Php 如何获得多维数组的多个交点?
首字母:Php 如何获得多维数组的多个交点?,php,arrays,floyd-warshall,Php,Arrays,Floyd Warshall,首字母: Array ( [0] => Array ( [0] => a [1] => b ) [1] => Array ( [0] => a [1] => c ) [2] => Array ( [0] => c
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
)
)
初始数组的前三项相互关联,但最后一项不相关。我认为这可以通过使用弗洛伊德·沃沙尔算法来解决。请帮助我获得结果。不知道floyd warshall algorhitm是什么,也不确定您的优化需求,但我做了以下几点:
$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
$array=array(
['a','b'],
[a',c'],
[c'、[b'],
['d','e']
);
$result=array();
foreach($array as$itemOriginal){//传递原始数组中的每个数组
$passed=false;
foreach($result as&$itemResult){//传递新数组中的每个数组(开始时为空)
foreach($itemOriginal作为$item){//传递原始数组中的每个项
if(in_数组($item,$itemResult)){//检查该项是否在已传输到新数组的较早传递数组中
$itemResult=array_unique(array_merge($itemResult,$itemOriginal));//如果其中一个项等于
$passed=true;//无需检查当前原始数组中的其他项
打破
}
}
if($passed==true)//无需在新数组中查找任何原始项
打破
}
if($passed==false)//对于这种情况,选中的原始项都不在新数组中
$result[]=$itemOriginal;
}
回声';
打印(结果)//检查一下
Stack overflow(堆栈溢出)问题预计会显示出一些自己解决问题的努力。@Dagon,对不起,伙计-我回答了:-[@Reloecc我有一大堆工作要做,我可以免费发给你吗?;)array_merge不在乎口是心非,我用array_Uniques解决了这个问题谢谢你的回答。太好了,谢谢你@Reloecc@user3757515很高兴为您提供帮助。最好对您的更改进行解释,这样就不会太明显了。@cubejokey我没有做任何更改,你在说什么?:/抱歉,那么请解释一下你的代码,如果这更有意义的话。
$result = array();
foreach ($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;
}
}
}