Python Program For Sorting A Linked List Of 0s, 1s And 2s By Changing Links

Given a linked list of 0s, 1s and 2s, sort it.
Examples:

Input: 2->1->2->1->1->2->0->1->0
Output: 0->0->1->1->1->1->2->2->2
The sorted Array is 0, 0, 1, 1, 1, 1, 2, 2, 2.

Input: 2->1->0
Output: 0->1->2
The sorted Array is 0, 1, 2

Method 1: There is a solution discussed in below post that works by changing data of nodes. 
Sort a linked list of 0s, 1s and 2s
The above solution does not work when these values have associated data with them. 
For example, these three represent three colors and different types of objects associated with the colors and sort the objects (connected with a linked list) based on colors.

Method 2: In this post, a new solution is discussed that works by changing links.
Approach: Iterate through the linked list. Maintain 3 pointers named zero, one, and two to point to current ending nodes of linked lists containing 0, 1, and 2 respectively. For every traversed node, we attach it to the end of its corresponding list. Finally, we link all three lists. To avoid many null checks, we use three dummy pointers zeroD, oneD, and twoD that work as dummy headers of three lists.

Python3

# Python3 Program to sort a linked list
# 0s, 1s or 2s by changing links
import math

# Link list node
class Node:
    def __init__(self, data):
        self.data = data
        self.next = None

#Node* newNode( data)

# Sort a linked list of 0s, 1s
# and 2s by changing pointers.
def sortList(head):
    if (head == None or
        head.next == None):
        return head

    # Create three dummy nodes to point
    # to beginning of three linked lists.
    # These dummy nodes are created to
    # avoid many None checks.
    zeroD = Node(0)
    oneD = Node(0)
    twoD = Node(0)

    # Initialize current pointers for
    # three lists and whole list.
    zero = zeroD
    one = oneD
    two = twoD

    # Traverse list
    curr = head
    while (curr):
        if (curr.data == 0):
            zero.next = curr
            zero = zero.next
            curr = curr.next
        elif(curr.data == 1):
            one.next = curr
            one = one.next
            curr = curr.next
        else:
            two.next = curr
            two = two.next
            curr = curr.next

    # Attach three lists
    zero.next = (oneD.next) if (oneD.next)
                            else (twoD.next)
    one.next = twoD.next
    two.next = None

    # Updated head
    head = zeroD.next

    # Delete dummy nodes
    return head

# Function to create and return
# a node
def newNode(data):

    # Allocating space
    newNode = Node(data)

    # Inserting the required data
    newNode.data = data
    newNode.next = None
    return newNode

# Function to print linked list
def prList(node):
    while (node != None):
        print(node.data, end = " ")
        node = node.next

# Driver Code
if __name__=='__main__':

    # Creating the list 1.2.4.5
    head = newNode(1)
    head.next = newNode(2)
    head.next.next = newNode(0)
    head.next.next.next = newNode(1)

    print("Linked List Before Sorting")
    prList(head)

    head = sortList(head)

    print("Linked List After Sorting")
    prList(head)
# This code is contributed by Srathore

Output :

Linked List Before Sorting
1  2  0  1  
Linked List After Sorting
0  1  1  2  

Complexity Analysis:

• Time Complexity: O(n) where n is a number of nodes in linked list. 
  Only one traversal of the linked list is needed.
• Auxiliary Space: O(1). 
  As no extra space is required.

Please refer complete article on Sort a linked list of 0s, 1s and 2s by changing links for more details!