Question

初始：

Array
(
    [0] => Array
        (
            [0] => a
            [1] => b
        )

    [1] => Array
        (
            [0] => a
            [1] => c
        )

    [2] => Array
        (
            [0] => c
            [1] => b
        )
    [3] => Array
        (
            [0] => d
            [1] => e
        )        
)

结果：

Array
(
    [0] => Array
        (
            [0] => a
            [1] => b
            [2] => c
        )
    [1] => Array
        (
            [0] => d
            [1] => e
        )        
)

初始数组的前三项彼此相关但最后一项不相关。我认为可以通过使用Floyd-Warshall Algoritm来解决。请帮我拿到结果。

Answer 1

不知道什么是floyd-warshall algorhitm，并且不确定你的最佳需求，但我做到了这一点：

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

$result = array();
foreach($array as $itemOriginal){ //passing every array from the original array
   $passed = false;

   foreach($result as &$itemResult){ //passing every array from the new array (empty in the start)
      foreach($itemOriginal as $item){ //passing every item from original arrays
         if(in_array($item, $itemResult)){ //checking if the item is in one of earlier passed array transfered into new array already
            $itemResult = array_unique(array_merge($itemResult,  $itemOriginal)); //merging items into new array if one of their items equals
            $passed = true; //no need to check another item from the current original array
            break;
         }
      }

      if($passed == true) //no need to find any of original items in new array
         break;
   }

   if($passed == false) //for case the none of checked original items are in new array
      $result[] = $itemOriginal;
}

echo '<pre>';
print_r($result); //to check it

Answer 2

$result = array();
foreach ($array as $item) {
    // If we're just getting started, get started
    if (count($result) === 0) {
        $result[] = $item;
    // Otherwise, look for merge opportunities
    } else {
        $merged = false;
        // Loop existing items
        foreach ($result as $k => $resultItem) {
            // If there's a match, merge & break
            if (count(array_intersect($resultItem, $item)) > 0) {
                $result[$k] = array_merge($resultItem, $item);
                $merged = true;
                break;
            }
        }
        // If no match was found, create a new element
        if (!$merged) {
            $result[] = $item;
        }
    }
}

如何获得多维数组的多个交叉点？

2 个答案: