Given a linked list which is sorted based on absolute values. Sort the list based on actual values.
Input : 1 -> -10 output: -10 -> 1 Input : 1 -> -2 -> -3 -> 4 -> -5 output: -5 -> -3 -> -2 -> 1 -> 4 Input : -5 -> -10 Output: -10 -> -5 Input : 5 -> 10 output: 5 -> 10
Source : Amazon Interview
A simple solution is to traverse the linked list from beginning to end. For every visited node, check if it is out of order. If it is, remove it from its current position and insert at correct position. This is implementation of insertion sort for linked list and time complexity of this solution is O(n*n).
A better solution is to sort the linked list using merge sort. Time complexity of this solution is O(n Log n).
An efficient solution can work in O(n) time. An important observation is, all negative elements are present in reverse order. So we traverse the list, whenever we find an element that is out of order, we move it to the front of linked list.
Below is the implementation of above idea.
# Python3 program to sort a linked list,
# already sorted by absolute values
# Linked list Node
def __init__(self, d):
self.data = d
self.next = None
self.head = None
# To sort a linked list by actual values.
# The list is assumed to be sorted by
# absolute values.
def sortedList(self, head):
# Initialize previous and
# current nodes
prev = self.head
curr = self.head.next
# Traverse list
while(curr != None):
# If curr is smaller than prev,
# then it must be moved to head
if(curr.data < prev.data): # Detach curr from linked list prev.next = curr.next # Move current node to beginning curr.next = self.head self.head = curr # Update current curr = prev # Nothing to do if current element # is at right place else: prev = curr # Move current curr = curr.next return self.head # Inserts a new Node at front of the list def push(self, new_data): # 1 & 2: Allocate the Node & # Put in the data new_node = Node(new_data) # 3. Make next of new Node as head new_node.next = self.head # 4. Move the head to point to new Node self.head = new_node # Function to print linked list def printList(self, head): temp = head while (temp != None): print(temp.data, end = " ") temp = temp.next print() # Driver Code llist = SortList() # Constructed Linked List is # 1->2->3->4->5->6->7->8->8->9->null
print(“Original List :”)
start = llist.sortedList(llist.head)
print(“Sorted list :”)
# This code is contributed by
# Prerna Saini
Original list : 0 -> 1 -> -2 -> 3 -> 4 -> 5 -> -5 Sorted list : -5 -> -2 -> 0 -> 1 -> 3 -> 4 -> 5
This article is contributed by Rahul Titare. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
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