Merge Sort on Doubly Linked List

Medium Accuracy: 60.43% Submissions: 9631 Points: 4

Given Pointer/Reference to the head of a doubly linked list of N nodes, the task is to Sort the given doubly linked list using Merge Sort in both non-decreasing and non-increasing order.

Example 1:

Input:
N = 8
value[] = {7,3,5,2,6,4,1,8}
Output:
1 2 3 4 5 6 7 8
8 7 6 5 4 3 2 1
Explanation: After sorting the given
linked list in both ways, resultant
matrix will be as given in the first
two line of output, where first line
is the output for non-decreasing
order and next line is for non-
increasing order.

Example 2:

Input:
N = 5
value[] = {9,15,0,-1,0}
Output:
-1 0 0 9 15
15 9 0 0 -1
Explanation: After sorting the given
linked list in both ways, the
resultant list will be -1 0 0 9 15
in non-decreasing order and 
15 9 0 0 -1 in non-increasing order.

Your Task:
The task is to complete the function sortDoubly() which sorts the doubly linked list. The printing is done automatically by the driver code.



#User function Template for python3


'''

class Node:

def __init__(self, data):

self.data = data

self.next = None

self.prev = None


'''

#Function to sort the given doubly linked list using Merge Sort.

def split(head):

 

    slow = head

    fast = head.next

 

    # Advance 'fast' by two nodes, and advance 'slow' by single node

    while fast:

        fast = fast.next

        if fast:

            slow = slow.next

            fast = fast.next

 

    return slow

 

 

# Recursive function to merge nodes of two sorted lists together

# into a single sorted list

def merge(a, b):

 

    # Base cases

    if a is None:

        return b

 

    if b is None:

        return a

 

    # Pick either a or b, and recur

    if a.data <= b.data:

        a.next = merge(a.next, b)

        a.next.prev = a

        a.prev = None

        return a

    else:

        b.next = merge(a, b.next)

        b.next.prev = b

        b.prev = None

        return b

 

 

# Function to sort a doubly linked list using merge sort algorithm

def sortDoubly(head):

 

    # base case: 0 or 1 node

    if head is None or head.next is None:

        return head

 

    # split head into 'a' and 'b' sublists

    a = head

 

    slow = split(head)

    b = slow.next

    slow.next = None

 

    # recursively sort the sublists

    a = sortDoubly(a)

    b = sortDoubly(b)

 

    # merge the two sorted lists together

    head = merge(a, b)

    return head

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