如何在PHP中的所有變體中劃分字符串

Question

我很抱歉我的英語。 我需要在所有變化中劃分字符串的函數，順序和長度保持不變。

輸入 'ABC'

輸出 'ABC/A，BC/AB，C/A，B，C'

或

輸入 'RRD'

輸出「RRD/R，RD/RR，d/R，R，d」

謝謝

Answer 1

檢查這個 '美麗' 的代碼，漂亮的腦TRAI對我來說:)
它絕對需要一些優化，但作品完美。
注意：array_reverse（）和strrev（）可以被移除，但是這樣順序看起來更好。

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); 
} 

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 
) 

另一個實施例

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

我做了類似安妮的事情，除了我使用了Adrian Akison here的變體類。

正如在Anne的解決方案中，我找到了1和0的所有變體（Variations.cs），長度比原始字符串小1。這是爲了抵消逗號的考慮。請注意，稍後，我會將1作爲逗號替換，將0作爲空格替換。

「位」 字符串111 = 「,,,」 或101 = 「」 等

所以，得到使用某些字符一定大小的所有變體，在該殼體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; 
    } 

接下來，我把這些「位」的字符串，它們與我原來的順序轉換成逗號和空格和合並。在我的情況下，我有另一個步驟，我使用enum（未顯示）將數字解碼爲字母：

Ex。 「位」字符串= 101 =將'，，'的分隔符添加到原始序列1234 ='1,2 3,4'

Ex。 「位」 串= 111 =的deliminators ',,,' 加到1234 =原始序列 '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; 
    }