Question

我有一个数组：

$cards = [
    [
        "from" => "Barcelona",
        "to" => "Gerona Airport",        
    ],
    [
        "from" => "Stockholm",
        "to" => "New York JFK",          
    ],
    [
        "from" => "Gerona Airport",
        "to" => "Stockholm",          
    ],
    [
        "from" => "Madrid",
        "to" => "Barcelona",          
    ]
];

我需要对其进行排序以进行持续旅行，其中from与之前的to匹配，或者如果不匹配则设置为开始，例如：

Array (
    [0] => Array ( [from] => Madrid [to] => Barcelona )  
    [1] => Array ( [from] => Barcelona [to] => Gerona Airport )
    [2] => Array ( [from] => Gerona Airport [to] => Stockholm ) 
    [3] => Array ( [from] => Stockholm [to] => New York JFK ) 
)

我正在尝试使用usort：

usort($cards, function ($a, $b) {
          return ( $a["to"] === $b["from"] ) ? -1 : 1;
      });

print_r($cards);

但它并没有按照描述的方式排序。有任何想法吗？

Answer 1

希望有一个更好的解决方案，但是如果你排序的次数与数组中的元素一样多，并返回相等或更大，则它可以工作;即使你不知道开头或开头不是第一个元素。由于usort遍历和比较的方式，返回相等或更少不起作用：

foreach($cards as $card) {
    usort($cards, function ($a, $b) {
                      return ( $a["to"] === $b["from"] ) ? 0 : 1;
                  });
}

它可以使用较少的迭代，但将每个迭代与所有其他迭代相比似乎是安全的。

Answer 2

您可以采用递归方法并指定要开始的城市。

默认情况下，您使用doSort($cards, $result);从马德里开始。

如果您希望从＃34; Gerona Airport＆＃34;继续旅行，您可以使用doSort($cards, $result, "Gerona Airport");启动它

$cards = [
    [
        "from" => "Barcelona",
        "to" => "Gerona Airport",
    ],
    [
        "from" => "Stockholm",
        "to" => "New York JFK",
    ],
    [
        "from" => "Gerona Airport",
        "to" => "Stockholm",
    ],
    [
        "from" => "Madrid",
        "to" => "Barcelona",
    ]
];
$result = [];

function doSort($cards, &$result, $start = "Madrid") {
    foreach ($cards as $key => $card) {
        if ($card["from"] === $start) {
            $result[] = $card;
            doSort($cards, $result, $card["to"]);
        }
    }
}
doSort($cards, $result);
print_r($result);

Php output demo

那会给你：

Array
(
    [0] => Array
        (
            [from] => Madrid
            [to] => Barcelona
        )

    [1] => Array
        (
            [from] => Barcelona
            [to] => Gerona Airport
        )

    [2] => Array
        (
            [from] => Gerona Airport
            [to] => Stockholm
        )

    [3] => Array
        (
            [from] => Stockholm
            [to] => New York JFK
        )

)

Answer 3

这是一种方式。通过迭代找到开始并搜索路径：

// 1. Find the start
$start = null;
$starts = array_column($cards,'from');
$ends = array_column($cards,'to');
foreach ($cards as $card) {
    if (!in_array($card['from'], $ends)) $start=$card;
}
// 2. Find way
$way=[$start];
while (count($cards)>1) {
    $found = false ;
    foreach ($cards as $k => $card) {
        if ($card['from'] == $start['to']) {
            $way[] = $card;
            $start = $card;
            unset($cards[$k]);
            $found = true ;
            break 1;
        }
    }
    if (!$found) { echo "Impossible to use all cards.\n" ; break ; }
}
print_r($way);

输出：

    Array (
    [0] => Array ([from] => Madrid [to] => Barcelona )
    [1] => Array ([from] => Barcelona [to] => Gerona Airport )
    [2] => Array ([from] => Gerona Airport [to] => Stockholm )
    [3] => Array ([from] => Stockholm [to] => New York JFK )
)

通过继续值

3 个答案: