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Php根据另一个数组定义的顺序对数组进行排序_Php - Fatal编程技术网

Php根据另一个数组定义的顺序对数组进行排序

Php根据另一个数组定义的顺序对数组进行排序,php,Php,给定两个数组A1[]和A2[],对A1进行排序,使元素之间的相对顺序与A2中的顺序相同。对于A2中不存在的元素,最后按排序顺序追加 Input: A1[] = {2, 1, 2, 5, 7, 1, 9, 3, 6, 8, 8} A2[] = {2, 1, 8, 3} Output: A1[] = {2, 2, 1, 1, 8, 8, 3, 5, 6, 7, 9} 尝试:- $sorted = array(); foreach($a1 as $key => $value)

给定两个数组A1[]和A2[],对A1进行排序,使元素之间的相对顺序与A2中的顺序相同。对于A2中不存在的元素,最后按排序顺序追加

Input: A1[] = {2, 1, 2, 5, 7, 1, 9, 3, 6, 8, 8}
       A2[] = {2, 1, 8, 3}
Output: A1[] = {2, 2, 1, 1, 8, 8, 3, 5, 6, 7, 9}
尝试:-

$sorted = array();

foreach($a1 as $key => $value) {
    if(in_array($value, $a2)) {
        $sorted[array_search($value, $a1)] = $value;
    }
}

这可以通过为每个回路执行:

$arr1 = array(2, 1, 2, 5, 7, 1, 9, 3, 6, 8, 8); // array to be sorted
$arr2 = array(2, 1, 8, 3); // refrence array for sort logic
// Output: A1[] = {2, 2, 1, 1, 8, 8, 3, 5, 6, 7, 9}

$sortarr = array(); // array to store final sorted values
foreach ($arr2 as $a) {
    foreach ($arr1 as $k => $b) {
        if($b==$a) {
            $sortarr[]=$b;
            unset($arr1[$k]);
        }
    }

}

$finalarr = array_merge($sortarr, $arr1);

print_r($finalarr);

我尝试了此代码并成功:

<?php
/*
Input: A1[] = {2, 1, 2, 5, 7, 1, 9, 3, 6, 8, 8}
       A2[] = {2, 1, 8, 3}
Output: A1[] = {2, 2, 1, 1, 8, 8, 3, 5, 6, 7, 9}

*/
$a1=array(2, 1, 2, 5, 7, 1, 9, 3, 6, 8, 8);
$a2=array(2, 1, 8, 3);
$a3=array();

sort($a1);//order array
//order array a3 with a2 value....
for($i=0;$i<sizeof($a2);$i++){
  for($j=0;$j<sizeof($a1);$j++){
    if($a1[$j]==$a2[$i]){
      array_push($a3,$a2[$i]);
      //if exsist value i change value in a1 in x
      $a1[$j]="x";
    }
  }
}
//write in a3 the next number not present in a2
for($i=0;$i<sizeof($a1);$i++){
  if($a1[$i]<>"x"){
    array_push($a3, $a1[$i]);
  }
}
print_r($a3);
//out:Array ( [0] => 2 [1] => 2 [2] => 1 [3] => 1 [4] => 8 [5] => 8 [6] => 3 [7] => 5 [8] => 6 [9] => 7 [10] => 9 )
 ?>


希望这有帮助

您可以这样使用
usort

$k = array_flip($a2); // Create an associative array for the second array
usort($a1, function($a, $b) use ($k) {
    return isset($k[$a]) ? (isset($k[$b]) ? $k[$a]-$k[$b] : -1) : (isset($k[$b]) ? 1 : $a-$b);
});

其他使用嵌套循环的解决方案的时间复杂度为O(n²),而此解决方案的时间复杂度为O(nlogn)。

到目前为止您尝试了什么?以上编辑请检查。