Sort 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: Here 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 colours and different types of objects associated with the colours and sort the objects (connected with a linked list) based on colours.
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.
Implementation:
C++
// CPP Program to sort a linked list 0s, 1s // or 2s by changing links #include <bits/stdc++.h> /* Link list node */ struct Node { int data; struct Node* next; }; Node* newNode( int data); // Sort a linked list of 0s, 1s and 2s // by changing pointers. Node* sortList(Node* head) { if (!head || !(head->next)) return head; // Create three dummy nodes to point to beginning of // three linked lists. These dummy nodes are created to // avoid many null checks. Node* zeroD = newNode(0); Node* oneD = newNode(0); Node* twoD = newNode(0); // Initialize current pointers for three // lists and whole list. Node *zero = zeroD, *one = oneD, *two = twoD; // Traverse list Node* curr = head; while (curr) { if (curr->data == 0) { zero->next = curr; zero = zero->next; } else if (curr->data == 1) { one->next = curr; one = one->next; } else { two->next = curr; two = two->next; } curr = curr->next; } // Attach three lists zero->next = (oneD->next) ? (oneD->next) : (twoD->next); one->next = twoD->next; two->next = NULL; // Updated head head = zeroD->next; // Delete dummy nodes delete zeroD; delete oneD; delete twoD; return head; } // Function to create and return a node Node* newNode( int data) { // allocating space Node* newNode = new Node; // inserting the required data newNode->data = data; newNode->next = NULL; } /* Function to print linked list */ void printList( struct Node* node) { while (node != NULL) { printf ( "%d " , node->data); node = node->next; } printf ( "\n" ); } /* Driver program to test above function*/ int main( void ) { // Creating the list 1->2->4->5 Node* head = newNode(1); head->next = newNode(2); head->next->next = newNode(0); head->next->next->next = newNode(1); printf ( "Linked List Before Sorting\n" ); printList(head); head = sortList(head); printf ( "Linked List After Sorting\n" ); printList(head); return 0; } // This code is contributed by Aditya Kumar (adityakumar129) |
C
// C Program to sort a linked list 0s, 1s // or 2s by changing links #include <stdio.h> #include <stdlib.h> /* Link list node */ typedef struct Node { int data; struct Node* next; } Node; Node* newNode( int data); // Sort a linked list of 0s, 1s and 2s // by changing pointers. Node* sortList(Node* head) { if (!head || !(head->next)) return head; // Create three dummy nodes to point to beginning of // three linked lists. These dummy nodes are created to // avoid many null checks. Node* zeroD = newNode(0); Node* oneD = newNode(0); Node* twoD = newNode(0); // Initialize current pointers for three // lists and whole list. Node *zero = zeroD, *one = oneD, *two = twoD; // Traverse list Node* curr = head; while (curr) { if (curr->data == 0) { zero->next = curr; zero = zero->next; } else if (curr->data == 1) { one->next = curr; one = one->next; } else { two->next = curr; two = two->next; } curr = curr->next; } // Attach three lists zero->next = (oneD->next) ? (oneD->next) : (twoD->next); one->next = twoD->next; two->next = NULL; // Updated head head = zeroD->next; // Delete dummy nodes free (zeroD); free (oneD); free (twoD); return head; } // Function to create and return a node Node* newNode( int data) { // allocating space Node* newNode = (Node*) malloc ( sizeof (Node)); // inserting the required data newNode->data = data; newNode->next = NULL; } /* Function to print linked list */ void printList( struct Node* node) { while (node != NULL) { printf ( "%d " , node->data); node = node->next; } printf ( "\n" ); } /* Driver program to test above function*/ int main( void ) { // Creating the list 1->2->4->5 Node* head = newNode(1); head->next = newNode(2); head->next->next = newNode(0); head->next->next->next = newNode(1); printf ( "Linked List Before Sorting\n" ); printList(head); head = sortList(head); printf ( "Linked List After Sorting\n" ); printList(head); return 0; } // This code is contributed by Aditya Kumar (adityakumar129) |
Java
// Java Program to sort a linked list 0s, 1s // or 2s by changing links public class Sort012 { // Sort a linked list of 0s, 1s and 2s // by changing pointers. public static Node sortList(Node head) { if (head== null || head.next== null ) { return head; } // Create three dummy nodes to point to // beginning of three linked lists. These // dummy nodes are created to avoid many // null checks. Node zeroD = new Node( 0 ); Node oneD = new Node( 0 ); Node twoD = new Node( 0 ); // Initialize current pointers for three // lists and whole list. Node zero = zeroD, one = oneD, two = twoD; // Traverse list Node curr = head; while (curr!= null ) { if (curr.data == 0 ) { zero.next = curr; zero = zero.next; curr = curr.next; } else if (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!= null ) ? (oneD.next) : (twoD.next); one.next = twoD.next; two.next = null ; // Updated head head = zeroD.next; return head; } // function to create and return a node public static Node newNode( int data) { // allocating space Node newNode = new Node(data); newNode.next = null ; return newNode; } /* Function to print linked list */ public static void printList(Node node) { while (node != null ) { System.out.print(node.data+ " " ); node = node.next; } } public static void main(String args[]) { Node head = new Node( 1 ); head.next = new Node( 2 ); head.next.next = new Node( 0 ); head.next.next.next = new Node( 1 ); System.out.println(" Linked List Before Sorting"); printList(head); head = sortList(head); System.out.println(" \nLinked List After Sorting"); printList(head); } } class Node { int data; Node next; Node( int data) { this .data=data; } } //This code is contributed by Gaurav Tiwari |
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 printList(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" ) printList(head) head = sortList(head) print ( "\nLinked List After Sorting" ) printList(head) # This code is contributed by Srathore |
C#
// C# Program to sort a linked list 0s, 1s // or 2s by changing links using System; public class Sort012 { // Sort a linked list of 0s, 1s and 2s // by changing pointers. public static Node sortList(Node head) { if (head == null || head.next == null ) { return head; } // Create three dummy nodes to point to // beginning of three linked lists. These // dummy nodes are created to avoid many // null checks. Node zeroD = new Node(0); Node oneD = new Node(0); Node twoD = new Node(0); // Initialize current pointers for three // lists and whole list. Node zero = zeroD, one = oneD, two = twoD; // Traverse list Node curr = head; while (curr != null ) { if (curr.data == 0) { zero.next = curr; zero = zero.next; curr = curr.next; } else if (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 != null ) ? (oneD.next) : (twoD.next); one.next = twoD.next; two.next = null ; // Updated head head = zeroD.next; return head; } // function to create and return a node public static Node newNode( int data) { // allocating space Node newNode = new Node(data); newNode.next = null ; return newNode; } /* Function to print linked list */ public static void printList(Node node) { while (node != null ) { Console.Write(node.data + " " ); node = node.next; } } // Driver code public static void Main() { Node head = new Node(1); head.next = new Node(2); head.next.next = new Node(0); head.next.next.next = new Node(1); Console.WriteLine( "Linked List Before Sorting" ); printList(head); head = sortList(head); Console.WriteLine( "\nLinked List After Sorting" ); printList(head); } } public class Node { public int data; public Node next; public Node( int data) { this .data = data; } } // This code is contributed by PrinciRaj1992 |
Javascript
<script> // JavaScript Program to sort a // linked list 0s, 1s // or 2s by changing links class Node { constructor(val) { this .data = val; this .next = null ; } } // Sort a linked list of 0s, 1s and 2s // by changing pointers. function sortList(head) { if (head == null || head.next == null ) { return head; } // Create three dummy nodes to point to // beginning of three linked lists. These // dummy nodes are created to afunction many // null checks. var zeroD = new Node(0); var oneD = new Node(0); var twoD = new Node(0); // Initialize current pointers for three // lists and whole list. var zero = zeroD, one = oneD, two = twoD; // Traverse list var curr = head; while (curr != null ) { if (curr.data == 0) { zero.next = curr; zero = zero.next; curr = curr.next; } else if (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 != null ) ? (oneD.next) : (twoD.next); one.next = twoD.next; two.next = null ; // Updated head head = zeroD.next; return head; } // function to create and return a node function newNode(data) { // allocating space var newNode = new Node(data); newNode.next = null ; return newNode; } /* Function to print linked list */ function printList(node) { while (node != null ) { document.write(node.data + " " ); node = node.next; } } var head = new Node(1); head.next = new Node(2); head.next.next = new Node(0); head.next.next.next = new Node(1); document.write( "Linked List Before Sorting<br/>" ); printList(head); head = sortList(head); document.write( "<br/>Linked List After Sorting<br/>" ); printList(head); // This code contributed by gauravrajput1 </script> |
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.
Thanks to Musarrat_123 for suggesting above solution in a comment here.
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