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:
- Separate two lists.
- Reverse the one with descending order
- 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!