Similar to this: Is there any way to do n-level nested loops in Java?
I want to create a recursive function, which generates N nested loops, where the indicies depend on the depth of the loop. So basically, I want to do this recursively:
// N = 3, so we want three nested loops
for(int i1 = 0; i1 < max; i1++){
for(int i2 = i1+1; i2 < max; i2++){
for(int i3 = i2+1; i3 < max; i3++){
int value1 = getValue(i1);
int value2 = getValue(i2);
int value3 = getValue(i3);
doSomethingWithTheValues( ... );
}
}
}
I've looked at the answers in the other question, and tried to modify the answer (by oel.neely), but without luck. My guess is that it only needs a small modification, but right now, I'm just confusing myself!
Its C#, but should be easily convertable to Java:
class ImmutableStack<T>
{
public readonly T Head;
public readonly ImmutableStack<T> Tail;
public ImmutableStack(T head, ImmutableStack<T> tail)
{
this.Head = head;
this.Tail = tail;
}
public static ImmutableStack<T> Cons(T head, ImmutableStack<T> tail)
{
return new ImmutableStack<T>(head, tail);
}
public static ImmutableStack<T> Reverse(ImmutableStack<T> s)
{
ImmutableStack<T> res = null;
while (s != null)
{
res = Cons(s.Head, res);
s = s.Tail;
}
return res;
}
}
class Program
{
static void AwesomeRecursion(int toDepth, int start, int max, ImmutableStack<int> indices)
{
if (toDepth < 0)
{
throw new ArgumentException("toDepth should be >= 0");
}
else if (toDepth == 0)
{
Console.Write("indices: ");
indices = ImmutableStack<int>.Reverse(indices);
while (indices != null)
{
Console.Write("{0}, ", indices.Head);
indices = indices.Tail;
}
Console.WriteLine();
}
else
{
for (int i = start; i < max; i++)
{
AwesomeRecursion(toDepth - 1, i + 1, max, ImmutableStack<int>.Cons(i, indices));
}
}
}
static void Main(string[] args)
{
AwesomeRecursion(4, 1, 10, null);
Console.WriteLine("Done");
Console.ReadKey(true);
}
}
We keep the indices on an immutable stack since it makes backtracking so much easier than mutable stacks or queues.
Related
i am a cse student who takes data structures course. Trying to implement binary search algorithm to my SinglyLinkedList class, somehow i've failed. Could you check it what's wrong please ?
The related method;
I've debugged and it just enters the loops this side: else if(temp.getElement() > target)
public int binarySearchLinkedList(SinglyLinkedList<E> list, E target) {
int left = 0;
int right = list.getSize();
while (left <= right) {
int mid = (left + right) / 2;
Node<E> temp = head;
for (int i = 0; i < mid - 1; i++) {
temp = temp.next;
}
if (temp.getElement() instanceof Number && target instanceof Number) {
if (Integer.parseInt(temp.getElement().toString()) == Integer.parseInt(target.toString())) {
return mid;
} else if (Integer.parseInt(temp.getElement().toString()) > Integer.parseInt(target.toString())) {
left = mid + 1;
} else {
right = mid - 1;
}
}
}
All class for better understanding;
public class SinglyLinkedList<E> {
private static class Node<E> {
private E element;
private Node<E> next;
public Node(E e, Node<E> n) {
element = e;
next = n;
}
private E getElement() {
return element;
}
private Node<E> getNext() {
return next;
}
private void setNext(Node<E> n) {
next = n;
}
}
private Node<E> head;
private Node<E> tail;
private int size;
public SinglyLinkedList() {
};
public int getSize() {
return size;
}
public void append(E e) {
if (head == null) {
head = new Node<E>(e, null);
tail = head;
size++;
return;
}
Node<E> temp = head;
while (temp != tail) {
temp = temp.next;
}
temp.setNext(tail = new Node<E>(e, null));
size++;
return;
}
public int binarySearchLinkedList(SinglyLinkedList<E> list, E target) {
int left = 0;
int right = list.getSize();
while (left <= right) {
int mid = (left + right) / 2;
Node<E> temp = head;
for (int i = 0; i < mid - 1; i++) {
temp = temp.next;
}
if (temp.getElement() instanceof Number && target instanceof Number) {
if (Integer.parseInt(temp.getElement().toString()) == Integer.parseInt(target.toString())) {
return mid;
} else if (Integer.parseInt(temp.getElement().toString()) > Integer.parseInt(target.toString())) {
left = mid + 1;
} else {
right = mid - 1;
}
}
}
return -1;
}
public String toString() {
StringBuilder sb = new StringBuilder();
Node<E> temp = head;
while (temp != tail) {
sb.append(temp.getElement()).append(", ");
temp = temp.next;
if (temp == tail) {
sb.append(temp.getElement());
}
}
return sb.toString();
}
}
And the main method;
public static void main(String[] args) {
SinglyLinkedList<Integer> list = new SinglyLinkedList<>();
list.append(10);
list.append(20);
list.append(30);
list.append(40);
list.append(50);
list.append(60);
list.append(70);
list.append(80);
list.append(90);
list.append(100);
System.out.println(list);
System.out.println(list.binarySearchLinkedList(list, 30));
}
It returns;
10, 20, 30, 40, 50, 60, 70, 80, 90, 100
-1
I'm guessing this line is the biggest issue:
for (int i = 0; i < mid - 1; i++) {
Consider what happens if mid is 1. The loop won't execute, because it is not the case that 0 < 1-1. So the inspected node won't be the one you think it is. The actual node at index mid will never be inspected. So the outer loop will eventually exit, never having found the target. Presumably your method ends with return -1;.
Other issues include:
You're initializing right to an exclusive value, but treat it as inclusive elsewhere. Consider using int right = list.getSize() - 1;
You've declared a generic method, but implemented it only for Integer. One way around this would be to limit the generic type to one that supports Comparable - e.g., E extends Comparable<E>.
You're implementing a binary search in a linked list. In a linked list, linear search would be simpler and no less efficient. Or you could use a data structure that supports constant-time access by index, such as an array or ArrayList.
Current I have a home work question which says,
It is possible to make the heap sort algorithm more efficient by
writing a method that will order the entire list at once instead of
adding the elements one at a time.
However I can't figure out what exactly it means by "instead of adding elements one at a time", surely one has to building a heap first (which involves adding element from a unsorted list one by one), then remove the largest from the heap one at a time.
Here is my heap array:
import exceptions.exceptions.*;
public class ArrayHeap<T> extends ArrayBinaryTree<T> implements HeapADT<T> {
public ArrayHeap(){
super();
}
public void addElement (T element){
if (count==size())
expandCapacity();
tree[count] = element;
count++;
if (count > 1)
heapifyAdd();
}
private void heapifyAdd(){
int index = count - 1;
while ((index != 0) && (((Comparable)tree[index]).compareTo(tree[(index-1)/2]) < 0))
{
T temp = tree[index];
tree[index] = tree[(index-1)/2];
tree[(index-1)/2] = temp;
index = (index-1)/2;
}
}
public T removeMin(){
if (isEmpty())
throw new EmptyCollectionException ("Empty Heap");
T minElement = findMin();
tree[0] = tree[count-1];
heapifyRemove();
count--;
return minElement;
}
private void heapifyRemove()
{
T temp;
int node = 0;
int left = 1;
int right = 2;
int next;
if ((tree[left] == null) && (tree[right] == null))
next = count;
else if (tree[left] == null)
next = right;
else if (tree[right] == null)
next = left;
else if (((Comparable)tree[left]).compareTo(tree[right]) < 0)
next = left;
else
next = right;
while ((next < count) && (((Comparable)tree[next]).compareTo(tree[node]) < 0)){
temp = tree[node];
tree[node] = tree[next];
tree[next] = temp;
node = next;
left = 2*node + 1;
right = 2*(node+1);
if ((tree[left] == null) && (tree[right] == null))
next = count;
else if (tree[left] == null)
next = right;
else if (tree[right] == null)
next = left;
else if (((Comparable)tree[left]).compareTo(tree[right]) < 0)
next = left;
else
next = right;
}
}
public T findMin() {
if (isEmpty())
throw new EmptyCollectionException ("Empty Heap");
return tree[0];
}
}
Here is more HeapSort algorithm:
import ArrayHeap;
public class HeapSort<T>{
public T[] heapsort(T[] data, int min, int max){
ArrayHeap<T> temp = new ArrayHeap<T>();
for (int c = min; c <= max; c++){
temp.addElement(data[c]);
}
int count = min;
while(!(temp.isEmpty())){
T jj = temp.removeMin();
data[count] = jj;
count ++;
}
return data;
}
The most straight-forward way to perform heapsort is to use a separate heap and add all the elements to it, then the elements will be in order when we pop them out one by one. This is what "adding the elements one at a time" refers to in the statement, and this is what your implementation is doing: create a heap of type ArrayHeap and insert the elements of data to it, in the end pop the elements back to data.
A more efficient way (in terms of both space and time) is to perform in-place sorting, where we use the array to be sorted as the heap, rather than using additional memory for the heap, this is what "order the entire list at once" refers to. The steps of this implementation is as follow, we will order the elements in non-decreasing order:
We max-heapify the input array (i.e. we re-arrange the elements in the array so that it follows the max-heap property.
For i = n - 1 to 1:
Swap the 0-th element in the array with the i-th element.
Decrease the size of the heap by 1 (i.e. the heap should be of size i).
Perform the sift-down operation on the heap to restore the max-heap property.
Note that whenever the max-heap property holds, the top-most element in the heap is the largest element, so at the start of the k-th iteration (k = n - i here) the 0-th element is the k-largest element, and we place is in the correct position in the array by swapping.
Note that step 1 can be done in O(n), and in step 2 there are O(n) iterations and each sift-down operation takes time O(log(n)), so the overall time complexity is O(n log(n)).
Below is an implementation in Java for your reference:
import java.util.Random;
public class HeapSort {
public static void main(String[] args) {
for (int i = 1; i <= 10; i++) {
System.out.println(String.format("Iteration number %d%n", i));
Integer[] array = randomIntArray(10, 0, 100);
System.out.println(String.format("Array before sorting: [%s]", toStr(array)));
heapSort(array);
System.out.println(String.format("Array after sorting: [%s]", toStr(array)));
System.out.println("================================================================");
}
}
private static <T extends Comparable<T>> T[] heapSort(T[] array) {
maxHeapify(array, array.length);
for (int i = array.length - 1; i > 0; i--) {
swap(array, 0, i);
siftDown(array, i, 0);
}
return array;
}
private static <T extends Comparable<T>> void maxHeapify(T[] array, int heapSize) {
for (int i = getParentIdx(heapSize - 1); i >= 0; i--) {
siftDown(array, heapSize, i);
}
}
private static <T extends Comparable<T>> void siftDown(T[] array, int heapSize, int idx) {
final int length = Math.min(array.length, heapSize) - 1;
if (idx > length || idx < 0) throw new IllegalArgumentException("Index out of range");
while (true) {
int maxIdx = idx;
int leftChildIdx = getLeftChildIdx(idx);
int rightChildIdx = getRightChildIdx(idx);
if (leftChildIdx <= length && array[maxIdx].compareTo(array[leftChildIdx]) < 0) maxIdx = leftChildIdx;
if (rightChildIdx <= length && array[maxIdx].compareTo(array[rightChildIdx]) < 0) maxIdx = rightChildIdx;
if (idx != maxIdx) {
swap(array, idx, maxIdx);
idx = maxIdx;
} else {
return;
}
}
}
private static int getParentIdx(int idx) {
return (idx - 1) / 2;
}
private static int getLeftChildIdx(int idx) {
return idx * 2 + 1;
}
private static int getRightChildIdx(int idx) {
return idx * 2 + 2;
}
private static <T> void swap(T[] array, int i, int j) {
T tmp = array[i];
array[i] = array[j];
array[j] = tmp;
}
private static <T> String toStr(T[] array) {
StringBuilder sb = new StringBuilder();
for (T element : array) {
sb.append(element + ", ");
}
return sb.substring(0, sb.length() - 2);
}
private static Integer[] randomIntArray(int size, int lowerBound, int upperBound) {
Integer[] result = new Integer[size];
Random random = new Random();
int diff = upperBound - lowerBound + 1;
for (int i = 0; i < size; i++) result[i] = lowerBound + random.nextInt(diff);
return result;
}
}
This question already has answers here:
Algorithm to find next greater permutation of a given string
(14 answers)
Closed 8 years ago.
I am currently making a Permutation class for java. One of my methods for this class, is advance(), where the computer will take the array, and then display all permutations of the array.
So, for example, if I give the array {0,1,2,3,4,5}, or the number 6, it should give me from 012345.....543210.
Here is the code I have so far:
import java.util.*;
public class Permutation extends java.lang.Object {
public static int[] permutation;
public static int[] firstPerm;
public static int[] lastPerm;
public static int length;
public static int count;
public static void main(String[] args) {
// TODO Auto-generated method stub
}
public Permutation(int n) {
length = n;
permutation = new int[length];
for (int i = 0; i < length; i++) {
permutation[i] = i;
}
}
public Permutation(int[] perm) {
length = perm.length;
permutation = new int[length];
boolean[] t = new boolean[length];
for (int i = 0; i < length; i++) {
if (perm[i] < 0 || perm[i] >= length) {
throw new IllegalArgumentException("INVALID ELEMENT");
}
if (t[perm[i]]) {
throw new IllegalArgumentException("DUPLICATE VALUES");
}
t[perm[i]] = true;
permutation[i] = perm[i];
}
}
public void advance() {
}
public int getElement(int i) {
return permutation[i];
}
public boolean isFirstPerm() {
firstPerm = new int[permutation.length];
for (int i = 0; i < permutation.length; i++) {
firstPerm[i] = permutation[i];
}
Arrays.sort(firstPerm);
if (Arrays.equals(firstPerm, permutation)) {
return true;
} else {
return false;
}
}
public boolean isLastPerm() {
lastPerm = new int[firstPerm.length];
for (int i = 0; i < firstPerm.length; i++) {
lastPerm[i] = firstPerm[firstPerm.length - 1 - i];
}
if (Arrays.equals(permutation, lastPerm)) {
return true;
} else {
return false;
}
}
public static Permutation randomPermutation(int n) {
if (n <= 0) {
throw new IllegalArgumentException("INVALID NUMBER");
} else {
length = n;
permutation = new int[length];
for (int i = 0; i < length; i++) {
permutation[i] = i;
}
Collections.shuffle(Arrays.asList(permutation));
return new Permutation(permutation);
}
}
public void reset() {
Arrays.sort(permutation);
}
public boolean isValid(int[] perm) {
boolean[] t = new boolean[length];
for (int i = 0; i < length; i++) {
if (perm[i] < 0 || perm[i] >= length) {
return false;
}
if (t[perm[i]]) {
return false;
}
}
return true;
}
public int[] toArray() {
return permutation;
}
public String toString() {
StringBuffer result = new StringBuffer();
for (int i = 0; i < permutation.length; i++) {
result.append(permutation[i]);
}
String perms = result.toString();
return perms;
}
public static long totalPermutations(int n) {
count = 1;
for (int i = 1; i <= n; i++) {
count = count * i;
}
return count;
}
}
As you can see, the advance() method is the last thing I need to do, but I can't figure it out. Any help will be grand.
One of methods you can employ is:
Fix the first element and recursively find all permutations of rest of the array.
Then change the first elements by trying each of the remaining elements.
Base case for recursion is when you travel the entire length to get 0 element array. Then, either print it or add it to a List which you can return at the end.
public void advance() {
int[] temp = Arrays.copyOf(arr, arr.length);
printAll(0,temp);
}
private void printAll(int index,int[] temp) {
if(index==n) { //base case..the end of array
//print array temp here
}
else {
for(int i=index;i<n;i++) {//change the first element stepwise
swap(temp,index,i);//swap to change
printAll(index+1, temp);//call recursively
swap(temp,index,i);//swap again to backtrack
}
}
}
private void swap(int[] arr, int i, int j) {
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
The way your code looks right now, it sounds like you want to be able to control the permutation class externally, rather than only supporting the one operation of printing all the permutations in order.
Here's an example of how to calculate a permutation.
import java.io.IOException;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
public class Test {
public static int factorial(int x) {
int f = 1;
while (x > 1) {
f = f * x;
x--;
}
return f;
}
public static List<Integer> permute(List<Integer> list, int iteration) {
if (list.size() <= 1) return list;
int fact = factorial(list.size() - 1);
int first = iteration / fact;
List<Integer> copy = new ArrayList<Integer>(list);
Integer head = copy.remove(first);
int remainder = iteration % fact;
List<Integer> tail = permute(copy, remainder);
tail.add(0, head);
return tail;
}
public static void main(String[] args) throws IOException {
List<Integer> list = Arrays.asList(4, 5, 6, 7);
for (int i = 0; i < 24; i++) {
System.out.println(permute(list, i));
}
}
}
Just to elaborate, the idea behind the code is to map an integer (iteration) to a particular permutation (ordering of the list). We're treating it as a base-n representation of the permutation where each digit represents which element of the set goes in that position of the resulting permutation.
For example, if we're permuting (1, 2, 3, 4) then we know there are 4! permutations, and that "1" will be the first element in 3! of them, and it will be followed by all permutations of (2, 3, 4). Of those 3! permutations of the new set (2, 3, 4), "2" will be the first element in 2! of them, etc.
That's why we're using / and % to calculate which element goes into each position of the resulting permutation.
This should work, and it's pretty compact, only drawback is that it is recursive:
private static permutation(int x) {
if (x < 1) {
throw new IllegalArgumentException(x);
}
LinkedList<Integer> numbers = new LinkedList<>();
for (int i = 0; i < x; i++) {
numbers.add(i);
}
printPermutations(numbers, new LinkedList<>());
}
private static void printPermutations(
LinkedList<Integer> numbers, LinkedList<Integer> heads) {
int size = numbers.size();
for (int i = 0; i < size; i++) {
int n = numbers.getFirst();
numbers.removeFirst();
heads.add(n);
printPermutations(numbers, heads);
numbers.add(n);
heads.removeLast();
}
if (numbers.isEmpty()) {
String sep = "";
for (int n : heads) {
System.out.print(sep + n);
sep = " ";
}
System.out.println("");
}
}
I'm having trouble with a bubbleSort method for my very unique homework assignment.
We are supposed to use a sorting method of our choice to sort, get this, a linked list of int arrays. Not an ArrayList not just a LinkedList. It works like a linked list but the each Node contains an array of a capacity of 10 ints.
I am stuck on the sorting method. I chose bubbleSort just because it was used in the last assignment and I felt most familiar with it. Any tips for a better sorting method to try would be considered helpful as well.
Here is my code:
public void bubbleSort() {
current = head; // Start at the head ArrayNode
for (int i = 0; i < size; i++) { // iterate through each ArrayNode
currentArray = current.getArray(); // get the array in this ArrayNode
int in, out;
for (out = size-1; out > 1; out--) { // outer loop (backwards)
for (in = 0; in < out; in++) { // inner loop (forwards)
if (currentArray[in] > currentArray[in+1]) // out of order?
swap(in, in+1); // swap them!
}
}
current.setArray(currentArray);
current = current.getNext();
}
}// End bubbleSort() method
// A helper method for the bubble sort
private void swap(int one, int two) {
int temp = currentArray[one];
currentArray[one] = currentArray[two];
currentArray[two] = temp;
} // End swap() method
This is a picture example of what I am supposed to be doing.
I have found a solution with selectionsort. There are a few test values, just run it to see it.
I can provide further information if needed.
import java.util.ArrayList;
import java.util.Random;
public class ArrayedListSort {
int listsize = 5; // how many nodes
int maxValue = 99; // the highest value (0 to this)
int nodeSize = 3; // size for every node
public static void main(String[] args) {
// run non static
new ArrayedListSort().runTest();
}
/**
* Log function.
*/
public void log(Object s) {
System.out.println(s);
}
public void logNoBR(Object s) {
System.out.print(s);
}
/**
* Output of list we have.
*/
public void logMyList(ArrayList<ListNode> listNode, String name) {
log("=== LOG OUTPUT " + name + " ===");
for ( int i=0; i < listNode.size(); i++) {
logNoBR(" node <" + i + ">");
logNoBR(" (");
for (int j=0; j < listNode.get(i).getSize(); j++) {
if ( j != (listNode.get(i).getSize()-1)) // if not last add ","
logNoBR( listNode.get(i).getValueAt(j) + "," );
else
logNoBR( listNode.get(i).getValueAt(j) );
}
log(")");
}
log("=====================================\n");
}
public void runTest() {
// create example List
ArrayList<ListNode> myList = new ArrayList<ListNode>();
// fill the nodes with random values
for ( int i = 0; i < listsize; i++) {
myList.add(new ListNode(nodeSize));
for (int j=0; j < nodeSize; j++) {
int randomValue = new Random().nextInt(maxValue);
myList.get(i).addValue(randomValue);
}
}
logMyList(myList, "myList unsorted"); // to see what we have
// now lets sort it
myList = sortListNode(myList);
logMyList(myList, "myList sorted"); // what we have after sorting
}
/**
* Selectionsort
*/
public ArrayList<ListNode> sortListNode(ArrayList<ListNode> myList) {
ArrayList<ListNode> retList = new ArrayList<ListNode>();
for ( int i = 0; i < listsize; i++) {
retList.add(new ListNode(nodeSize));
}
int lastSmallest = myList.get(0).getValueAt(0);
while ( !myList.isEmpty() ) {
int lastJ=0, lastI=0;
for ( int i = 0; i < myList.size(); i++) {
for (int j=0; j < myList.get(i).getSize(); j++) {
if ( myList.get(i).getValueAt(j) <= lastSmallest ) {
lastSmallest = myList.get(i).getValueAt(j);
lastJ = j;
lastI = i;
//log("Found smallest element at <"+i+","+j+"> (" + lastSmallest + ")");
}
}
}
myList.get(lastI).removeValue(lastJ);
if ( myList.get(lastI).getSize() == 0 )
myList.remove(lastI);
// add value to new list
for ( int i = 0; i < listsize; i++) {
if ( retList.get(i).getSize() < retList.get(i).getMaxSize() ) {
retList.get(i).addValue(lastSmallest);
break;
}
}
lastSmallest = Integer.MAX_VALUE;
}
return retList;
}
public class ListNode {
private ArrayList<Integer> values = new ArrayList<Integer>();
private int maxSize;
public ListNode(int maxSize) {
this.maxSize = maxSize;
}
public ArrayList<Integer> getValues() {
return values;
}
public int getMaxSize() {
return maxSize;
}
public int getSize() {
return values.size();
}
public int getValueAt(int position) {
if ( position < values.size())
return values.get(position);
else
throw new IndexOutOfBoundsException();
}
public void addValue(int value) {
values.add(value);
}
public void removeValue(int position) {
if ( position < values.size()) {
values.remove(position);
} else
throw new IndexOutOfBoundsException();
}
}
}
Here we go. The trivial solution consist in extracting all the elements of each array node and store them in a single big array, sort that big array (using Bubble Sort, Quick Sort, Shell Sort, etc.) and finally reconstruct the linked list of arrays with the the sorted values. I am almost sure that is not exactly what are you looking for.
If you want to sort the numbers in place, I can think of the following algorithm:
As others have commented, you need a comparison function that determines if a node A goes before a node B. The following algorithm use the first idea but for each pair of nodes, e.g. A->[3, 9, 7] and B->[1, 6, 8] becomes [1, 3, 6, 7, 8, 9] and finally A->[1,3, 6] and B->[7, 8, 9]. If we apply this rearrangement for each possible pair will end up with a sorted linked list of arrays (I have no proof, though).
A = head;
while (A.hasNext()) {
arrayA = A.getArray();
B = A.getNext();
while (B.hasNext()) {
arrayB = B.getArray();
// concatenate two arrays
int[] C = new int[arrayA.size() + arrayB.size()];
int i;
for (i = 0; i < arrayA.size(); i++)
C[i] = arrayA[i];
for ( ; i < arrayA.size() + arrayB.size(); i++)
C[i] = arrayB[i-arrayA.size()];
// sort the new arrays using agains Bubble sort or any
// other method, or Arrays.sort()
Arrays.sort(C);
// now return the sorted values to the two arrays
for (i = 0; i < arrayA.size(); i++)
arrayA[i] = C[i];
for (i = 0; i < arrayB.size(); i++)
arrayB[i] = C[i+arrayA.size()];
}
}
This is kind of pseudo code because I haven't worked with linked lists in Java but I think you get the idea.
NOTE: I haven't tested the code, it may contain horrors.
How can i implement a recursive binary search in an int array using only 1 parameter in java ?
it tried but my code doesn't work. I implemented a class which its instances are objects having arrays and a count variable to detect how many elements are their in the array. any idea how can i implement the recursive binary search using only 1 parameter ?
public class LinearSortedArray {
int count;
int[] a;
public LinearSortedArray() {
count = 0;
}
public LinearSortedArray(int size) {
count = 0;
a = new int[size];
}
public static int[] copyingMethod(int startPoint, int endPoint,
LinearSortedArray arrayObj) {
int[] copyingArray = new int[endPoint - startPoint];
int j = startPoint;
for (int i = 0; i < copyingArray.length; i++) {
copyingArray[i] = arrayObj.a[j];
j++;
}
return copyingArray;
}
public int binarySearchRec(int x) {
if (count == 0) {
return -1;
}
int pivot = count / 2;
LinearSortedArray newArrayObj;
if (x > a[pivot]) {
newArrayObj = new LinearSortedArray(count - pivot);
newArrayObj.count = newArrayObj.a.length;
newArrayObj.a = copyingMethod(pivot, count, this);
for (int i = 0; i < newArrayObj.a.length; i++) {
System.out.print(newArrayObj.a[i]);
System.out.print(" ");
}
System.out.println();
return pivot + newArrayObj.binarySearchRec(x);
} else if (x < a[pivot]) {
newArrayObj = new LinearSortedArray(pivot);
newArrayObj.count = newArrayObj.a.length;
newArrayObj.a = copyingMethod(0, pivot, this);
for (int i = 0; i < newArrayObj.a.length; i++) {
System.out.print(newArrayObj.a[i]);
System.out.print(" ");
}
System.out.println();
return newArrayObj.binarySearchRec(x);
} else {
return pivot;
}
}
}
P.S.: The arrays are already sorted
Binary search really requires a range and a target value -- so if you're only passing one parameter, this has to be the target and this must encapsulate the array & range.
public class ArraySegment {
protected int[] array;
protected int boundLo;
protected int boundHi;
public class ArraySegment (int[] array) {
// entire array.
this( array, 0, array.length);
}
public class ArraySegment (int[] array, int lo, int hi) {
this.array = array;
this.boundLo = lo;
this.boundHi = hi;
}
public int binarySearch (int target) {
if (boundHi <= boundLo) {
return -1; // Empty; not found.
}
int pivot = (boundLo + boundHi) / 2;
int pivotEl = array[ pivot];
if (target == pivotEl) {
return pivot; // Found!
}
if (target < pivotEl) {
// recurse Left of pivot.
ArraySegment sub = new ArraySegment( array, boundLo, pivot);
return sub.binarySearch( target);
} else {
// recurse Right of pivot.
ArraySegment sub = new ArraySegment( array, pivot, boundHi);
return sub.binarySearch( target);
}
}
}
It's a little bit questionable what kind of result you should return -- there isn't a good answer with the question posed like this, as an "integer index" kinda defeats the purpose of the ArraySegment/ range wrapper, and returning an ArraySegment containing only the found value is also fairly useless.
PS: You really shouldn't be copying the array or it's contents, just passing round references to ranges on that array. Like java.lang.String is a range on a character array.
You could contrive a single-parameter by using the Value Object Pattern, where you pass one "wrapper" object, but the object has many fields.
For example:
class SearchParams {
int target;
int start;
int end;
SearchParams(t, s, e) {
target = t;
start = s;
end = e'
}
}
int search(SearchParams params) {
// some impl
return search(new SearchParams(params.target, a, b));
}
Technically, this is one parameter. Although it may not be in the spirit of the rules.