Given a linked list that is sorted based on absolute values. Sort the list based on actual values.
Examples:
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 it at the correct position. This is the implementation of insertion sort for linked list and the 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 the linked list.
Below is the implementation of the above idea.
ever we find an element that is out of order, we move it to the front of the linked list.
Below is the implementation of the above idea.
Python3
# Python3 program to sort a linked list, # already sorted by absolute values # Linked list Node class Node: def __init__(self, d): self.data = d self.next = None class SortList: def __init__(self): 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 llist.push(-5) llist.push(5) llist.push(4) llist.push(3) llist.push(-2) llist.push(1) llist.push(0) print("Original List :") llist.printList(llist.head) start = llist.sortedList(llist.head) print("Sorted list :") llist.printList(start) # This code is contributed by Prerna Saini |
Output:
Original list : 0 -> 1 -> -2 -> 3 -> 4 -> 5 -> -5 Sorted list : -5 -> -2 -> 0 -> 1 -> 3 -> 4 -> 5
Time Complexity: O(N)
Auxiliary Space: O(1)
Please refer complete article on Sort linked list which is already sorted on absolute values for more details!
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