Java, this task is solved only through a set of if?

Question

Java

I need to remove from the array all numbers that are greater than neighboring ones.
For example
5 3 8 1 becomes 3 ( 5 is greater than 3 ) 1 ( 8 is greater than 3 and 1 ).
18, 1, 3, 6, 7, -5 becomes 1, 3, 6, -5, etc.

But faced with such difficulty, when index > 0 and < array.length-1, then we compare the previous number and the next one with the current one.
When index 0 is only next
when index array.length-1 is only previous.

It turns out that a big if with three conditions is the only way to complete the task?
Or is there a more elegant approach?

public static int[] removeMax(int[] array) {
        int[] localMax = new int[array.length];
        int s = -1;
        for (int i = 0; i < array.length; i++) {

            if ((i == 0 && array[i] > array[i + 1]) ||
                    (i != 0 && i != array.length - 1 && array[i] > array[i + 1] && array[i] > array[i - 1]) ||
                    (i == array.length-1 && array[i-1] < array[i])) {
                continue;
            }
            localMax[++s] = array[i];
        }
        return Arrays.copyOf(localMax, s+1);
    }

Here is an example of my if, maybe I misunderstand something and there is an easier way.

Answer 1

for (int i = 0; i < array.length; i++) {
    double previous = i == 0 ? Double.POSITIVE_INFINITY : array[i - 1];
    double next = i < array.length ? array[i + 1] : Double.POSITIVE_INFINITY;
    int current = array[i];

    if (current > previous || current > next)
        continue;

    localMax[++s] = current;
}