8

可能重复:
检查一个数组是否包含另一个数组的所有元素

我之前在 Stackoverflow 上发布过类似的内容,但答案并不能完全让我满意。这就是为什么我再次发布问题,但一直在改变问题。

有些人帮助我构建了一个函数,该函数检查array($GroupOfEight[$i])作为多维元素的元素是否array($GroupOfEight)等于另一个数组($stackArray),而不考虑数组中的数字顺序。

但是,我需要检查的是所提到的是否在多维中array($stackArray)包含任何其他元素,这意味着 main可以包含比.array($GroupOfEight[$i])array($GroupOfEight)array($stackArray)subarrays($GroupOfEight[$i])

这是我迄今为止收集的一个工作代码,但需要修改为我想要的版本:

 <?php
    $GroupOfEight = array (
                          array(0,1,3,2,4,5,7,6),
                          array(4,5,6,7,15,12,13,14),
                          array(12,13,15,14,8,9,11,10),
                          array(2,6,14,10,3,7,15,11),
                          array(1,3,5,7,13,15,9,11),
                          array(0,4,12,8,1,5,13,9),
                          array(0,1,3,2,8,9,11,10)
                );


    $stackArray = array(0,4,12,1,9,8,5,13,9,2,5,2,10);
    /*$stackArray gets value with POST Method by URL parameter.
    This is just the example. As you see this array contains 
    $GroupOfEight[4], and also it contains many other numbers.*/


    /* The function given below checks if $stackArray equals any
    of the subarrays of $GroupOfEight. However, we want to check
    if $stackArray caontains any of the subarrays of function.
    If it does, function should return the index number, if it
    doesnt it should return -1.*/
    function searcheight($stackArray,$GroupOfEight){
        for($i=0; $i<count($GroupOfEight);$i++){


  $containsSearch = (count(array_intersect($stackArray,$GroupOfEight[$i])) == count($stackArray) && count(array_intersect($stackArray,$GroupOfEight[$i])) == count($GroupOfEight[$i]));
        if($containsSearch){
            return $i; //This specifies which index in GroupOfEight contains a matching array
        }
    }
    return -1;
}

// Calling the function that is given above.
echo searcheight($stackArray,$GroupOfEight);
?>

任何合乎逻辑的想法或解决方案将不胜感激。谢谢。

4

2 回答 2

9

这个很快:

function contains_array($array){
    foreach($array as $value){
        if(is_array($value)) {
          return true;
        }
    }
    return false;
}
于 2012-10-17T14:51:35.373 回答
3

你可以试试

$GroupOfEight = array(
        array(0,1,3,2,4,5,7,6),
        array(4,5,6,7,15,12,13,14),
        array(12,13,15,14,8,9,11,10),
        array(2,6,14,10,3,7,15,11),
        array(1,3,5,7,13,15,9,11),
        array(0,4,12,8,1,5,13,9),
        array(0,1,3,2,8,9,11,10));

$stackArray = array(0,4,12,1,9,8,5,13,9,2,5,2,10);

function searcheight($stackArray, $GroupOfEight) {
    $list = array();
    for($i = 0; $i < count($GroupOfEight); $i ++) {
        $intercept = array_intersect($GroupOfEight[$i], $stackArray);
        $len = count($intercept);
        if ($len % 4 == 0) {
            $list[$i] = $len;
        }
    }
    arsort($list);
    if (empty($list))
        return - 1;
    return key($list);
}
echo searcheight($stackArray, $GroupOfEight);

输出

5
于 2012-10-17T15:21:35.790 回答