Question

有什么可以帮助我获得输出：

a：b a：c 广告 a：e b：c b：d b：e c：d c：e d：e a：b：c a：b：d a：b：e a：c：d a：c：e a：d：e b：c：d b：c：e b：d：e c：d：e A B C D a：b：c：e a：b：d：e a：c：d：e b：c：d：e a：b：c：d：e

数据数组$ n =数组（＆＃39; a＆＃39;，＆＃39; b＆＃39;，＆＃39; c＆＃39;，＆＃39; d＆＃39;，＆＃39 ; E＆＃39）;

我尝试过的代码如下：

for($a=0;$a<count($n);$a++)
{
    for($b=$a+1;$b<count($n);$b++)
    {
        echo $n[$a].' : '.$n[$b].'<br />';  
    }
}

for($a=0;$a<count($n);$a++)
{
    for($b=$a+1;$b<count($n);$b++)
    {
        for($c=$b+1;$c<count($n);$c++)
        {
            echo $n[$a].' : '.$n[$b].' : '.$n[$c].'<br />';     
        }       
    }
}

for($a=0;$a<count($n);$a++)
{
    for($b=$a+1;$b<count($n);$b++)
    {
        for($c=$b+1;$c<count($n);$c++)
        {
            for($d=$c+1;$d<count($n);$d++)
            {
                echo $n[$a].' : '.$n[$b].' : '.$n[$c].' : '.$n[$d].'<br />';            
            }           
        }       
    }
}

for($a=0;$a<count($n);$a++)
{
    for($b=$a+1;$b<count($n);$b++)
    {
        for($c=$b+1;$c<count($n);$c++)
        {
            for($d=$c+1;$d<count($n);$d++)
            {
                for($e=$d+1;$e<count($n);$e++)
                {
                    echo $n[$a].' : '.$n[$b].' : '.$n[$c].' : '.$n[$d].' : '.$n[$e].'<br />';               
                }               
            }           
        }       
    }
}

但我觉得代码太长了请帮忙简化，

感谢

Answer 1

<?php

 $n = array ('a', 'b', 'c', 'd', 'e');

 function getPerm($array, $key)
 {
   foreach( $array as $arrkey=>$value )
   {
      if($arrkey != $key)
      {
        echo " $array[$key]:$array[$arrkey] ";
      }
   }
 }


foreach($n as $key=>$value)
{
  getPerm($n, $key);
}

?>

请参阅键盘http://codepad.org/uTaW8ssx

Answer 2

试试这个：

<?php
function getCombinations($base,$n){

$baselen = count($base);
if($baselen == 0){
    return;
}
    if($n == 1){
        $return = array();
        foreach($base as $b){
            $return[] = array($b);
        }
        return $return;
    }else{
        //get one level lower combinations
        $oneLevelLower = getCombinations($base,$n-1);

        //for every one level lower combinations add one element to them that the last element of a combination is preceeded by the element which follows it in base array if there is none, does not add
        $newCombs = array();

        foreach($oneLevelLower as $oll){

            $lastEl = $oll[$n-2];
            $found = false;
            foreach($base as  $key => $b){
                if($b == $lastEl){
                    $found = true;
                    continue;
                    //last element found

                }
                if($found == true){
                        //add to combinations with last element
                        if($key < $baselen){

                            $tmp = $oll;
                            $newCombination = array_slice($tmp,0);
                            $newCombination[]=$b;
                            $newCombs[] = array_slice($newCombination,0);
                        }

                }
            }

        }

    }

    return $newCombs;


}

echo "<pre>";
print_r(getCombinations(array('a','b','c','d','e'),2));
?>

参考：Php recursion to get all possibilities of strings

Answer 3

爪哇：

ArrayList<String> setProduct (ArrayList<String> a, ArrayList<String> b)
{
    ArrayList<String> prod = new ArrayList<String>();

    for (String s : a)
    {
        for (String t : b)
        {
            prod.add(s + ":" + t);
        }
    }

    return prod;
}

PHP：

<?php
function setProduct($a, $b)
{
    $arr = array();

    foreach ($a as $s)
    {
        foreach ($b as $t)
        {
            $arr[] = $s . ":" . $t;
        }
    }

    return $arr;
}
?>

数组数据的组合

3 个答案: