从1-10开始，找到下一个缺失的数字？ [关闭]

Say i have the following array: 1,2,3,6,8

I need to find that the next missing number is 4

What is the easiest and quickest way to perform this ?

Iterating the array and checking in_array() should do it.

$missing = null;

for ($i = 1; $i <= 10; $i++) {
    if (!in_array($i, $myArray)) {
        $missing = $i;
        break;
    }
}

if (!is_null($missing)) {
    // hit
}

$missing = array_diff(range(min($myArray), max($myArray)), $myArray);
$nextMissing = (!empty($missing)) ? array_shift($missing) : max($myArray) + 1;

The Question is worded poorly but from my understanding you have 1,2,3 and then you need to check that next value should be 4

$prev = array[0];
for($i = 1; $i < count(array); $i++){
   if(array[$i] != prev+1) return (prev+1) 
}

I would do it like this (if the array is sorted):

$array = array(1, 2, 3, 6, 8);
$size = count($array);
$missing = 0;
$firstval = $array[0];

for ($i = 0; $i < $size; ++$i) {
    if ($firstval++ != $array[$i]) {
        $missing = $firstval - 1;
        break;
    }
}

if ($missing != 0) echo $missing.' is missing'.PHP_EOL;

Which gives:

4 is missing