Python Program For Sorting 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: 

Python

# 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 two dummy nodes and
        # initialize 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,
            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 code
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

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.

Please refer complete article on Sort a linked list that is sorted alternating ascending and descending orders? for more details!