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Question

I need help with my homework

My question is: How to complete the method findMaxItem below to return a pointer to the largest element of the linked list given by the parameter ptr, or return null if the list is empty. The solution must be a recursive one.

This my my answer but is not correct:

public static Node findMaxItem (Node ptr) {
    if (ptr == null) {
        return null;

    } else {
        if (ptr.next.data > ptr.data) {
            Node n = ptr.next;


            if (ptr.next.data >= n.data) {

                n = ptr.next;
                //ptr = n.next;
                findMaxItem(n);
            } else {

                ptr = ptr.next;
                findMaxItem(ptr);
            }
            ptr = n.next;
            findMaxItem(ptr);
            ptr = n;
        }
    //   ptr = ptr.next;
     //  findMaxItem(ptr);
    }
    return ptr;
}

Thanks for help everyone :)

Answer 1

You've at least got the first part right: if the list is empty, return null . This is the base case of the recursion.

    if (ptr == null) return null;

You're meant to use recursion for the rest of the cases. But how?

If the list is not empty, you can divide it into two parts: the first item, and the rest of the list. Would it help you "find the max item" in the rest of the list?

    Node restOfTheList = ptr.next;
    Node maxOfRestOfTheList = findMax(restOfTheList);

Now you have the first item, referenced by ptr , and the item with the greatest value in the rest of the list. If you compare the data in those, you can figure out which has the greatest value. And which ever it is, that is the node with the largest item. Does that make sense?

    if data of "ptr" is greater than data of "maxOfRestOfTheList": return ptr
    else: return maxOfRestOfTheList;

You'll also have to deal with the case where there is only one item in the list. In that situtation restOfTheList is empty.

Answer 2

One possible solution is to create a method that receive the array with the values -1 and the last value. Then check if the next last element is greater that the one you have or not, if it is, the new value is the current maximum remove it and call the same method by using the new array (with the removed value) and the current maximum.

The base case will be when there is only one element in the array, take or compare it with the last maximum value you passed to the recuerdo be función and return the greater value of these two. This last will be the maximum in a recursive way.

I have no possibility to write the code right now but in meta-code it should be something like this

private int finalMax
public void maxRecursive(arrayList<int>List, Int tempMax){
  //Base case
   Int newTempMax = tempMax
    If (tempMax >list.get(list.size-1)){
       newTempMax=list.get(list.size()-1)
    }
    If (list.size()>1){
      List.remove(list.size()-1)
      maxRecursive(list, newTemMax)
    }else{
       finalMax=newTempMax
    }
 }

As I said this is nothing more than pseudo-code that looks adapted to java, I cannot son anything better right now but AI think that you can get the Idea from here.

Best!