问使用php查找一次出现的数组中的元素

EN

<?php

$array = [1, 2, 3, 4, 4, 1, 2, 5, 5, 11, 11];

$array_count_values = array_count_values($array);// get how many times a value appreas inside array

foreach($array_count_values as $key=>$val){ // now iterate over this newly created array
   if($val ==1){ // if count is 1
     echo $key. " in array come only one time.\n"; // this means value appears only one time inside array
   }
}

<?php

$array = [1, 2, 3, 4, 4, 1, 2, 5, 5, 11, 11,13]; // increased one value to show you the output

$array_count_values = array_count_values($array);

$single_time_comming_values_array = [];
foreach($array_count_values as $key=>$val){
   if($val ==1){
     $single_time_comming_values_array[] =  $key;
   }
}

print_r($single_time_comming_values_array);

<?php
function appearedOnce($arr)
{
  $result = 0;

       for($i=0; $i<sizeof($arr); $i++)
       {
          $result =  $result ^ $arr[$i];   

       }
    return $result;
}
$num = array(1, 2, 3, 4, 4, 1, 2, 5, 5, 11, 11);
print_r(appearedOnce($num)."\n")
?>

<?php

$arr =  [1, 2, 3, 4, 4, 1, 2, 5, 5, 11, 11];

sort($arr);

for($i = 0, $max = count($arr); $i < $max; $i++){

    // is single number last element in array?
    if($i == count($arr)-1)
        $singleNum = $arr[$i];

    // If adjacent elements the same, skip  
    if($i < count($arr)-1 && $arr[$i] == $arr[$i+1]){
        $i++;
    }
    else
    {
        // found single number.
        $singleNum = $arr[$i];
     }
}
var_dump($singleNum);