如何判断以逗号分隔的数字列表是否遵循数字的自然顺序

Question

I have a comma delimited list of numbers which i am converting into an array and what i want to know about the list of numbers is if the numbers listed obey a natural ordering of numbers,you know,have a difference of exactly 1 between the next and the previous.

If its true the list obeys the natural ordering,i want to pick the first number of the list and if not the list obeys not the natural order,i pick the second.

This is my code.

<?php
error_reporting(0);
/**
Analyze numbers

Condition 1

if from number to the next has a difference of 1,then pick the first number in the list

Condition 2

if from one number the next,a difference of greater than 1 was found,then pick next from first

Condition 3

if list contains only one number,pick the number
*/
$number_picked = null;

$a = '5,7,8,9,10';
$b = '2,3,4,5,6,7,8,9,10';
$c = '10';

$data = explode(',', $b);
$count = count($data);

foreach($data as $index => $number)
{
/**
If array has exactly one value
*/
if($count == 1){
echo 'number is:'.$number;
exit();
}

  $previous = $data[($count+$index-1) % $count]; 

  $current = $number;
  $next = $data[($index+1) % $count];

  $diff = ($next - $previous);

  if($diff == 1){
  $number_picked = array_values($data)[0];
  echo $number_picked.'correct';
  }
  elseif($diff > 1){
  $number_picked = array_values($data)[1];
  echo $number_picked.'wrong';
  }
}
?>

The problem i am having is to figure out how to test the difference for all array elements.

Answer 1

Try using a function to solve this... Like so:

<?php
error_reporting(0);
/**
Analyze numbers

Condition 1

if from number to the next has a difference of 1,then pick the first number in the list

Condition 2

if from one number the next,a difference of greater than 1 was found,then pick next from first

Condition 3

if list contains only one number,pick the number
*/
$number_picked = null;

$a = '5,7,8,9,10';
$b = '2,3,4,5,6,7,8,9,10';
$c = '10';

function test($string) {    
    $data = explode(',', $string);
    if(count($data) === 1){
        return 'number is:'.$number;
    }

    foreach($data as $index => $number)
    {

        $previous = $data[($count+$index-1) % $count]; 

        $current = $number;
        $next = $data[($index+1) % $count];

        $diff = ($next - $previous);

        if($diff == 1){
            $number_picked = array_values($data)[0];
            return $number_picked.'correct';
        }
        elseif($diff > 1){
            $number_picked = array_values($data)[1];
            return $number_picked.'wrong';
        }
    }
}

echo test($a);
echo test($b);
echo test($c);
?>

Answer 2

You already know how to explode the list, so I'll skip that. You already handle a single item, so I'll skip that as well.

What is left, is checking the rest of the array. Basically; there's two possible outcome values: either the first element or the second. So we'll save those two first:

$outcome1 = $list[0];
$outcome2 = $list[1];

Next, we'll loop over the items. We'll remember the last found item, and make sure that the difference between the new and the old is 1. If it is, we continue. If it isn't, we abort and immediately return $outcome2.

If we reach the end of the list without aborting, it's naturally ordered, so we return $outcome1.

$lastNumber = null;
foreach( $items as $number ) {
  if($lastNumber === null || $number - $lastNumber == 1 ) {
    // continue scanning
    $lastNumber = $number;
  }
  else {
    // not ordened
    return $outcome2;
  }
}
return $outcome1; // scanned everything; was ordened.

(Note: code not tested)

Answer 3

To avoid the headache of accessing the previous or next element, and deciding whether it still is inside the array or not, use the fact that on a natural ordering the item i and the first item have a difference of i.

Also the corner case you call condition 3 is easier to handle outside the loop than inside of it. But easier still, the way we characterize a natural ordered list holds for a 1-item list :

$natural = true;
for($i=1; $i<$count && $natural; $i++)
    $natural &= ($data[$i] == $data[0] + $i)

$number = $natural ? $data[0] : $data[1];

For $count == 1 the loop is never entered and thus $natural stays true : you select the first element.

Answer 4

No loops are needed, a little bit of maths will help you here. Once you have your numbers in an array:

$a = explode(',', '5,7,8,9,10');

pass them to this function:-

function isSequential(array $sequence, $diff = 1)
{
    return $sequence[count($sequence) - 1] === $sequence[0] + ($diff * (count($sequence) - 1));
}

The function will return true if the numbers in the array follow a natural sequence. You should even be able to adjust it for different spacings between numbers, eg 2, 4, 6, 8 , etc using the $diff parameter, although I haven't tested that thoroughly.

See it working .

Keep in mind that this will only work if your list of numbers is ordered from smallest to largest.