Rearrange a linked list such that all even and odd positioned nodes are together
Rearrange a linked list in such a way that all odd position nodes are together and all even positions node are together,
Examples:
Input: 1->2->3->4 Output: 1->3->2->4 Input: 10->22->30->43->56->70 Output: 10->30->56->22->43->70
The important thing in this question is to make sure that all below cases are handled
1) Empty linked list
2) A linked list with only one node
3) A linked list with only two nodes
4) A linked list with an odd number of nodes
5) A linked list with an even number of nodes
The below program maintains two pointers ‘odd’ and ‘even’ for current nodes at odd and even positions respectively. We also store the first node of even linked list so that we can attach the even list at the end of odd list after all odd and even nodes are connected together in two different lists.
C++
// C++ program to rearrange a linked list in such a // way that all odd positioned node are stored before // all even positioned nodes #include<bits/stdc++.h> using namespace std; // Linked List Node class Node { public : int data; Node* next; }; // A utility function to create a new node Node* newNode( int key) { Node *temp = new Node; temp->data = key; temp->next = NULL; return temp; } // Rearranges given linked list such that all even // positioned nodes are before odd positioned. // Returns new head of linked List. Node *rearrangeEvenOdd(Node *head) { // Corner case if (head == NULL) return NULL; // Initialize first nodes of even and // odd lists Node *odd = head; Node *even = head->next; // Remember the first node of even list so // that we can connect the even list at the // end of odd list. Node *evenFirst = even; while (1) { // If there are no more nodes, then connect // first node of even list to the last node // of odd list if (!odd || !even || !(even->next)) { odd->next = evenFirst; break ; } // Connecting odd nodes odd->next = even->next; odd = even->next; // If there are NO more even nodes after // current odd. if (odd->next == NULL) { even->next = NULL; odd->next = evenFirst; break ; } // Connecting even nodes even->next = odd->next; even = odd->next; } return head; } // A utility function to print a linked list void printlist(Node * node) { while (node != NULL) { cout << node->data << "->" ; node = node->next; } cout << "NULL" << endl; } // Driver code int main( void ) { Node *head = newNode(1); head->next = newNode(2); head->next->next = newNode(3); head->next->next->next = newNode(4); head->next->next->next->next = newNode(5); cout << "Given Linked List\n" ; printlist(head); head = rearrangeEvenOdd(head); cout << "Modified Linked List\n" ; printlist(head); return 0; } // This is code is contributed by rathbhupendra |
C
// C program to rearrange a linked list in such a // way that all odd positioned node are stored before // all even positioned nodes #include<bits/stdc++.h> using namespace std; // Linked List Node struct Node { int data; struct Node* next; }; // A utility function to create a new node Node* newNode( int key) { Node *temp = new Node; temp->data = key; temp->next = NULL; return temp; } // Rearranges given linked list such that all even // positioned nodes are before odd positioned. // Returns new head of linked List. Node *rearrangeEvenOdd(Node *head) { // Corner case if (head == NULL) return NULL; // Initialize first nodes of even and // odd lists Node *odd = head; Node *even = head->next; // Remember the first node of even list so // that we can connect the even list at the // end of odd list. Node *evenFirst = even; while (1) { // If there are no more nodes, then connect // first node of even list to the last node // of odd list if (!odd || !even || !(even->next)) { odd->next = evenFirst; break ; } // Connecting odd nodes odd->next = even->next; odd = even->next; // If there are NO more even nodes after // current odd. if (odd->next == NULL) { even->next = NULL; odd->next = evenFirst; break ; } // Connecting even nodes even->next = odd->next; even = odd->next; } return head; } // A utility function to print a linked list void printlist(Node * node) { while (node != NULL) { cout << node->data << "->" ; node = node->next; } cout << "NULL" << endl; } // Driver code int main( void ) { Node *head = newNode(1); head->next = newNode(2); head->next->next = newNode(3); head->next->next->next = newNode(4); head->next->next->next->next = newNode(5); cout << "Given Linked List\n" ; printlist(head); head = rearrangeEvenOdd(head); cout << "\nModified Linked List\n" ; printlist(head); return 0; } |
Java
// Java program to rearrange a linked list // in such a way that all odd positioned // node are stored before all even positioned nodes class GfG { // Linked List Node static class Node { int data; Node next; } // A utility function to create a new node static Node newNode( int key) { Node temp = new Node(); temp.data = key; temp.next = null ; return temp; } // Rearranges given linked list // such that all even positioned // nodes are before odd positioned. // Returns new head of linked List. static Node rearrangeEvenOdd(Node head) { // Corner case if (head == null ) return null ; // Initialize first nodes of even and // odd lists Node odd = head; Node even = head.next; // Remember the first node of even list so // that we can connect the even list at the // end of odd list. Node evenFirst = even; while ( 1 == 1 ) { // If there are no more nodes, // then connect first node of even // list to the last node of odd list if (odd == null || even == null || (even.next) == null ) { odd.next = evenFirst; break ; } // Connecting odd nodes odd.next = even.next; odd = even.next; // If there are NO more even nodes // after current odd. if (odd.next == null ) { even.next = null ; odd.next = evenFirst; break ; } // Connecting even nodes even.next = odd.next; even = odd.next; } return head; } // A utility function to print a linked list static void printlist(Node node) { while (node != null ) { System.out.print(node.data + "->" ); node = node.next; } System.out.println( "NULL" ) ; } // Driver code public static void main(String[] args) { Node head = newNode( 1 ); head.next = newNode( 2 ); head.next.next = newNode( 3 ); head.next.next.next = newNode( 4 ); head.next.next.next.next = newNode( 5 ); System.out.println( "Given Linked List" ); printlist(head); head = rearrangeEvenOdd(head); System.out.println( "Modified Linked List" ); printlist(head); } } // This code is contributed by Prerna saini |
Python3
# Python3 program to rearrange a linked list # in such a way that all odd positioned # node are stored before all even positioned nodes # Linked List Node class Node: def __init__( self , d): self .data = d self . next = None class LinkedList: def __init__( self ): self .head = None # A utility function to create # a new node def newNode( self , key): temp = Node(key) self . next = None return temp # Rearranges given linked list # such that all even positioned # nodes are before odd positioned. # Returns new head of linked List. def rearrangeEvenOdd( self , head): # Corner case if ( self .head = = None ): return None # Initialize first nodes of # even and odd lists odd = self .head even = self .head. next # Remember the first node of even list so # that we can connect the even list at the # end of odd list. evenFirst = even while ( 1 = = 1 ): # If there are no more nodes, # then connect first node of even # list to the last node of odd list if (odd = = None or even = = None or (even. next ) = = None ): odd. next = evenFirst break # Connecting odd nodes odd. next = even. next odd = even. next # If there are NO more even nodes # after current odd. if (odd. next = = None ): even. next = None odd. next = evenFirst break # Connecting even nodes even. next = odd. next even = odd. next return head # A utility function to print a linked list def printlist( self , node): while (node ! = None ): print (node.data, end = "") print ( "->" , end = "") node = node. next print ( "NULL" ) # Function to insert a new node # at the beginning def push( self , new_data): new_node = Node(new_data) new_node. next = self .head self .head = new_node # Driver code ll = LinkedList() ll.push( 5 ) ll.push( 4 ) ll.push( 3 ) ll.push( 2 ) ll.push( 1 ) print ( "Given Linked List" ) ll.printlist(ll.head) start = ll.rearrangeEvenOdd(ll.head) print ( "\nModified Linked List" ) ll.printlist(start) # This code is contributed by Prerna Saini |
C#
// C# program to rearrange a linked list // in such a way that all odd positioned // node are stored before all even positioned nodes using System; class GfG { // Linked List Node class Node { public int data; public Node next; } // A utility function to create a new node static Node newNode( int key) { Node temp = new Node(); temp.data = key; temp.next = null ; return temp; } // Rearranges given linked list // such that all even positioned // nodes are before odd positioned. // Returns new head of linked List. static Node rearrangeEvenOdd(Node head) { // Corner case if (head == null ) return null ; // Initialize first nodes of even and // odd lists Node odd = head; Node even = head.next; // Remember the first node of even list so // that we can connect the even list at the // end of odd list. Node evenFirst = even; while (1 == 1) { // If there are no more nodes, // then connect first node of even // list to the last node of odd list if (odd == null || even == null || (even.next) == null ) { odd.next = evenFirst; break ; } // Connecting odd nodes odd.next = even.next; odd = even.next; // If there are NO more even nodes // after current odd. if (odd.next == null ) { even.next = null ; odd.next = evenFirst; break ; } // Connecting even nodes even.next = odd.next; even = odd.next; } return head; } // A utility function to print a linked list static void printlist(Node node) { while (node != null ) { Console.Write(node.data + "->" ); node = node.next; } Console.WriteLine( "NULL" ) ; } // Driver code public static void Main() { Node head = newNode(1); head.next = newNode(2); head.next.next = newNode(3); head.next.next.next = newNode(4); head.next.next.next.next = newNode(5); Console.WriteLine( "Given Linked List" ); printlist(head); head = rearrangeEvenOdd(head); Console.WriteLine( "Modified Linked List" ); printlist(head); } } /* This code is contributed PrinciRaj1992 */ |
Javascript
<script> // Javascript program to rearrange a linked list // in such a way that all odd positioned // node are stored before all even positioned nodes // Linked List Node class Node { constructor() { this .data = 0; this .next = null ; } } // A utility function to create a new node function newNode(key) { var temp = new Node(); temp.data = key; temp.next = null ; return temp; } // Rearranges given linked list // such that all even positioned // nodes are before odd positioned. // Returns new head of linked List. function rearrangeEvenOdd(head) { // Corner case if (head == null ) return null ; // Initialize first nodes of even and // odd lists var odd = head; var even = head.next; // Remember the first node of even list so // that we can connect the even list at the // end of odd list. var evenFirst = even; while (1 == 1) { // If there are no more nodes, // then connect first node of even // list to the last node of odd list if (odd == null || even == null || (even.next) == null ) { odd.next = evenFirst; break ; } // Connecting odd nodes odd.next = even.next; odd = even.next; // If there are NO more even nodes // after current odd. if (odd.next == null ) { even.next = null ; odd.next = evenFirst; break ; } // Connecting even nodes even.next = odd.next; even = odd.next; } return head; } // A utility function to print a linked list function printlist(node) { while (node != null ) { document.write(node.data + "->" ); node = node.next; } document.write( "NULL<br/>" ); } // Driver code var head = newNode(1); head.next = newNode(2); head.next.next = newNode(3); head.next.next.next = newNode(4); head.next.next.next.next = newNode(5); document.write( "Given Linked List<br/>" ); printlist(head); head = rearrangeEvenOdd(head); document.write( "Modified Linked List<br/>" ); printlist(head); // This code contributed by gauravrajput1 </script> |
Output:
Given Linked List 1->2->3->4->5->NULL Modified Linked List 1->3->5->2->4->NULL
Time complexity: O(n) since using a loop to traverse the list, where n is the size of the linked list
Auxiliary Space: O(1)
This article is contributed by Harsh Parikh.Please see here another code contributed by Gautam Singh.If you like GeeksforGeeks and would like to contribute, you can also write an article and mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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