Question

我有一个像 -

这样的多维数组

array(
    "C"  => array('A','A1','A2','B',B1'),
    "C1" => array('A','A1','A2','B',B1'),
    "A"  => array('B','B1','C',C1'),
    "B1" => array('A','A1','A2','C',C1'),
    "B1" => array('A','A1','A2','C',C1'),
    "A2" => array('B','B1','C',C1'),
    "A1" => array('B','B1','C',C1')
);

我想要一个脚本，它会为每个键分配给定数组的值，

结果将像 -

C   A/A1/A2/B/B1    -   A  //I can assign any value from A/A1/A2/B/B1
C1  A/A1/A2/B/B1    -   A1 //Once A is assigned for the above key then C1 will get value from A1/A2/B/B1 skipping the 'A' which is assigned to key 'C'
A   B/B1/C/C1       -   B  //Now A will get value from B/B1/C/C1
B1  A/A1/A2/C/C1    -   A2 //B1 will get value from A2/C/C1
B   A/A1/A2/C/C1    -   C  //Likewise all the values will be assigned uniquely
A2  B/B1/C/C1       -   B1
A1  B/B1/C/C1       -   C1

条件是不会将相同的值分配给多个键，即所有键都具有不同的值。（检查以上结果以了解究竟需要分配的内容）

Answer 1

$data = array(
    "C"  => array('A','A1','A2','B','B1'),
    "C1" => array('A','A1','A2','B','B1'),
    "A"  => array('B','B1','C','C1'),
    "B1" => array('A','A1','A2','C','C1'),
    "B" => array('A','A1','A2','C','C1'),
    "A2" => array('B','B1','C','C1'),
    "A1" => array('B','B1','C','C1')
);

$assigned = array();
$result = array();
foreach($data as $k => $v) {
   for($i=0;$i<count($v);$i++) {
     if (!in_array($v[$i],$assigned)) {
        $assigned[] = $v[$i];
        $result[$k] = $v[$i];
        break;
     }
   }
}
var_dump($result);

输出：

array(7) {
  ["C"]=>
  string(1) "A"
  ["C1"]=>
  string(2) "A1"
  ["A"]=>
  string(1) "B"
  ["B1"]=>
  string(2) "A2"
  ["B"]=>
  string(1) "C"
  ["A2"]=>
  string(2) "B1"
  ["A1"]=>
  string(2) "C1"
}

这个解决方案并不好;）如果你将C1定义为数组（'A'），那么C1将没有值，以防止你可以尝试检查是否有可能在不同的配置中分配值。

$data = array(
    "C"  => array('A','A1'),
    "C1" => array('A'),
    "A"  => array('B','B1','C','C1'),
    "B1" => array('A','A1','A2','C','C1'),
    "B" => array('A','A1','A2','C','C1'),
    "A2" => array('B'),
    "A1" => array('B','B1','C','C1')
);

$assigned = array();
$keys = array_keys($data);
for($i=0;$i<count($keys);$i++) {
    getValue($data, $assigned, $keys, $i);  
}
function getValue($data, &$assigned, $keys, &$i, $x = null) {
   $j = $x === null ? 0 : array_search($x, $data[$keys[$i]]) +1;
   for($j; $j<count($data[$keys[$i]]); $j++) {
      $v = $data[$keys[$i]][$j];
      if(!in_array($v, $assigned)) {
         $assigned[$keys[$i]] = $v;
         return;
      }
   }
   $i--;
   if ($i < 0) {
       throw new Exception('no solutions');
   }
   $val = $assigned[$keys[$i]];
   $assigned[$keys[$i]] = null; 
   getValue($data, $assigned, $keys, $i, $val);
}
var_dump($assigned);

我想要一个php算法来解决以下问题

1 个答案: