Question

我正在尝试浏览字典并确定所述字典中的哪些单词遵循Java中的某个字母顺序。例如，模式i，n和g将包含连枷 ing ，同时允许其他23个字母中的任何一个分隔i ，n和g，但仍必须采用相同的顺序（即名为isong的单词仍符合条件，因为它遵循i，{{1} }， THEN n

但不是“对齐”因为它跟随“i”，“g”，然后是“n”）。作为参考，“words.txt”是没有空格的字典文件。另外，我的教授不允许使用正则表达式。我要做的是修复findWords并读取文件，并将所有单词放入g中名为“words”的String数组中。这是我的代码：

readDictionary

Answer 1

你可以使用java正则表达式来帮助你做到这一点。只需按照以下方式构建一个模式，即可搜索输入字符串是否依次包含“i”，“n”和“g”：

.*i.*n.*g.*

''正则表达式表示任何字符，'*'表示0到无穷大发生

以下代码快照供您参考使用java regex

public boolean matchPattern(String pattern, String inputString){
    Pattern pattern = 
                Pattern.compile(pattern);

                Matcher matcher = 
                pattern.matcher(inputString);

                boolean found = false;
                while (matcher.find()) {
                   return true;
                }

      return false;
}

使用此方法，只需调用它：

   boolean foundIng = matchPattern(".*i.*n.*g.*" , "sample Input String ");

Answer 2

import java.util.ArrayList;

static void findWords (String[] inputWords, char[] inputset)
{
    ArrayList[] words = new ArrayList<String>(); // list of words that match the pattern. Lists* are of dynamic length, unlike their cousins, the array, which are fixed length. This one is typecasted to only be able to contain Strings
    char[] ing = inputset; // the pattern

    for (String s : inputWords)
    { // a for-each** loop through inputWords. Incidentally, in your original code, inputWords was called 'words', the same as your resulting list.
        int wi = 0; // index in the 'ing' array that we are up to
        for (char c : s.toCharArray())
        { // loop through every char in each word...
            if (c == ing[wi])
            { // ... then check if the current character is the next one in the pattern...
                wi++; // ... and if so, proceed to the next character in the pattern to match against
                if (wi == ing.length)
                { // We've run out of characters to check against, so we know that the word matches the pattern
                    words.add(s); // add the word to the list of words that match the pattern
                    break; // break out of the checking loop, so that the first loop proceeds onto the next word in the dictionary
                }
            }
        }
    }
    // Do something with your final list of words here
    // String[] wordsList = words.toArray(new String[words.size()]); // Uncomment this line if you want the words list to be an array, instead of a list
}

* http://docs.oracle.com/javase/7/docs/api/java/util/ArrayList.html是ArrayLists上的oracle javadoc页面 **有关for-each循环的详细信息，请参阅http://docs.oracle.com/javase/1.5.0/docs/guide/language/foreach.html

在字典.txt文件中查找字符模式

2 个答案: