Integraconnect Company ONLINE TEST QUESTION

Find Length of a Linked List (Iterative and Recursive)
Write a C function to count number of nodes in a given singly linked list.

Iterative Solution

1) Initialize count as 0
2) Initialize a node pointer, current = head.
3) Do following while current is not NULL
a) current = current -> next
b) count++;
4) Return count

// Java program to count number of nodes in a linked list

/* Linked list Node*/
class Node
{
int data;
Node next;
Node(int d) { data = d; next = null; }
}

// Linked List class
class LinkedList
{
Node head; // head of list

/* Inserts a new Node at front of the list. */
public void push(int new_data)
{
/* 1 & 2: Allocate the Node &
Put in the data*/
Node new_node = new Node(new_data);

/* 3. Make next of new Node as head */
new_node.next = head;

/* 4. Move the head to point to new Node */
head = new_node;
}

/* Returns count of nodes in linked list */
public int getCount()
{
Node temp = head;
int count = 0;
while (temp != null)
{
count++;
temp = temp.next;
}
return count;
}

/* Drier program to test above functions. Ideally
this function should be in a separate user class.
It is kept here to keep code compact */
public static void main(String[] args)
{
/* Start with the empty list */
LinkedList llist = new LinkedList();
llist.push(1);
llist.push(3);
llist.push(1);
llist.push(2);
llist.push(1);

System.out.println(“Count of nodes is ” +
llist.getCount());
}
}
Run on IDE

Output:
count of nodes is 5

Recursive Solution

int getCount(head)
1) If head is NULL, return 0.
2) Else return 1 + getCount(head->next)

Find Length of a Linked List (Iterative and Recursive)
Write a C function to count number of nodes in a given singly linked list

Iterative Solution

1) Initialize count as 0
2) Initialize a node pointer, current = head.
3) Do following while current is not NULL
a) current = current -> next
b) count++;
4) Return count

// Java program to count number of nodes in a linked list

/* Linked list Node*/
class Node
{
int data;
Node next;
Node(int d) { data = d; next = null; }
}

// Linked List class
class LinkedList
{
Node head; // head of list

/* Inserts a new Node at front of the list. */
public void push(int new_data)
{
/* 1 & 2: Allocate the Node &
Put in the data*/
Node new_node = new Node(new_data);

/* 3. Make next of new Node as head */
new_node.next = head;

/* 4. Move the head to point to new Node */
head = new_node;
}

/* Returns count of nodes in linked list */
public int getCount()
{
Node temp = head;
int count = 0;
while (temp != null)
{
count++;
temp = temp.next;
}
return count;
}

/* Drier program to test above functions. Ideally
this function should be in a separate user class.
It is kept here to keep code compact */
public static void main(String[] args)
{
/* Start with the empty list */
LinkedList llist = new LinkedList();
llist.push(1);
llist.push(3);
llist.push(1);
llist.push(2);
llist.push(1);

System.out.println(“Count of nodes is ” +
llist.getCount());
}
}

Merge Sort for Linked Lists
Merge sort is often preferred for sorting a linked list. The slow random-access performance of a linked list makes some other algorithms (such as quicksort) perform poorly, and others (such as heapsort) completely impossible.

Let head be the first node of the linked list to be sorted and headRef be the pointer to head. Note that we need a reference to head in MergeSort() as the below implementation changes next links to sort the linked lists (not data at the nodes), so head node has to be changed if the data at original head is not the smallest value in linked list.

MergeSort(headRef)
1) If head is NULL or there is only one element in the Linked List
then return.
2) Else divide the linked list into two halves.
FrontBackSplit(head, &a, &b); /* a and b are two halves */
3) Sort the two halves a and b.
MergeSort(a);
MergeSort(b);
4) Merge the sorted a and b (using SortedMerge() discussed here)
and update the head pointer using headRef.
*headRef = SortedMerge(a, b);

// Java program to illustrate merge sorted
// of linkedList

public class linkedList
{
node head = null;
// node a,b;
static class node
{
int val;
node next;

public node(int val)
{
this.val = val;
}
}

node sortedMerge(node a, node b)
{
node result = null;
/* Base cases */
if (a == null)
return b;
if (b == null)
return a;

/* Pick either a or b, and recur */
if (a.val <= b.val)
{
result = a;
result.next = sortedMerge(a.next, b);
}
else
{
result = b;
result.next = sortedMerge(a, b.next);
}
return result;

}

node mergeSort(node h)
{
// Base case : if head is null
if (h == null || h.next == null)
{
return h;
}

// get the middle of the list
node middle = getMiddle(h);
node nextofmiddle = middle.next;

// set the next of middle node to null
middle.next = null;

// Apply mergeSort on left list
node left = mergeSort(h);

// Apply mergeSort on right list
node right = mergeSort(nextofmiddle);

// Merge the left and right lists
node sortedlist = sortedMerge(left, right);
return sortedlist;
}

// Utility function to get the middle of the linked list
node getMiddle(node h)
{
//Base case
if (h == null)
return h;
node fastptr = h.next;
node slowptr = h;

// Move fastptr by two and slow ptr by one
// Finally slowptr will point to middle node
while (fastptr != null)
{
fastptr = fastptr.next;
if(fastptr!=null)
{
slowptr = slowptr.next;
fastptr=fastptr.next;
}
}
return slowptr;
}

void push(int new_data)
{
/* allocate node */
node new_node = new node(new_data);

/* link the old list off the new node */
new_node.next = head;

/* move the head to point to the new node */
head = new_node;
}

// Utility function to print the linked list
void printList(node headref)
{
while (headref != null)
{
System.out.print(headref.val + ” “);
headref = headref.next;
}
}

public static void main(String[] args)
{

linkedList li = new linkedList();
/*
* Let us create a unsorted linked lists to test the functions Created
* lists shall be a: 2->3->20->5->10->15
*/
li.push(15);
li.push(10);
li.push(5);
li.push(20);
li.push(3);
li.push(2);
System.out.println(“Linked List without sorting is :”);
li.printList(li.head);

// Apply merge Sort
li.head = li.mergeSort(li.head);
System.out.print(“\n Sorted Linked List is: \n”);
li.printList(li.head);
}
}

Given only a pointer to a node to be deleted in a singly linked list, how do you delete it?
A simple solution is to traverse the linked list until you find the node you want to delete. But this solution requires pointer to the head node which contradicts the problem statement.

Fast solution is to copy the data from the next node to the node to be deleted and delete the next node. Something like following.

struct Node *temp = node_ptr->next;
node_ptr->data = temp->data;
node_ptr->next = temp->next;
free(temp);
Program:
// Java program to del the node in which only a single pointer
// is known pointing to that node

class LinkedList {

static Node head;

static class Node {

int data;
Node next;

Node(int d) {
data = d;
next = null;
}
}

void printList(Node node) {
while (node != null) {
System.out.print(node.data + ” “);
node = node.next;
}
}

void deleteNode(Node node) {
Node temp = node.next;
node.data = temp.data;
node.next = temp.next;
System.gc();

}

// Driver program to test above functions
public static void main(String[] args) {
LinkedList list = new LinkedList();
list.head = new Node(1);
list.head.next = new Node(12);
list.head.next.next = new Node(1);
list.head.next.next.next = new Node(4);
list.head.next.next.next.next = new Node(1);

System.out.println(“Before Deleting “);
list.printList(head);

/* I m deleting the head itself.
You can check for more cases */
list.deleteNode(head);
System.out.println(“”);
System.out.println(“After deleting “);
list.printList(head);
}
}

FOR MORE DETAILS CHECK THESE LINKS
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https://www.geeksforgeeks.org/commonly-asked-data-structure-interview-questions-set-1/

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