Sort a linked list that is sorted alternating ascending and descending orders?

Given a Linked List. The Linked List is in alternating ascending and descending orders. Sort the list efficiently. 

Example:

Input List: 10 -> 40 -> 53 -> 30 -> 67 -> 12 -> 89 -> NULL
Output List: 10 -> 12 -> 30 -> 40 -> 53 -> 67 -> 89 -> NULL

Input List: 1 -> 4 -> 3 -> 2 -> 5 -> NULL
Output List: 1 -> 2 -> 3 -> 4 -> 5 -> NULL

Simple Solution:

Approach: The basic idea is to apply to merge sort on the linked list. 
The implementation is discussed in this article: Merge Sort for linked List.

Complexity Analysis:

• Time Complexity: The merge sort of linked list takes O(n log n) time. In the merge sort tree, the height is log n. Sorting each level will take O(n) time. So time complexity is O(n log n).
• Auxiliary Space: O(n log n), In the merge sort tree the height is log n. Storing each level will take O(n) space. So space complexity is O(n log n).

Efficient Solution:

Approach:

1. Separate two lists.
2. Reverse the one with descending order
3. Merge both lists.

Diagram:

Below are the implementations of the above algorithm: 

C++

// C++ program to sort a linked
// list that is alternatively
// sorted in increasing and decreasing order
#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);

// This is the main function that sorts the
// linked list
void sort(Node** head)
{
    // Split the list into lists
    Node *Ahead, *Dhead;
    splitList(*head, &Ahead, &Dhead);

    // Reverse the descending linked list
    reverselist(Dhead);

    // Merge the two linked lists
    *head = mergelist(Ahead, Dhead);
}

// A utility function to create a new node
Node* newNode(int key)
{
    Node* temp = new Node;
    temp->data = key;
    temp->next = NULL;
    return temp;
}

// A utility 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;
}

// A utility function to print a linked list
void printlist(Node* head)
{
    while (head != NULL) {
        cout << head->data << " ";
        head = head->next;
    }
    cout << endl;
}

// A utility 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 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* 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;
}

// Driver program to test above function
int main()
{
    Node* head = newNode(10);
    head->next = newNode(40);
    head->next->next = newNode(53);
    head->next->next->next = newNode(30);
    head->next->next->next->next = newNode(67);
    head->next->next->next->next->next = newNode(12);
    head->next->next->next->next->next->next = newNode(89);

    cout << "Given Linked List is " << endl;
    printlist(head);

    sort(&head);

    cout << "Sorted Linked List is " << endl;
    printlist(head);

    return 0;
}

Java

// Java program to sort a
// linked list that is alternatively
// sorted in increasing and decreasing order
import java.io.*;
public class LinkedList {
    Node head; // head of list

    /* Linked list Node*/
    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;
    }

    /* Function to print linked list */
    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 dummyHead = new Node(0);
        Node temp = dummyHead;
        while(head1 != null && head2 != null) {
            if (head1.data <= head2.data) {
                temp.next = head1;
                head1 = head1.next;
            } else {
                temp.next = head2;
                head2 = head2.next;
            }
            temp = temp.next;
        }

        temp.next = (head1 != null) ? head1 : head2;

        return dummyHead.next;
    }

    // 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[])
    {
        LinkedList llist = new LinkedList();
        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();
    }

} /* This code is contributed by Rajat Mishra */

Python3

# Python program to sort a linked list that is alternatively
# sorted in increasing and decreasing order
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:
    # 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
    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

# Driver program
llist = LinkedList()
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)

print ('Given linked list')
llist.printList()

llist.sort()

print ('Sorted linked list')
llist.printList()

# This code is contributed by BHAVYA JAIN

C#

// C# program to sort a linked list that is alternatively
// sorted in increasing and decreasing order
using System;
class LinkedList {
    Node head; // head of list

    /* Linked list Node*/
    public class Node {
        public int data;
        public Node next;
        public 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;
    }

    /* Function to print linked list */
    void printList()
    {
        Node temp = head;
        while (temp != null) {
            Console.Write(temp.data + " ");
            temp = temp.next;
        }
        Console.WriteLine();
    }

    // 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 code */
    public static void Main(String[] args)
    {
        LinkedList llist = new LinkedList();
        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);

        Console.WriteLine("Given linked list");
        llist.printList();

        llist.sort();

        Console.WriteLine("Sorted linked list");
        llist.printList();
    }
}

/* This code is contributed by Arnab Kundu */

Javascript

<script>
// javascript program to sort a
// linked list that is alternatively
// sorted in increasing and decreasing order

    var head; // head of list

    /* Linked list Node */
     class Node {
            constructor(val) {
                this.data = val;
                this.next = null;
            }
     }

    function newNode(key) {
        return new Node(key);
    }

    /*
     * This is the main function that sorts the linked list.
     */
    function sort() {
        /*
         * Create 2 dummy nodes and initialise as heads of linked lists
         */
        var 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 */
    function reverseList(Dhead) {
        var current = Dhead;
        var prev = null;
        var next;
        while (current != null) {
            next = current.next;
            current.next = prev;
            prev = current;
            current = next;
        }
        Dhead = prev;
        return Dhead;
    }

    /* Function to print linked list */
    function printList() {
        var temp = head;
        while (temp != null) {
            document.write(temp.data + " ");
            temp = temp.next;
        }
        document.write();
    }

    // A utility function to merge
    // two sorted linked lists
    function mergeList(head1,  head2) {
        // Base cases
        if (head1 == null)
            return head2;
        if (head2 == null)
            return head1;

        var 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
    function splitList(Ahead,  Dhead) {
        var ascn = Ahead;
        var dscn = Dhead;
        var 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 */


        head = newNode(10);
        head.next = newNode(40);
        head.next.next = newNode(53);
        head.next.next.next = newNode(30);
        head.next.next.next.next = newNode(67);
        head.next.next.next.next.next = newNode(12);
        head.next.next.next.next.next.next = newNode(89);

        document.write("Given linked list<br/>");
        printList();

        sort();

        document.write("<br/>Sorted linked list<br/>");
        printList();

// This code contributed by aashish1995
</script>

Output

Given Linked List is 
10 40 53 30 67 12 89 
Sorted Linked List is 
10 12 30 40 53 67 89 

Complexity Analysis:

• Time Complexity: O(n). 
  One traversal is needed to separate the list and reverse them. The merging of sorted lists takes O(n) time.
• Auxiliary Space: O(1). 
  No extra space is required.

Thanks to Gaurav Ahirwar for suggesting this method.