如何在 PHP 的所有变体中分割字符串

Question

我为我的英语感到抱歉。 我需要所有变体中的除法字符串函数，顺序和长度保持不变。

输入 'abc'

输出 'abc / a,bc / ab,c / a,b,c'

或

输入 'rrd'

输出 'rrd / r,rd / rr,d / r,r,d'

谢谢

Answer 1

检查这个“漂亮”的代码，对我来说是很好的大脑训练器:)
它确实需要一些优化，但效果很完美。
注意： array_reverse() 和 strrev() 可以删除，但这样顺序看起来更好。

function TheFunction($s) {
    for($i=strlen($s)-1;$i>0;$i--) $h .= '1';
    $z = str_replace('1','0',$h);
    for($i=bindec($h);$i>=0;$i--) $array[] = strrev(substr_replace($z, decbin($i), strlen($z)-strlen(decbin($i))));
    foreach($array as $value){
        $value = str_replace(array('0','1'),array(' ',','),$value);
        $string = '';
        for($i=0;$i<strlen($s)-1;$i++) $string .= $s[$i].$value[$i];
        $string .= $s[strlen($s)-1];
        $results[] = str_replace(' ','',$string);
    }
    return array_reverse($results);
}

示例

print_r(TheFunction('Anne'));

返回

Array
(
    [0] => Anne
    [1] => A,nne
    [2] => An,ne
    [3] => A,n,ne
    [4] => Ann,e
    [5] => A,nn,e
    [6] => An,n,e
    [7] => A,n,n,e
)

另一个示例

print_r(TheFunction('Stack'));

返回：

Array
(
    [0] => Stack
    [1] => S,tack
    [2] => St,ack
    [3] => S,t,ack
    [4] => Sta,ck
    [5] => S,ta,ck
    [6] => St,a,ck
    [7] => S,t,a,ck
    [8] => Stac,k
    [9] => S,tac,k
    [10] => St,ac,k
    [11] => S,t,ac,k
    [12] => Sta,c,k
    [13] => S,ta,c,k
    [14] => St,a,c,k
    [15] => S,t,a,c,k
)

Answer 2

我做了与 Anne 类似的事情，只是我使用了 Adrian Akison 的 Variations 类 这里。

正如 Anne 的解决方案一样，我找到了 1 和 0 的所有变体 (Variations.cs)，长度 1 小于原始字符串。 这是为了抵消逗号的.请注意，稍后我会将 1 替换为逗号，将 0 替换为空格。

“位”字符串 111 =“,,,”或 101 =“,,”等

因此，使用某些字符获取特定大小的所有变体，在本例中为 1 和 0：

    private HashSet<string> fetchBinaryVariations(int size)
    {
        HashSet<string> returnVal = new HashSet<string>();
        String variationResultItem = string.Empty;
        string[] oneZero = { "1", "0" };

        /* Generate all variations of 1's and 0's given size using Adrian Akison's handy dandy variations class */
        Variations<string> variationsList = new Variations<string>(oneZero.ToList<string>(), size, GenerateOption.WithRepetition);
        Console.WriteLine("Total Variations: {0}", variationsList.Count());

        foreach (List<string> variationItem in variationsList)
        {
            variationResultItem = String.Join("", variationItem);
            returnVal.Add(variationResultItem);
            // Console.WriteLine("Variation: {0}", variationResultItem);
        }
        return returnVal;
    }

接下来，我采用这些“Bit”字符串，将它们转换为逗号和空格，并将它们与我的原始序列合并。就我而言，我还有另一个步骤，使用枚举将数字解码为字母（未显示）：

例如。 “Bit”字符串 = 101 = 将 ', ,' 分隔符添加到 1234 = '1,2 3,4' 的原始序列中

例如。 “Bit”字符串 = 111 = 分隔符 ',,,' 添加到原始序列 1234 = '1,2,3,4'

    private Dictionary<string, string> processDeliminatorsWithInputSequence(string sequence, HashSet<string> binaryVariations)
    {
        Dictionary<string, string> returnVal = new Dictionary<string, string>();
        string message = string.Empty, variationWithDelim = string.Empty, finalString = string.Empty;
        StringBuilder characterContainer = null;
        int satisfiedCnt = 0, unsatisfiedCnt = 0;

        foreach (string variation in binaryVariations)
        {
            variationWithDelim = variation.Replace('0', ' ').Replace('1', ','); // 0's are spaces and 1's are commas
            characterContainer = new StringBuilder();
            for (int i = 0; i < sequence.Length - 1; i++)
            {
                characterContainer.Append(sequence[i]); // Original Input
                characterContainer.Append(variationWithDelim[i]); // Append with space or comma
            }
            characterContainer.Append(sequence[sequence.Length - 1]); // Need to append last character from original input - offset again
            characterContainer.Replace(" ", ""); // Clean up empty spaces in final string

            finalString = decodeToAlphabet(characterContainer); // converat numerals to their alpha equivelant
            if (finalString != null)
                returnVal.Add(characterContainer.ToString(), finalString); // Add original encoding and decoded strings to hastable
            else
                unsatisfiedCnt++;

            satisfiedCnt = returnVal.Count();
        }

        message = String.Format("Input Sequence: {0}\r\nInput Binary Variations: {1}\r\n", sequence, binaryVariations.Count());
        message += String.Format("Valid Alphabet Sequence Variations: {0}\r\nInvalid Alphabet Sequence Variations: {1}", satisfiedCnt, unsatisfiedCnt);
        result.Messsage = message;
        Console.WriteLine(message);

        return returnVal;
    }