Return a min & max value and their positions within an arraylist using recursion

Question

I need to write a recursive function that returns the largest and smallest elements in the ArrayList and the corresponding indices. To return these four items, I need to return an object that is an instance of an inner class called MinMaxObject that has four defined private variables: max, min, maxPos, minPos of types double, double, int and int.

I've gotten to here on my own and I need to start the recursion but I have no idea how to get going. If someone could point me in the right direction I should be able to pick it up.

public static void main(String args[]) {

    Scanner console = new Scanner(System.in);
    System.out.print("Please enter a file name");
    String fileName = console.next();

    try {
        File fileRef = new File(fileName);
        Scanner tokens = new Scanner(fileRef);
        double input = tokens.nextDouble();

        ArrayList<Double> list = new ArrayList<Double>();

        while (tokens.hasNextLine()) {
            list.add(input);
        }

        System.out.println("For Loop");
        for (int counter = 0; counter < list.size(); counter++) {
            System.out.println(list.get(counter));
        }

    } catch (FileNotFoundException e) {
        e.printStackTrace();
    }


// 2 functions, one initialize & one to call the original recursion
    //can start at beggining and recurse to the end or vice versa
    //updates the min and max as it goes
    //update minPos and maxPos as well
}

public class MinMaxObject {
    private double min;
    private double max;
    private int minPos;
    private int maxPos;

public MinMaxObject(double newMin, double newMax, int newMinPos, int newMaxPos){
    min = newMin;
    max = newMax;
    minPos = newMinPos;
    maxPos = newMaxPos;
}

public double getMin(){
    return min;
}

public void setMin(double newMin){
    min = newMin;
}

public double getMax(){
    return max;
}

public void setMax(double newMax){
    max = newMax;
}

public int getMinPos(){
    return minPos;
}

public void setMinPos(int newMinPos){
    minPos = newMinPos;
}
public int getMaxPos(){
    return maxPos;
}

public void setMaxPos(int newMaxPos){
    maxPos = newMaxPos;
}

Answer 1

It sounds like recursion is a part of the assignment.

The simplest way is to iterate through the arraylist, but since you need recursion it's a little more complex. Since this is homework, and because I'm lazy, I'm not going to give you compilable Java code. This is a close approximation. I'm also not going to give you the whole method, cause you should figure the rest out from here.

private int min = 0;
private int max = 0;
findMinMax(int index, List<int> list)
{
    //in recursion always start your method with a base-case
    if(index >= list.Count)
        return ;

    //find min and max here

    //this is called tail recursion, it's basically the same as a loop
    findMinMax(index + 1, list);
}

Answer 2

Thanks @ScubaSteve! ...I outline little more/detailed:

void findMinMax(int idx, List items, MinMaxObject result) {
    // this is good:
    if (idx >= items.size()) {
        return;
    }
    // check whether list[idx] is min or max: compare with & store to 'result'

    // recursion:
    findMinMax(idx + 1, items, result);
}

The initial/final/outer invocation would be:

// initialize with high min, low max and "out of range" indices
// ...alternatively in default constructor/field declaration.
MinMaxObject result = new MinMaxObject(Integer.MAX_VALUE, Integer.MIN_VALUE, -1, -1);
// start recursion at index 0:
findMinMax(0, items, result);

---

A recursion alternative would be: To recourse not only the "head" of the list (idx : 0 ... size - 1), but also the "tail" like:

void findMinMax(int head, int tail, List items, MinMaxObject result) {
    if(head > tail) {
        // done!
        return;
    }
    // check items[head] and items[tail]... store to result
    // iterate:
    return findMinMax(head + 1, tail - 1, items, result);
}

the "alternative" halves the count of recursions (but doubles the "inline complexity").

---

Another "divide & conquer" approach:

void findMinMax(int left, int right, items, result) {
    if(left > right) {
        return;// alternatively do nothing
    } else if (left == right) {
       //check items[left] ...
       // return;
    } else { // left < right
        int mid = (right - left) / 2;
        findMinMax(left, mid, items, result);
        findMinMax(mid + 1, right, items, result);
        // return;
    }
}    

...with (same) O(n) complexity.