正则表达式 – 是否有可能为任意字符串的字谜生成(紧凑)正则表达

我认为这个 javascript代码将根据您的规范工作.正则表达式长度将随输入的长度线性增加.它生成一个正则表达式,使用正向前瞻来匹配输入字符串的字谜.正则表达式的前瞻部分确保测试输入字符串中存在所有字符而忽略它们的顺序,匹配部分确保测试输入字符串的长度与输入字符串的长度相同(为其构??造正则表达式).

function anagramRegexGenerator(input) {
    var lookaheadPart = '';
    var matchingPart = '^';
    var positiveLookaheadPrefix='(?=';
    var positiveLookaheadSuffix=')';
    var inputCharacterFrequencyMap = {}
    for ( var i = 0; i< input.length; i++ )
    {
        if (!inputCharacterFrequencyMap[input[i]]) {
            inputCharacterFrequencyMap[input[i]] = 1
        } else {
            ++inputCharacterFrequencyMap[input[i]];
        }
    }
    for ( var j in inputCharacterFrequencyMap) {
        lookaheadPart += positiveLookaheadPrefix;
        for (var k = 0; k< inputCharacterFrequencyMap[j]; k++) {
            lookaheadPart += '.*';
            if (j == ' ') {
                lookaheadPart += 's';
            } else {
                lookaheadPart += j;
            }
            matchingPart += '.';
        }
        lookaheadPart += positiveLookaheadSuffix;
    }
    matchingPart += '$';
    return lookaheadPart + matchingPart;
}

样本输入和输出如下

anagramRegexGenerator('aaadaaccc')
//generates the following string.
"(?=.*a.*a.*a.*a.*a)(?=.*d)(?=.*c.*c.*c)^.........$"
anagramRegexGenerator('abcdef ghij'); 
//generates the following string.
"(?=.*a)(?=.*b)(?=.*c)(?=.*d)(?=.*e)(?=.*f)(?=.*s)(?=.*g)(?=.*h)(?=.*i)(?
=.*j)^...........$" 
//test run returns true
/(?=.*a)(?=.*b)(?=.*c)(?=.*d)(?=.*e)(?=.*f)(?=.*s)(?=.*g)(?=.*h)(?=.*i)(?
=.*j)^...........$/.test('acdbefghij ')
//or using the RegExp object
//this returns true
new RegExp(anagramRegexGenerator('abcdef ghij')).test('acdbefghij ') 
//this returns false
new RegExp(anagramRegexGenerator('abcdef ghij')).test('acdbefghijj')