java - 单词相似度的正则表达式 "n letter difference"

Question

假设我有一个这样的词； 值得。我想搜索该单词的所有 1 个字母差异组合。 aert、ert、meat、mmert、merst、merts 等均适用。所以我的正则表达式就像

[a-z]{0,2}ert OR m[a-z]{0,2}rt OR me[a-z]{0,2}t OR mer[a-z]{0,2}

因此，对于 n 个字母的差异，我只需将 2 替换为 n-1，并且您无法获得所有组合。

我的问题是这样的；有没有更短的方法来编写这个正则表达式？

谢谢

Answer 1

请检查此解决方案，我已在下面测试了此代码。似乎有效。

/**
 * Then function will return list of the words matched with nth_difference
 *
 * @param pattern search pattern
 * @param data input data
 * @param nth_difference difference
 * @return
 */
static List<String> getNthDifferenceWords(String pattern, String[] data, int nth_difference) {
    Map<Character, Integer> frequencyTable = new HashMap<>();
    List<String> matchedWords = new ArrayList<>();
    //Code complexity : O(n)
    for (int i = 0; i < pattern.length(); ++i) {
        frequencyTable.put(pattern.charAt(i), 1);
    }
    //Code complexity : O(m) where m is size of entire input;
    for (String input : data) {
        int matchCounter = 0;
        for (int j=0; j<input.length(); ++j){
            if(frequencyTable.containsKey(input.charAt(j))){
                ++matchCounter;
            }
        }
        //System.out.println("matched=" + matchCounter);
        if(input.length() <= pattern.length() && (matchCounter == pattern.length() - nth_difference)){
                matchedWords.add(input);
        }
        if((input.length() - pattern.length() == 1) && (matchCounter >= input.length() - nth_difference)){
            matchedWords.add(input);
        }
   }
    return matchedWords;
}


    public static void main(String[] args) {
    int nth_difference = 1;
    String pattern = "mert";
    String[] data = new String[]{"aert", "ert", "meat", "mmert", "merst", "merts","meritos"};
    System.out.println(getNthDifferenceWords(pattern,data,nth_difference));

    nth_difference = 2;
    pattern = "merit";
    data = new String[]{"aert", "ert", "meat", "mmert", "merst", "merts","demerit","merito", "meritos"};
    System.out.println(getNthDifferenceWords(pattern,data,nth_difference));
}