);
    // String[] output = wordsContainingPhrase(words1, substring);
    // String[] output = wordsContainingAll(words1, letters);
    String[] output = wordsInBoth(words1, words2);

    for (int i = 0; i < output.length; i++) {
        System.out.println("Words: " + i + " " + output[i]);
    }
}

public static String[] wordsInBoth(String[] words1, String[] words2) {
    // method that finds and returns common words in two arrays
    String[] returnWords;
    int countWords = 0;

    for (int i = 0; i < words1.length; i++) {
        for (int j = 0; j < words2.length; j++) {
            if (words1[i].equalsIgnoreCase(words2[j])) {
                countWords++;
            }
        }
    }

    returnWords = new String[countWords];
    countWords = 0;

    for (int i = 0; i < words1.length; i++) {
        for (int j = 0; j < words2.length; j++) {
            if (words1[i].equalsIgnoreCase(words2[j]) 
                    && !words1[i].equalsIgnoreCase(returnWords[countWords])) {
                returnWords[countWords] = words1[i];
                countWords++;
            }
        }
    }

    return returnWords;
}

解决方法:

一种可能性是存储在HashSet中找到的单词,这些单词不会添加重复项.

// method that finds and returns common words in two arrays
public static String[] wordsInBoth(String[] words1, String[] words2) { 

    Set<String> returnWords = new HashSet<String>();

    for (int i = 0; i < words1.length; i++) {
        for (int j = 0; j < words2.length; j++) {
            if (words1[i].equalsIgnoreCase(words2[j]))
                returnWords.add(words1[i]);
        }
    }

    return returnWords.toArray(new String[returnWords.size()]);
}