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Computing the size of a linked list using recursion/helper function - Java

Beginner here using Java (first year student), and am unable to get the below function to work. The goal is to use recursion and a helper function to compute the size of a singly linked list. When running the code against test lists, it keeps returning List changed to [].
I'm struggling in general with Java, so any help is appreciated. Thank you
public class MyLinked {
static class Node {
public Node(double item, Node next) {
this.item = item;
this.next = next;
}
public double item;
public Node next;
}
int N;
Node first;
public int sizeForward() {
return sizeForwardHelper(first);
}
public int sizeForwardHelper(Node n) {
Node current = first;
if (current == null) {
return 0;
} else {
first = first.next;
return sizeForward() + 1;
}
}
I believe I have the first portion set up to return 0 if there are no elements in the List. I believe it's the second part that isn't setting up correctly?
Thanks

Because it’s important for your learning to not spoonfeed you, I’ll describe an approach rather than provide code.
Use this fact:
The length of the list from any given node to the end is 1 plus the length measured from the next node (if there is one).
Usually (as would work here), recursive functions take this form:
If the terminating condition is true, return some value
Otherwise, return some value plus the recursively calculated value
When writing a recursive function, first decide on the terminating condition. In this case, n == null is the obvious choice, and you’d return 0, because you’ve run off the end of the list and the length of nothing (ie no node) is nothing. This also handles the empty list (when first is null) without any special code.
Otherwise, return 1 (the length of one node) plus the length of next.
Put that all together and you’ll have your answer.
——
Hint: The body of the recursive helper method can be coded using one short line if you use a ternary expression.

Instead of calling your wrapper function call your helper function recursively. Try the following:
public int sizeForward () {
return sizeForwardHelper (first);
}
public int sizeForwardHelper(Node n) {
if (n == null) // base case
return 0;
return sizeForwardHelper(n.next) + 1; // count this node + rest of list
}

Your method that computes the size of the list actually modifies the list in the process (with first = first.next; you set the first element to the next, and since there is a recursion, the first element always end up being null which is equivalent to an empty list with your design). Your method will work once, but your list will be empty afterwards.
To illustrate this, I added a print next to the instruction first = first.next; and wrote the following main:
public static void main(String[] args) {
Node n2 = new Node(2d, null);
Node n1 = new Node(1d, n2);
Node n = new Node(0, n1);
MyLinked l = new MyLinked(n);
System.out.println("The first element is: "+l.first.item);
System.out.println("The size is: "+l.sizeForward());
System.out.println("The first element is: "+l.first);
}
It yields:
The first element is: 0.0
first is set to 1.0
first is set to 2.0
first is set to null
The size is: 3
The first element is: null
Clearly, you should not modify the list while computing its size. The helper method should return 0 if the node is null (empty list), and 1 plus the size of the rest of the list otherwise. Here is the code.
public int sizeForwardHelper(Node n) {
if (n == null)
return 0;
else
return sizeForwardHelper(n.next) +1;
}
The goal of the arg free method sizeForward() is just to call the helper. The helper should not use it though.

Implementing an equals() method to compare contents of two 'bag' objects

I am working on a school assignment. The objective is to practice GUI's, clone() methods, and using/ modifying existing code. I am trying to write an equals method in the way the instructor desires-- by using a clone of the object, removing items from the bag (returns boolean based on success or failure to remove).
The bag is represented in an array, and should return true in cases such as {1,2,3} and {3,2,1}, ie order does not matter, only the number of each number present in the arrays.
Here is the issue
It works in most cases, however there is a bug in cases where the bags contain numbers as such: {1,1,2} and {1,2,2} and other similar iterations. It is returning true instead of false.
I believe it has something to do with the remove() method we are supposed to use. If i understand it correctly, it is supposed to put the value at the 'end' of the array and decrease the manyItems counter (this is a variable for number of items in the array, because array.length is by default in the constructor 10.)
The code is largely written by another person. We had to import the existing files and write new methods to complete the task we were given. I have all the GUI part done so i will not include that class, only the used methods in the IntArrayBag class.
A second pair of eyes would be helpful. Thanks.
public class IntArrayBag implements Cloneable
{
// Invariant of the IntArrayBag class:
// 1. The number of elements in the bag is in the instance variable
// manyItems, which is no more than data.length.
// 2. For an empty bag, we do not care what is stored in any of data;
// for a non-empty bag, the elements in the bag are stored in data[0]
// through data[manyItems-1], and we don�t care what�s in the
// rest of data.
private int[ ] data;
private int manyItems;
public IntArrayBag( )
{
final int INITIAL_CAPACITY = 10;
manyItems = 0;
data = new int[INITIAL_CAPACITY];
}
public IntArrayBag clone( )
{ // Clone an IntArrayBag object.
IntArrayBag answer;
try
{
answer = (IntArrayBag) super.clone( );
}
catch (CloneNotSupportedException e)
{ // This exception should not occur. But if it does, it would probably
// indicate a programming error that made super.clone unavailable.
// The most common error would be forgetting the "Implements Cloneable"
// clause at the start of this class.
throw new RuntimeException
("This class does not implement Cloneable");
}
answer.data = data.clone( );
return answer;
}
public int size( )
{
return manyItems;
}
public boolean remove(int target)
{
int index; // The location of target in the data array.
// First, set index to the location of target in the data array,
// which could be as small as 0 or as large as manyItems-1; If target
// is not in the array, then index will be set equal to manyItems;
for (index = 0; (index < manyItems) && (target != data[index]); index++)
// No work is needed in the body of this for-loop.
;
if (index == manyItems)
// The target was not found, so nothing is removed.
return false;
else
{ // The target was found at data[index].
// So reduce manyItems by 1 and copy the last element onto data[index].
manyItems--;
data[index] = data[manyItems];
return true;
}
}
//I added extra variables that are not needed to try to increase readability,
//as well as when i was trying to debug the code originally
public boolean equals(Object obj){
if (obj instanceof IntArrayBag){
IntArrayBag canidate = (IntArrayBag) obj; // i know this can be changed, this was required
IntArrayBag canidateTest = (IntArrayBag) canidate.clone(); //this was created
//as a clone because it was otherwise referring to the same memory address
//this caused items to be removed from bags when testing for equality
IntArrayBag test = (IntArrayBag) this.clone();
//fast check to see if the two objects have the same number of items,
//if they dont will return false and skip the item by item checking
if (test.size() != canidateTest.size())
return false;
//the loop will go through every element in the test bag it will
//then remove the value that is present at the first index of the test bag
for (int i = 0; (i < (test.size()) || i < (canidateTest.size())); i++){
int check = test.data[i];
//remove() returns a boolean so if the value is not present in each bag
//then the conditional will be met and the method will return false
boolean test1 = test.remove(check);
boolean test2 = canidateTest.remove(check);
if (test1 != test2)
return false;
}//end for loop
// if the loop goes through every element
//and finds every value was true it will return true
return true;
}//end if
else
return false;
}//end equals
}

I cannot see the big picture, as I havent coded GUIs in Java before, however, as far as comparing 2 int[] arrays, I would sort the arrays before the comparison. This will allow you to eliminate problem cases like the one you stated ( if sorting is possible), then apply something like:
while(array_1[index]==array_2[index] && index<array_1.length)
{index++;}
and find where did the loop break by checking the final value of index

Is it explicitly stated to use clone? You can achieve it easily by overriding the hashCode() for this Object.
You can override the hashCode() for this object as follows:
#Override
public int hashCode() {
final int prime = 5;
int result = 1;
/* Sort Array */
Arrays.sort(this.data);
/* Calculate Hash */
for(int d : this.data) {
result = prime * result + d;
}
/* Return Result */
return result;
}
#Override
public boolean equals(Object obj) {
if (this == obj) return true;
if (obj == null || this.getClass() != obj.getClass()){
return false;
}
return false;
}
If you want to continue using your implementation for equals to compare test and CandidateTest then also you can compute unique hashes and make decision based on the results.
Here is the code snippet:
/* Assuming that you have put size comparison logic on top
and the two objects are of same size */
final int prime = 31;
int testResult = 1;
int candidateTestResult = 1;
for(int i = 0; i < test.size(); i++) {
testResult = prime * testResult + test.data[i];
candidateTestResult = prime * candidateTestResult + candidateTest.data[i];
}
/* Return Result */
return testResult == candidateTestResult;

I believe the problem is in this line:
for (int i = 0; (i < (test.size()) || i < (canidateTest.size())); i++){
The problem here is that test and canidateTest are the clones that you made, and you are removing elements from those bags. And any time you remove an element from the bag, the size will decrease (because you decrease manyItems, and size() returns manyItems). This means you're only going to go through half the array. Suppose the original size is 4. Then, the first time through the loop, i==0 and test.size()==4; the second time, i==0 and test.size()==3; the third time, i==2 and test.size()==2, and you exit the loop. So you don't look at all 4 elements--you only look at 2.
You'll need to decide: do you want to go through the elements of the original array, or the elements of the clone? If you go through the elements of the clone, you actually never need to increment i. You can always look at test.data[0], since once you look at it, you remove it, so you know test.data[0] will be replaced with something else. In fact, you don't need i at all. Just loop until the bag size is 0, or until you determine that the bags aren't equal. On the other hand, if you go through the elements of this.data (i.e. look at this.data[i] or just data[i]), then make sure i goes all the way up to this.size().
(One more small point: the correct spelling is "candidate".)

Maybe you should try SET interface
view this in detail ：http://www.tutorialspoint.com/java/java_set_interface.htm
A set object cannot contains duplicate elements, so it's suitable for your assignment than build your own class.
For example：[1,1,2] and [1,2,2]
you can use this to test whether they are equal
arr1 = {1,1,2}
arr2 = {1,2,2}
Set<Integer> set = new HashSet<Integer>();
for(int i : arr1){//build set of arr1
if(set.contains(i)==false){
set.add(i)
}
}
for(int i:arr2){
if(set.contains(i)==false){
System.out.println('not equal');
break;
}
}
Hope this is helpful.

Is it possible to get next element in the Stream?

I am trying to converting a for loop to functional code. I need to look ahead one value and also look behind one value. Is it possible using streams?
The following code is to convert the Roman text to numeric value.
Not sure if reduce method with two/three arguments can help here.
int previousCharValue = 0;
int total = 0;
for (int i = 0; i < input.length(); i++) {
char current = input.charAt(i);
RomanNumeral romanNum = RomanNumeral.valueOf(Character.toString(current));
if (previousCharValue > 0) {
total += (romanNum.getNumericValue() - previousCharValue);
previousCharValue = 0;
} else {
if (i < input.length() - 1) {
char next = input.charAt(i + 1);
RomanNumeral nextNum = RomanNumeral.valueOf(Character.toString(next));
if (romanNum.getNumericValue() < nextNum.getNumericValue()) {
previousCharValue = romanNum.getNumericValue();
}
}
if (previousCharValue == 0) {
total += romanNum.getNumericValue();
}
}
}

No, this is not possible using streams, at least not easily. The stream API abstracts away from the order in which the elements are processed: the stream might be processed in parallel, or in reverse order. So "the next element" and "previous element" do not exist in the stream abstraction.
You should use the API best suited for the job: stream are excellent if you need to apply some operation to all elements of a collection and you are not interested in the order. If you need to process the elements in a certain order, you have to use iterators or maybe access the list elements through indices.

I haven't see such use case with streams, so I can not say if it is possible or not. But when I need to use streams with index, I choose IntStream#range(0, table.length), and then in lambdas I get the value from this table/list.
For example
int[] arr = {1,2,3,4};
int result = IntStream.range(0, arr.length)
.map(idx->idx>0 ? arr[idx] + arr[idx-1]:arr[idx])
.sum();

By the nature of the stream you don't know the next element unless you read it. Therefore directly obtaining the next element is not possible when processing current element. However since you are reading current element you obiously know what was read before, so to achieve such goal as "accesing previous element" and "accessing next element", you can rely on the history of elements which were already processed.
Following two solutions are possible for your problem:
Get access to previously read elements. This way you know the current element and defined number of previously read elements
Assume that at the moment of stream processing you read next element and that current element was read in previous iteration. In other words you consider previously read element as "current" and currently processed element as next (see below).
Solution 1 - implemenation
First we need a data structure which will allow keeping track of data flowing through the stream. Good choice could be an instance of Queue because queues by their nature allows data flowing through them. We only need to bound the queue to the number of last elements we want to know (that would be 3 elements for your use case). For this we create a "bounded" queue keeping history like this:
public class StreamHistory<T> {
private final int numberOfElementsToRemember;
private LinkedList<T> queue = new LinkedList<T>(); // queue will store at most numberOfElementsToRemember
public StreamHistory(int numberOfElementsToRemember) {
this.numberOfElementsToRemember = numberOfElementsToRemember;
}
public StreamHistory save(T curElem) {
if (queue.size() == numberOfElementsToRemember) {
queue.pollLast(); // remove last to keep only requested number of elements
}
queue.offerFirst(curElem);
return this;
}
public LinkedList<T> getLastElements() {
return queue; // or return immutable copy or immutable view on the queue. Depends on what you want.
}
}
The generic parameter T is the type of actual elements of the stream. Method save returns reference to instance of current StreamHistory for better integration with java Stream api (see below) and it is not really required.
Now the only thing to do is to convert the stream of elements to the stream of instances of StreamHistory (where each next element of the stream will hold last n instances of actual objects going through the stream).
public class StreamHistoryTest {
public static void main(String[] args) {
Stream<Character> charactersStream = IntStream.range(97, 123).mapToObj(code -> (char) code); // original stream
StreamHistory<Character> streamHistory = new StreamHistory<>(3); // instance of StreamHistory which will store last 3 elements
charactersStream.map(character -> streamHistory.save(character)).forEach(history -> {
history.getLastElements().forEach(System.out::print);
System.out.println();
});
}
}
In above example we first create a stream of all letters in alphabet. Than we create instance of StreamHistory which will be pushed to each iteration of map() call on original stream. Via call to map() we convert to stream containing references to our instance of StreamHistory.
Note that each time the data flows through original stream the call to streamHistory.save(character) updates the content of the streamHistory object to reflect current state of the stream.
Finally in each iteration we print last 3 saved characters. The output of this method is following:
a
ba
cba
dcb
edc
fed
gfe
hgf
ihg
jih
kji
lkj
mlk
nml
onm
pon
qpo
rqp
srq
tsr
uts
vut
wvu
xwv
yxw
zyx
Solution 2 - implementation
While solution 1 will in most cases do the job and is fairly easy to follow, there are use cases were the possibility to inspect next element and previous is really convenient. In such scenario we are only interested in three element tuples (pevious, current, next) and having only one element does not matter (for simple example consider following riddle: "given a stream of numbers return a tupple of three subsequent numbers which gives the highest sum"). To solve such use cases we might want to have more convenient api than StreamHistory class.
For this scenario we introduce a new variation of StreamHistory class (which we call StreamNeighbours). The class will allow to inspect the previous and the next element directly. Processing will be done in time "T-1" (that is: the currently processed original element is considered as next element, and previously processed original element is considered to be current element). This way we, in some sense, inspect one element ahead.
The modified class is following:
public class StreamNeighbours<T> {
private LinkedList<T> queue = new LinkedList(); // queue will store one element before current and one after
private boolean threeElementsRead; // at least three items were added - only if we have three items we can inspect "next" and "previous" element
/**
* Allows to handle situation when only one element was read, so technically this instance of StreamNeighbours is not
* yet ready to return next element
*/
public boolean isFirst() {
return queue.size() == 1;
}
/**
* Allows to read first element in case less than tree elements were read, so technically this instance of StreamNeighbours is
* not yet ready to return both next and previous element
* #return
*/
public T getFirst() {
if (isFirst()) {
return queue.getFirst();
} else if (isSecond()) {
return queue.get(1);
} else {
throw new IllegalStateException("Call to getFirst() only possible when one or two elements were added. Call to getCurrent() instead. To inspect the number of elements call to isFirst() or isSecond().");
}
}
/**
* Allows to handle situation when only two element were read, so technically this instance of StreamNeighbours is not
* yet ready to return next element (because we always need 3 elements to have previos and next element)
*/
public boolean isSecond() {
return queue.size() == 2;
}
public T getSecond() {
if (!isSecond()) {
throw new IllegalStateException("Call to getSecond() only possible when one two elements were added. Call to getFirst() or getCurrent() instead.");
}
return queue.getFirst();
}
/**
* Allows to check that this instance of StreamNeighbours is ready to return both next and previous element.
* #return
*/
public boolean areThreeElementsRead() {
return threeElementsRead;
}
public StreamNeighbours<T> addNext(T nextElem) {
if (queue.size() == 3) {
queue.pollLast(); // remove last to keep only three
}
queue.offerFirst(nextElem);
if (!areThreeElementsRead() && queue.size() == 3) {
threeElementsRead = true;
}
return this;
}
public T getCurrent() {
ensureReadyForReading();
return queue.get(1); // current element is always in the middle when three elements were read
}
public T getPrevious() {
if (!isFirst()) {
return queue.getLast();
} else {
throw new IllegalStateException("Unable to read previous element of first element. Call to isFirst() to know if it first element or not.");
}
}
public T getNext() {
ensureReadyForReading();
return queue.getFirst();
}
private void ensureReadyForReading() {
if (!areThreeElementsRead()) {
throw new IllegalStateException("Queue is not threeElementsRead for reading (less than two elements were added). Call to areThreeElementsRead() to know if it's ok to call to getCurrent()");
}
}
}
Now, assuming that three elements were already read, we can directly access current element (which is the element going through the stream at time T-1), we can access next element (which is the element going at the moment through the stream) and previous (which is the element going through the stream at time T-2):
public class StreamTest {
public static void main(String[] args) {
Stream<Character> charactersStream = IntStream.range(97, 123).mapToObj(code -> (char) code);
StreamNeighbours<Character> streamNeighbours = new StreamNeighbours<Character>();
charactersStream.map(character -> streamNeighbours.addNext(character)).forEach(neighbours -> {
// NOTE: if you want to have access the values before instance of StreamNeighbours is ready to serve three elements
// you can use belows methods like isFirst() -> getFirst(), isSecond() -> getSecond()
//
// if (curNeighbours.isFirst()) {
// Character currentChar = curNeighbours.getFirst();
// System.out.println("???" + " " + currentChar + " " + "???");
// } else if (curNeighbours.isSecond()) {
// Character currentChar = curNeighbours.getSecond();
// System.out.println(String.valueOf(curNeighbours.getFirst()) + " " + currentChar + " " + "???");
//
// }
//
// OTHERWISE: you are only interested in tupples consisting of three elements, so three elements needed to be read
if (neighbours.areThreeElementsRead()) {
System.out.println(neighbours.getPrevious() + " " + neighbours.getCurrent() + " " + neighbours.getNext());
}
});
}
}
The output of this is following:
a b c
b c d
c d e
d e f
e f g
f g h
g h i
h i j
i j k
j k l
k l m
l m n
m n o
n o p
o p q
p q r
q r s
r s t
s t u
t u v
u v w
v w x
w x y
x y z
By StreamNeighbours class it is easier to track the previous/next element (because we have method with appropriate names), while in StreamHistory class this is more cumbersome since we need to manually "reverse" the order of the queue to achieve this.

As others stated, it's not feasable, to get next elements from within an iterated Stream.
If IntStream is used as a for loop surrogate, which merely acts as an index iteration provider, it's possible use its range iteration index just like with for; one needs to provide a means of skipping the next element on the next iteration, though, e. g. by means of an external skip var, like this:
AtomicBoolean skip = new AtomicBoolean();
List<String> patterns = IntStream.range(0, ptrnStr.length())
.mapToObj(i -> {
if (skip.get()) {
skip.set(false);
return "";
}
char c = ptrnStr.charAt(i);
if (c == '\\') {
skip.set(true);
return String.valueOf(new char[] { c, ptrnStr.charAt(++i) });
}
return String.valueOf(c);
})
It's not pretty, but it works.
On the other hand, with for, it can be as simple as:
List<String> patterns = new ArrayList();
for (char i=0, c=0; i < ptrnStr.length(); i++) {
c = ptrnStr.charAt(i);
patternList.add(
c != '\\'
? String.valueOf(c)
: String.valueOf(new char[] { c, ptrnStr.charAt(++i) })
);
}
EDIT:
Condensed code and added for example.

Recursively finding indexes of an INT in an ever-shrinking LispList

2nd-year Computer-Science student here, and as part of a set of exercises on recursion, we've been given some arbitrary problems to solve with LispLists. I'm stuck half-way through, so if anyone can point me in the right direction without explicitly giving me the answer, that would be great.
I need to find the positions of every instance of intToFind in the LispList listToCheck - the only conditions are that:
no additional arguments can be used
it has to be done recursively
For everyone who hasn't encountered LispLists - they don't have indexing, and the only methods you can call on them are:
.isEmpty() returns boolean
.head() returns the element at the 0th position
.tail() returns a LispList of all elements that aren't the head
.cons(value) adds value to the 'head' position - shifting everything else one down
There's also one method I wrote previously called:
recursiveCountLength(list) returns an int of the length of the passed LispList.
The list I've been testing on is: [2,3,4,2,5,12,2,5], so the result I'm looking for is [0,3,6] - with that out the way, here's what I've got so far (explanation of what I'm attempting after):
public static LispList<Integer>
recursivePositions(LispList<Integer> listToCheck, int intToFind)
{
if(listToCheck.isEmpty()) return listToCheck;
else {
// go through the array in its entirety once through,
// do everything else 'on the way back up'
LispList<Integer> positions = recursivePositions(listToCheck.tail(), intToFind);
//get the current length and current head
int currentInt = listToCheck.head();
int currentLength = recursiveCountLength(listToCheck);
//if a match is found, add the current length of the list to the list
if(currentInt == intToFind) return positions.cons(currentLength);
else return positions;
}
}
My current theory is that length of the array at each encounter of the int we're looking for (in this case 2) subtracted from the original length of the list (in this case 8) will give us the indexes.
2 first happens with a length of 8 (8-8 = index of 0, so indexes now [0]),
2 next happens with a length of 5 (8-5 = index of 3, so indexes now [0, 3]),
2 lastly happens at a length of 2 (8-2 = index of 6, so indexes now [0, 3, 6]).
The only problem is that I can't figure out how to get a static '8' - which leaves me to conclude that I'm approaching this in entirely the wrong way. Does anyone have any tips for me here? Any help would be hugely appreciated.

To clarify: a LispList is just a singly-linked-list (to differentiate from a Java LinkedList, which is double-linked).
Usually, you'd use a helper that carries information into the recursive calls, such as the current position and the positions already found (the current partial result).
LispList<Integer> positions (final int item, final LispList<Integer> list) {
return positionsAux( item, list, 0, new LispList<Integer>() );
}
private LispList<Integer> positionsAux (final int item,
final LispList<Integer> list,
final int position,
final LispList<Integer> result) {
if (list.isEmpty()) {
return result.reverse();
}
if (list.head().intValue() == item) {
result = result.cons(position);
}
return positionsAux( item, list.tail(), position + 1, result );
}
If that is not allowed, you need to carry the results backwards. If you assume that the recursive call has returned the correct result for your list.tail(), you need to add 1 to each found position to get the right result for your list. Then, you cons a 0 to the result if the current element matches. This version is less efficient than the first, because you traverse the current result list for every element of the input list (so it is O(n·m) instead of O(n), where n is the length of the input list and m the length of the result list).
LispList<Integer> positions (final int item, final LispList<Integer> list) {
if (list.isEmpty()) {
return new LispList<Integer>();
}
final LispList<Integer> tailResult = positions( item, list.tail() );
final LispList<Integer> result = tailResult.addToEach( 1 );
if (list.head().intValue() == item) {
return result.cons( 0 );
} else {
return result;
}
}
Implementing reverse() for the first version and addToEach(int) for the second is left as an exercise to the reader.

You need at least 3 arguments. Pretty sure adding more arguments is okay since he allows helper methods (because otherwise you could just make your method call the helper method which has more arguments).
Simplest way recursively:
public static LispList<Integer> positions(LispList<Integer> list, Integer key, Integer position){
if (list.isEmpty()) return LispList.empty();
if (list.head().equals(key)) return positions(list.tail(), key, position+1).cons(position);
return positions(list.tail(), key, position+1);
}

Getting a sum from ArrayList<Number> using non-loop recursion

I am working on this lab assignment for school and I was wondering if anyone could give me some advice. My instructor wants us to add different number objects up from an array list and display the result. He wants us to do the addition in a non-loop recursive format. I've done a similar problem like this before with just integers, but I cannot seem to get this solution because there is more than one primitive type this time. Here is the Code I have so far:
Main:
import java.math.BigInteger;
import java.util.ArrayList;
public class TestSummation {
public static void main(String[] args) {
ArrayList<Number> aList = new ArrayList<Number>();
aList.add(new Integer(4));
aList.add(new Double(3.423));
aList.add(new Float(99.032));
aList.add(new BigInteger("32432"));
double result = Summation.sum(aList);
System.out.println(result);
}
}
Class that holds the recursive method:
import java.util.ArrayList;
public class Summation {
public static double sum(ArrayList<Number> aList) {
if (aList.size() < 1) {
return 0;
} else {
return sum(aList);
}
}
}
Right now this code throws a StackOverflowException and I was wondering if anyone had advice that would help me. I definitely know I need to add more to it, but I feel I'm on the right track with this current section of code. I'm just hitting a bad roadblock right now. Thanks in advance for any advice!

Recursion always works by cons two different cases:
the base case which is used to end the recursion
the recursive case which is applied on a specific N-th step
By thinking about your problem you can consider as your base case either a list of size 1 or a list of size 0. Let's choose the first for simplicity: the sum of all values of a list of size 1 is the only value that it is contained.
Now let's look for the recursive case: suppose that the list has length N. What we know is that the sum of all elements of a list of N elements is the N-th element added to the sum of a list containing N-1 elements (by removing the N-th element). It's pretty self explicatory as most of recursive implementations.
As you can see the recursive step reduces the size of the list so that the algorithm step-by-step reaches the base case, this is not what is happening to your code.

Since you've done something similar with integers, I gather that the problem is how to deal with the multitude of number class types. Since you want a double result and all Number objects must implement a doubleValue() method, you can use that to build your recursion. To recurse, you need to take the first element's value and add it to the (recurive) summation of the sublist that starts with the second element:
public class Summation {
public static double sum(List<Number> aList) {
final int len = aList.size();
if (len == 0) {
return 0;
}
final double val = aList.get(0).doubleValue();
if (len == 1) {
return val;
}
// help future compilers recognize tail recursion
return val + sum(aList.sublist(1, len));
}
}

Well, without giving it all away, I'll tell you that your code is throwing the exception because you're not modifying the list at all, simply passing it back into your recursive method, which is resulting in an infinite loop. The bit you'll have to add involves modifying the list before passing it into the recursive call (and, of course, doing the actual summation).