how to find all occurrences of a number in an array recursivelly

guys, given the fact I have an array how can I find all occurrences of the number recursively. I have figure out how to do a linear search recursively, but that would exit once it finds the number... what modification do I need to make in order to find all occurrences of the number.

Code :

---

int[] nums = {20, 25, 29, 55, 25}
 
//search with recursion
    public static int search(int[] a, int n, int i)
    {
        if(i == a.length-1)
            return -1;
        else if(a[i] == n)
            return i;
        else
            return search(a, n, ++i);        
    }
 
search(nums, 25, 0);

---

Are you wanting to simply count them or return an array of all of their indexes?

just count them...

thought about it for a bit, and came up with answer, may not be the most efficient. If someone got a better implementation.. post

Code :

---

static int count = 0, i = 0;
    public static int displayOccu(int[] a, int n)
    {
        if(i == a.length)
            return count;
        else 
        {
            if(a[i] == n)
            {
                i++;
                count++;
                return displayOccu(a, n);
            }
            else
            {
                i++;
                return displayOccu(a, n);
            }            
        }              
     }

---

There is a much better way of doing it, by taking the count variable out. This should also do it:

Code java:

---

public static int displayOccu(int[] a, int n, int i)
{
	if(i == a.length)
		return 0;
	else if(a[i] == n)
		return 1+displayOcc(a,n,i+1);
	else
		return displayOcc(a,n,i+1);
}

---

This code is very similar to examples of summation with recursion. If the number is a hit, we add one and check the rest, otherwise we just check the rest. When it reaches the end of the array, it returns 0 to the previous call, and then adds together each 1 as the recursive call returns to where it was called.
I haven't compiled or tested the code, but in theory it should work.