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

Namachila October 14, 2025

0 0 2 minutes read

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

Input: 1 -> 1 -> 2 -> 0 -> 2 -> 0 -> 1 -> NULL
Output: 0 -> 0 -> 1 -> 1 -> 1 -> 2 -> 2 -> NULL

Input: 1 -> 1 -> 2 -> 1 -> 0 -> NULL 
Output: 0 -> 1 -> 1 -> 1 -> 2 -> NULL

Source: Microsoft Interview | Set 1

Following steps can be used to sort the given linked list.

Traverse the list and count the number of 0s, 1s, and 2s. Let the counts be n1, n2, and n3 respectively.
Traverse the list again, fill the first n1 nodes with 0, then n2 nodes with 1, and finally n3 nodes with 2.

Below image is a dry run of the above approach:

Below is the implementation of the above approach:

Python

# Python program to sort a linked list  
# of 0, 1 and 2 
class LinkedList(object): 
    def __init__(self): 
  
         # Head of list 
         self.head = None
  
    # Linked list Node 
    class Node(object): 
        def __init__(self, d): 
            self.data = d 
            self.next = None
  
    def sortList(self): 
  
        # Initialise count of 0 1 
        # and 2 as 0 
        count = [0, 0, 0] 
  
        ptr = self.head 
  
        # Count total number of '0', '1'  
        # and '2' 
        # count[0] will store total number  
        # of '0's 
        # count[1] will store total number  
        # of '1's 
        # count[2] will store total number    
        # of '2's   
        while ptr != None: 
            count[ptr.data] += 1
            ptr = ptr.next
  
        i = 0
        ptr = self.head 
  
        # Let say count[0] = n1, count[1] = n2  
        # and count[2] = n3 
        # now start traversing list from head node, 
        # 1) fill the list with 0, till n1 > 0 
        # 2) fill the list with 1, till n2 > 0 
        # 3) fill the list with 2, till n3 > 0   
        while ptr != None: 
            if count[i] == 0: 
                i+=1
            else: 
                ptr.data = i 
                count[i] -= 1
                ptr = ptr.next
  
  
    # Utility functions 
    # 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 = self.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): 
        temp = self.head 
        while temp != None: 
            print str(temp.data), 
            temp = temp.next
        print '' 
  
# Driver code 
llist = LinkedList() 
llist.push(0) 
llist.push(1) 
llist.push(0) 
llist.push(2) 
llist.push(1) 
llist.push(1) 
llist.push(2) 
llist.push(1) 
llist.push(2) 
  
print "Linked List before sorting"
llist.printList() 
  
llist.sortList() 
  
print "Linked List after sorting"
llist.printList() 
# This code is contributed by BHAVYA JAIN 

Output:

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

Time Complexity: O(n) where n is the number of nodes in the linked list.
Auxiliary Space: O(1)

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

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