Last Updated on March 10, 2022 by Ria Pathak
Introduction
The linked list is one of the most important concepts and data structures to learn while preparing for interviews. Having a good grasp of Linked Lists can be a huge plus point in a coding interview.
Problem Statement
According to the problem statement, our task is to sort a linked list that is sorted alternating ascending and descending orders.
Problem Statement Understanding
Let’s try to understand the problem statement with of an example by referring competitive programming online course.
We have been given a linked list which is sorted alternating ascending and descending orders. We can understand this statement with an example:
1 → 40 → 5 → 30 → 6 → 12 → 20 → NULL
- 1, 5, 6, 20 these nodes are present in the linked list in ascending order, and 40, 30, 12 are present in descending order in the linked list. These nodes are present in alternating positions in the linked list.
- Our task is to sort the linked list such that after sorting, the final sorted linked list will be:
1 → 5 → 6 → 12 → 20 → 30 → 40 → NULL
Now I think from the above example, the problem statement is clear. So let’s see how we will approach it.
A straightforward approach is to sort the linked list using the merge sort algorithm, but the only problem is that here we are not utilizing the information given to us in the problem that the linked list is already sorted in alternating ascending and descending orders.
If we are using merge sort to sort the linked list the time complexity will be O(NlogN), so now we will try to think can we utilize the information given to us in the problem and will try to solve the problem in lower time complexity.
Before jumping to the approach, try to think how you can approach this problem. If stuck, no problem, we will thoroughly see how we can approach the problem in the next section.
Let’s move on to the approach section.
Approach
We have to divide the problem into three steps:
- The first step is to split the linked list into two linked lists. One list is for nodes in ascending order and the second is for nodes in descending order.
- After this step, We have to reverse the second linked list to change its order from descending to ascending.
- After this step our both linked lists are sorted in ascending order, Now merge both lists into a single list in such fashion that the final list will also remain sorted.
Algorithm
- We have to divide this problem into three steps.
- The first step is to split the linked list into two lists.
- Initialize two pointer variables of the type node.
- One pointer for nodes of ascending order and another is for nodes in descending order.
- Now we will have two lists, one is sorted in ascending order and another list is sorted in descending order.
- Now reverse the second list. This will make the second linked list sorted in ascending order.
- Now we have two sorted linked lists.
- Merge both the linked list.
- The final linked list after merging both the linked lists will be our resultant linked list.
Code Implementation
#include <bits stdc++.h="">
using namespace std;
// Linked list node
struct Node {
int data;
struct Node* next;
};
Node* mergelist(Node* head1, Node* head2);
void splitList(Node* head, Node** Ahead, Node** Dhead);
void reverselist(Node*& head);
// the function that sorts the
// linked list
void sort(Node** head){
// Split the linked list into lists
Node *Ahead, *Dhead;
splitList(*head, &Ahead, &Dhead);
// Reverse the descending ordered linked list into ascending order
reverselist(Dhead);
// Merge both linked lists
*head = mergelist(Ahead, Dhead);
}
// A function to create a new node
Node* newNode(int data)
{
Node* temp = new Node;
temp->data = data;
temp->next = NULL;
return temp;
}
// function to reverse a linked list
void reverselist(Node*& head){
Node *prev = NULL, *curr = head, *next;
while (curr) {
next = curr->next;
curr->next = prev;
prev = curr;
curr = next;
}
head = prev;
}
// function to print a linked list
void printlist(Node* head){
while (head != NULL) {
cout << head->data << " ";
head = head->next;
}
cout << endl;
}
// A function to merge two sorted linked lists
Node* mergelist(Node* head1, Node* head2)
{
// Base cases
if (!head1)
return head2;
if (!head2)
return head1;
Node* temp = NULL;
if (head1->data < head2->data) {
temp = head1;
head1->next = mergelist(head1->next, head2);
}
else {
temp = head2;
head2->next = mergelist(head1, head2->next);
}
return temp;
}
// This function alternatively splits
// a linked list into two lists.
// "Ahead" is pointer to head of ascending linked list
// "Dhead" is pointer to head of descending linked list
void splitList(Node* head, Node** Ahead, Node** Dhead){
// Create two dummy nodes to initialize
// heads of two linked list
*Ahead = newNode(0);
*Dhead = newNode(0);
Node* ascn = *Ahead;
Node* dscn = *Dhead;
Node* curr = head;
// Link alternate nodes
while (curr) {
// Link alternate nodes of ascending linked list
ascn->next = curr;
ascn = ascn->next;
curr = curr->next;
// Link alternate nodes of descending linked list
if (curr) {
dscn->next = curr;
dscn = dscn->next;
curr = curr->next;
}
}
ascn->next = NULL;
dscn->next = NULL;
*Ahead = (*Ahead)->next;
*Dhead = (*Dhead)->next;
}
int main()
{
Node* head = newNode(1);
head->next = newNode(9);
head->next->next = newNode(5);
head->next->next->next = newNode(8);
head->next->next->next->next = newNode(7);
cout << "Input linked list" << endl;
printlist(head);
sort(&head);
cout << "Final Sorted Linked List " << endl;
printlist(head);
return 0;
}
class Sort
{
Node head;
class Node
{
int data;
Node next;
Node(int d)
{
data = d;
next = null;
}
}
Node newNode(int key)
{
return new Node(key);
}
/* This is the main function that sorts
the linked list.*/
void sort()
{
/* Create 2 dummy nodes and initialise as
heads of linked lists */
Node Ahead = new Node(0), Dhead = new Node(0);
// Split the list into lists
splitList(Ahead, Dhead);
Ahead = Ahead.next;
Dhead = Dhead.next;
// reverse the descending list
Dhead = reverseList(Dhead);
// merge the 2 linked lists
head = mergeList(Ahead, Dhead);
}
/* Function to reverse the linked list */
Node reverseList(Node Dhead)
{
Node current = Dhead;
Node prev = null;
Node next;
while (current != null) {
next = current.next;
current.next = prev;
prev = current;
current = next;
}
Dhead = prev;
return Dhead;
}
void printList()
{
Node temp = head;
while (temp != null) {
System.out.print(temp.data + " ");
temp = temp.next;
}
System.out.println();
}
// A utility function to merge two sorted linked lists
Node mergeList(Node head1, Node head2)
{
// Base cases
if (head1 == null)
return head2;
if (head2 == null)
return head1;
Node temp = null;
if (head1.data < head2.data) {
temp = head1;
head1.next = mergeList(head1.next, head2);
}
else {
temp = head2;
head2.next = mergeList(head1, head2.next);
}
return temp;
}
// This function alternatively splits
// a linked list with head as head into two:
// For example, 10->20->30->15->40->7 is
// splitted into 10->30->40
// and 20->15->7
// "Ahead" is reference to head of ascending linked list
// "Dhead" is reference to head of descending linked list
void splitList(Node Ahead, Node Dhead)
{
Node ascn = Ahead;
Node dscn = Dhead;
Node curr = head;
// Link alternate nodes
while (curr != null) {
// Link alternate nodes in ascending order
ascn.next = curr;
ascn = ascn.next;
curr = curr.next;
if (curr != null) {
dscn.next = curr;
dscn = dscn.next;
curr = curr.next;
}
}
ascn.next = null;
dscn.next = null;
}
/* Driver program to test above functions */
public static void main(String args[])
{
Sort llist = new Sort();
llist.head = llist.newNode(10);
llist.head.next = llist.newNode(40);
llist.head.next.next = llist.newNode(53);
llist.head.next.next.next = llist.newNode(30);
llist.head.next.next.next.next = llist.newNode(67);
llist.head.next.next.next.next.next = llist.newNode(12);
llist.head.next.next.next.next.next.next = llist.newNode(89);
System.out.println("Given linked list");
llist.printList();
llist.sort();
System.out.println("Sorted linked list");
llist.printList();
}
}
class LinkedList(object):
def __init__(self):
self.head = None
# Linked list Node
class Node(object):
def __init__(self, d):
self.data = d
self.next = None
def newNode(self, key):
return self.Node(key)
# This is the main function that sorts
# the linked list.
def sort(self):
# Create 2 dummy nodes and initialise as
# heads of linked lists
Ahead = self.Node(0)
Dhead = self.Node(0)
# Split the list into lists
self.splitList(Ahead, Dhead)
Ahead = Ahead.next
Dhead = Dhead.next
# reverse the descending list
Dhead = self.reverseList(Dhead)
# merge the 2 linked lists
self.head = self.mergeList(Ahead, Dhead)
# Function to reverse the linked list
def reverseList(self, Dhead):
current = Dhead
prev = None
while current != None:
self._next = current.next
current.next = prev
prev = current
current = self._next
Dhead = prev
return Dhead
# Function to print linked list
def printList(self):
temp = self.head
while temp != None:
print (temp.data,end=" ")
temp = temp.next
print()
# A utility function to merge two sorted linked lists
def mergeList(self, head1, head2):
# Base cases
if head1 == None:
return head2
if head2 == None:
return head1
temp = None
if head1.data < head2.data:
temp = head1
head1.next = self.mergeList(head1.next, head2)
else:
temp = head2
head2.next = self.mergeList(head1, head2.next)
return temp
# This function alternatively splits a linked list with head
# as head into two:
# "Ahead" is reference to head of ascending linked list
# "Dhead" is reference to head of descending linked list
def splitList(self, Ahead, Dhead):
ascn = Ahead
dscn = Dhead
curr = self.head
# Link alternate nodes
while curr != None:
# Link alternate nodes in ascending order
ascn.next = curr
ascn = ascn.next
curr = curr.next
if curr != None:
dscn.next = curr
dscn = dscn.next
curr = curr.next
ascn.next = None
dscn.next = None
llist = LinkedList()
llist.head = llist.newNode(1)
llist.head.next = llist.newNode(9)
llist.head.next.next = llist.newNode(5)
llist.head.next.next.next = llist.newNode(8)
llist.head.next.next.next.next = llist.newNode(7)
print ('Given linked list')
llist.printList()
llist.sort()
print ('Sorted linked list')
llist.printList()
Output
Input Linked list:
1 9 5 8 7
Final Sorted Linked List:
1 5 7 8 9
Time Complexity: O(n), We have traversed through the linked list only once to separate the list and reverse the list. The merging of two sorted lists into a single linked list takes O(n) time.
Auxiliary Space: O(1), No extra space is required.
So, In this blog, we have learned to sort a linked list that is sorted alternating ascending and descending orders. If you want to solve more questions on Linked List, which are curated by our expert mentors at PrepBytes, you can follow this link Linked List.