PHP查找所有（某种程度上）数组的唯一组合-Java 学习之路

我整天都在看PHP数组排列/组合问题..但仍然无法弄明白：/

如果我有一个像这样的数组：

20 //key being 0    
20 //key being 1    
22 //key being 2    
24 //key being 3

我需要组合如：

20, 20, 22 //keys being 0 1 2    
20, 20, 24 //keys being 0 1 3    
20, 22, 24 //keys being 0 2 3
20, 22, 24 //keys being 1 2 3

我目前的代码给了我：

20, 22, 24

因为它不想重复20 ...但这就是我需要的！

这是我的代码 . 它直接来自Php recursion to get all possibilities of strings

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;


}

我一直在玩 ($b == $lastEl) 线，没有运气

===============

我已经看过的问题，并且与创建内存不足错误的OR不同！：

我已尝试使用12个项目的数组中的一些算法，并最终耗尽内存 . 然而，我目前使用的算法并没有给我一个内存不足的错误....但是......我需要那些重复！

6 回答

如果您不介意使用几个全局变量，可以在PHP中执行此操作（从JavaScript中的version翻译）：

<?PHP
$result = array(); 
$combination = array();

function combinations(array $myArray, $choose) {
  global $result, $combination;

  $n = count($myArray);

  function inner ($start, $choose_, $arr, $n) {
    global $result, $combination;

    if ($choose_ == 0) array_push($result,$combination);
    else for ($i = $start; $i <= $n - $choose_; ++$i) {
           array_push($combination, $arr[$i]);
           inner($i + 1, $choose_ - 1, $arr, $n);
           array_pop($combination);
         }
  }
  inner(0, $choose, $myArray, $n);
  return $result;
}

print_r(combinations(array(20,20,22,24), 3));
?>

OUTPUT：

Array ( [0] => Array ( [0] => 20 
                       [1] => 20 
                       [2] => 22 ) 
        [1] => Array ( [0] => 20 
                       [1] => 20 
                       [2] => 24 ) 
        [2] => Array ( [0] => 20 
                       [1] => 22 
                       [2] => 24 ) 
        [3] => Array ( [0] => 20 
                       [1] => 22 
                       [2] => 24 ) )

回复于 2024-04-24T08:01:59+08:00

梨包Math_Combinatorics使这种问题相当容易 . 它只需要相对较少的代码，简单直接，而且非常容易阅读 .

$ cat code/php/test.php
<?php
$input = array(20, 20, 22, 24);

require_once 'Math/Combinatorics.php';

$c = new Math_Combinatorics;
$combinations = $c->combinations($input, 3);
for ($i = 0; $i < count($combinations); $i++) {
  $vals = array_values($combinations[$i]);
  $s = implode($vals, ", ");
  print $s . "\n";
}
?>

$ php code/php/test.php
20, 20, 22
20, 20, 24
20, 22, 24
20, 22, 24

如果我必须将其打包为函数，我会做这样的事情 .

function combinations($arr, $num_at_a_time) 
{
    include_once 'Math/Combinatorics.php';

    if (count($arr) < $num_at_a_time) {
        $arr_count = count($arr);
        trigger_error(
            "Cannot take $arr_count elements $num_at_a_time " 
            ."at a time.", E_USER_ERROR
        );
    }

    $c = new Math_Combinatorics;
    $combinations = $c->combinations($arr, $num_at_a_time);

    $return = array();
    for ($i = 0; $i < count($combinations); $i++) {
        $values = array_values($combinations[$i]);
        $return[$i] = $values;
    }
    return $return;
}

这将返回一个数组数组 . 获取文本 . . .

<?php
  include_once('combinations.php');

  $input = array(20, 20, 22, 24);
  $output = combinations($input, 3);

  foreach ($output as $row) {
      print implode($row, ", ").PHP_EOL;
  }
?>
20, 20, 22
20, 20, 24
20, 22, 24
20, 22, 24

回复于 2024-04-24T08:01:59+08:00

想法很简单 . 假设您知道如何置换，那么如果您将这些排列保存在集合中，它就会变成组合 . 按定义设置处理重复值 . Set或HashSet的Php等效是SplObjectStorage，ArrayList是Array . 重写应该不难 . 我有一个Java实现：

public static HashSet<ArrayList<Integer>> permuteWithoutDuplicate(ArrayList<Integer> input){
          if(input.size()==1){
              HashSet<ArrayList<Integer>> b=new HashSet<ArrayList<Integer>>();
              b.add(input);
              return b;
          }
          HashSet<ArrayList<Integer>>ret= new HashSet<ArrayList<Integer>>();
          int len=input.size();
          for(int i=0;i<len;i++){
              Integer a = input.remove(i);
              HashSet<ArrayList<Integer>>temp=permuteWithoutDuplicate(new ArrayList<Integer>(input));
              for(ArrayList<Integer> t:temp)
                  t.add(a);
              ret.addAll(temp);
              input.add(i, a);
          }
          return ret;
      }

回复于 2024-04-24T08:01:59+08:00

为什么不使用二进制？至少它的简单和非常容易理解每行代码是什么样的？这是我在一个项目中为自己写的一个函数，我觉得它非常整洁！

function search_get_combos($array){
$bits = count($array); //bits of binary number equal to number of words in query;
//Convert decimal number to binary with set number of bits, and split into array
$dec = 1;
$binary = str_split(str_pad(decbin($dec), $bits, '0', STR_PAD_LEFT));
while($dec < pow(2, $bits)) {
    //Each 'word' is linked to a bit of the binary number.
    //Whenever the bit is '1' its added to the current term.
    $curterm = "";
    $i = 0;
    while($i < ($bits)){
        if($binary[$i] == 1) {
            $curterm[] = $array[$i]." ";
        }
        $i++;
    }
    $terms[] = $curterm;
    //Count up by 1
    $dec++;
    $binary = str_split(str_pad(decbin($dec), $bits, '0', STR_PAD_LEFT));
}
return $terms;
}

对于您的示例，此输出：

Array
(
    [0] => Array
        (
            [0] => 24 
        )
    [1] => Array
        (
            [0] => 22 
        )
    [2] => Array
        (
            [0] => 22 
            [1] => 24 
        )
    [3] => Array
        (
            [0] => 20 
        )
    [4] => Array
        (
            [0] => 20 
            [1] => 24 
        )
    [5] => Array
        (
            [0] => 20 
            [1] => 22 
        )
    [6] => Array
        (
            [0] => 20 
            [1] => 22 
            [2] => 24 
        )
    [7] => Array
        (
            [0] => 20 
        )
    [8] => Array
        (
            [0] => 20 
            [1] => 24 
        )
    [9] => Array
        (
            [0] => 20 
            [1] => 22 
        )
    [10] => Array
        (
            [0] => 20 
            [1] => 22 
            [2] => 24 
        )
    [11] => Array
        (
            [0] => 20 
            [1] => 20 
        )
    [12] => Array
        (
            [0] => 20 
            [1] => 20 
            [2] => 24 
        )
    [13] => Array
        (
            [0] => 20 
            [1] => 20 
            [2] => 22 
        )
    [14] => Array
        (
            [0] => 20 
            [1] => 20 
            [2] => 22 
            [3] => 24 
        )
)

回复于 2024-04-24T08:01:59+08:00

有同样的问题，发现了一个不同的，按位的，更快的解决方案：

function bitprint($u) {
    $s = array();
    for ($n=0; $u; $n++, $u >>= 1){
        if ($u&1){
            $s [] = $n;
        }
    }
    return $s;
}
function bitcount($u) {
    for ($n=0; $u; $n++, $u = $u&($u-1));
    return $n;
}
function comb($c,$n) {
    $s = array();
    for ($u=0; $u<1<<$n; $u++){
        if (bitcount($u) == $c){
            $s [] = bitprint($u);
        }
    }
    return $s;
}

这个生成从0到n-1的整数的所有大小m组合，因此例如m = 2，n = 3并且调用comb（2,3）将产生：

0 1
0 2
1 2

它为您提供索引位置，因此很容易通过索引指向数组元素 .

Edit: 输入梳子失败（30,5） . 不知道为什么，有人有什么想法吗？

回复于 2024-04-24T08:01:59+08:00

使用strrev和/ foreach循环清理Adi Bradfield的消化，并且只获得独特的结果 .

function search_get_combos($array = array()) {
sort($array);
$terms = array();

for ($dec = 1; $dec < pow(2, count($array)); $dec++) {
    $curterm = array();
    foreach (str_split(strrev(decbin($dec))) as $i => $bit) {
        if ($bit) {
            $curterm[] = $array[$i];
        }
    }
    if (!in_array($curterm, $terms)) {
        $terms[] = $curterm;
    }
}

return $terms;
}

回复于 2024-04-24T08:01:59+08:00

PHP查找所有（某种程度上）数组的唯一组合

6 回答

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