检查阵列的一半是否是回文

Question

for my school project, i need to write an algorithm which is gonna check if both halfs of array are palindromes, and if array has odd values, i should skip one in the middle. For example:

[1,2,1,6,4,2,5] should check if [1,2,1] is palindrome, and if [4,2,5] is palindrome, skipping the middle value cause its odd.

This is what i've wrote so far:

$somearray = ['1','1','1','1','1','2','1','1','2','1','1'];

$array_size = count($tablica);

function check_palindrome($arr, $start, $end)
{
    for($i=$start; $i < $end; $i++){
        if($arr[$i] != $arr[$start+$end-$i])
        {
            return false;
        }
    }
    return true;
}

if($array_size%2 == 0){
    $start1 = 0;
    $start2 = $array_size/2;
    $end1 = $array_size/2 - 1;
    $end2 = $array_size - 1;
} else {
    $start1 = 0;
    $start2 = round($array_size/2);
    $end1 = round($array_size/2, 0, PHP_ROUND_HALF_DOWN)-1;
    $end2 = $array_size -1;
}

$foo1 = check_palindrome($somearray, $start1, $end1);
$foo2 = check_palindrome($somearray, $start2, $end2);

if($foo1 && $foo2){
    echo 'Halfs are palindromes';
} else {
    echo 'Halfs are not palindromes';
}

And it works, but first most important part - my algorithm checks if first half is palindrome, then calling function twice i check second half - my teacher doesn't like that and tells me i should write it in one function, but i really don't have an idea how.

My second problem is how to improve my for loop, because for now it check half like this:

A=[1,2,4,2,1]

1=1, 2=2, 2=2, 1=1 if its true, but it should stop after checking 1=1 and 2=2

it's happening because of this statement in for loop: $i < $end but also don't know how to improve it :/

I really need your help. Thanks.

Answer 1

You can make your check_palindrome function more efficient by cutting the number of comparisons in half. For example, as it's currently written it would do the following for a 3-character string.

check index 0 against index 3
check index 1 against index 1
check index 3 against index 0

In an odd-length string, the middle character is by definition equal to itself, so there's no need to check it. And by the time you've reached the middle of the string on the left side, you've checked against all the characters on the right side.

A more efficient function would be:

function check_palindrome($arr, $start, $end)
{
    $length = ($end - $start + 1)/2;
    for($i=0; $i < length; $i++){
        if($arr[$start+$i] != $arr[$end-$i])
        {
            return false;
        }
    }
    return true;
}