Question

我正在尝试实施Levenshtein距离算法以找到最相似的匹配

列表就像

    List<String> values = new ArrayList<String>();
    values.add("*");
    values.add("3.63.161.2");
    values.add("3*");
    values.add("*2");
    values.add("3*1");
    values.add("3.1*");
    values.add("3.2*");
    values.add("3.3*");
    values.add("3.4*");
    values.add("3.5*");
    values.add("3.61*");
    values.add("3*0");
    values.add("2*");
    values.add("1*");
    values.add("3.62*");

 String stringToMatch = "3.63.162.2";

 for (String pattern : values) {
        String regex = pattern.replace(".", "\\.").replace("*", ".*");
        System.out.println("String ["+stringToMatch+"] matches with ["+pattern+"] :  "+ stringToMatch.matches(regex) +" and the difference is --> "+StringUtils.getLevenshteinDistance(stringToMatch, pattern));

    }

这是我期待的输出

String [3.63.162.2] matches with [*] :  true and the difference is --> 10
String [3.63.162.2] matches with [3.63.161.2] :  false and the difference is --> 1
String [3.63.162.2] matches with [3*] :  true and the difference is --> 9
String [3.63.162.2] matches with [*2] :  true and the difference is --> 9
String [3.63.162.2] matches with [3*1] :  false and the difference is --> 8
String [3.63.162.2] matches with [3.1*] :  false and the difference is --> 7
String [3.63.162.2] matches with [3.2*] :  false and the difference is --> 7
String [3.63.162.2] matches with [3.3*] :  false and the difference is --> 7
String [3.63.162.2] matches with [3.4*] :  false and the difference is --> 8
String [3.63.162.2] matches with [3.5*] :  false and the difference is --> 8
String [3.63.162.2] matches with [3.61*] :  false and the difference is --> 6
String [3.63.162.2] matches with [3*0] :  false and the difference is --> 9
String [3.63.162.2] matches with [2*] :  false and the difference is --> 9
String [3.63.162.2] matches with [1*] :  false and the difference is --> 9
String [3.63.162.2] matches with [3.62*] :  false and the difference is --> 6

所以最近的匹配字符串是

String [3.63.162.2] matches with [3*] :  true and the difference is --> 9
String [3.63.162.2] matches with [*2] :  true and the difference is --> 9

我想要的是给3 *作为最匹配，因为第一个字符匹配。

有没有办法使用相同的技术从左到右匹配字符串。

我知道我可以单独编写匹配的代码，但我也想看看其他方法。

使用Levenshtein距离算法找到最匹配的字符串

0 个答案: