在原始字符串的每個字母之後依次生成帶撇號的字符串

Question

如何迭代創建在每個字母后面有撇號的「Murrays」變體？最終結果應該是：

"m'rrays,mu'rrays,mur'rays,murr'ays,murra'ys,murray's" 

Answer 1

您想遍歷名稱，並用撇號重新打印嗎？請嘗試以下操作：

<?php 

    $string = "murrays"; 

    $array = str_split($string); 
    $length = count($array); 
    $output = ""; 

    for ($i = 0; $i < $length; $i++) { 
     for($j = 0; $j < $length; $j++) { 
     $output .= $array[$j]; 
     if ($j == $i) 
      $output.= "'"; 
     } 
     if ($i < ($length - 1)) 
     $output .= ","; 
    } 

    print $output; 

?> 

Answer 2

我的建議：

<?php      
function generate($str, $add, $separator = ',') 
{ 
    $split = str_split($str); 
    $total = count($split) - 1; 

    $new = ''; 
    for ($i = 0; $i < $total; $i++) 
    { 
     $aux = $split; 
     $aux[$i+1] = "'" . $aux[$i+1]; 
     $new .= implode('', $aux).$separator; 
    } 
    return $new; 
} 
echo generate('murrays', "'"); 
?> 

Answer 3

這裏的另一種解決方案：

$str = 'murrays'; 
$variants = array(); 
$head = ''; 
$tail = $str; 
for ($i=1, $n=strlen($str); $i<$n; $i++) { 
    $head .= $tail[0]; 
    $tail = substr($tail, 1); 
    $variants[] = $head . "'" . $tail; 
} 
var_dump(implode(',', $variants)); 

Answer 4

好，這就是爲什麼functionnal編程是這裏

此代碼的工作就OCAML和F＃ ，你可以很容易地使它運行在C＃

let generate str = 
    let rec gen_aux s index = 
     match index with 
      | String.length s -> [s] 
      | _ -> let part1 = String.substr s 0 index in 
        let part2 = String.substr s index (String.length s) in 
        (part1^"'"^part2)::gen_aux s (index + 1) 
    in gen_aux str 1;; 

generate "murrays";; 

此代碼返回原詞爲列表的末尾，你可以變通方法:)

Answer 5

在這裏你去：

$array = array_fill(0, strlen($string) - 1, $string); 
implode(',', array_map(create_function('$string, $pos', 'return substr_replace($string, "\'", $pos + 1, 0);'), $array, array_keys($array)));