Skip to content
Related Articles

Related Articles

C Program For Alternating Split Of A Given Singly Linked List- Set 1

Improve Article
Save Article
Like Article
  • Last Updated : 16 Dec, 2021

Write a function AlternatingSplit() that takes one list and divides up its nodes to make two smaller lists ‘a’ and ‘b’. The sublists should be made from alternating elements in the original list. So if the original list is 0->1->0->1->0->1 then one sublist should be 0->0->0 and the other should be 1->1->1.

Method 1(Simple): 
The simplest approach iterates over the source list and pull nodes off the source and alternately put them at the front (or beginning) of ‘a’ and b’. The only strange part is that the nodes will be in the reverse order that occurred in the source list. Method 2 inserts the node at the end by keeping track of the last node in sublists.


/* C Program to alternatively split a 
   linked list into two halves */
// Link list node 
struct Node
    int data;
    struct Node* next;
/* Pull off the front node of the 
   source and put it in dest */
void MoveNode(struct Node** destRef, 
              struct Node** sourceRef) ;
/* Given the source list, split its nodes 
   into two shorter lists. If we number the 
   elements 0, 1, 2, ... then all the even 
   elements should go in the first list, and 
   all the odd elements in the second. The 
   elements in the new lists may be in any order. */
void AlternatingSplit(struct Node* source, 
                      struct Node** aRef, 
                      struct Node** bRef) 
  /* Split the nodes of source to 
     these 'a' and 'b' lists */
  struct Node* a = NULL; 
  struct Node* b = NULL;
  struct Node* current = source;
  while (current != NULL) 
    // Move a node to list 'a' 
    MoveNode(&a, &t); 
    if (current != NULL) 
       // Move a node to list 'b' 
       MoveNode(&b, &t); 
  *aRef = a;
  *bRef = b;
/* Take the node from the front of the source, 
   and move it to the front of the dest. It is 
   an error to call this with the source list 
   Before calling MoveNode():
   source == {1, 2, 3}   
   dest == {1, 2, 3}      
   After calling MoveNode():
   source == {2, 3}      
   dest == {1, 1, 2, 3} */
void MoveNode(struct Node** destRef, 
              struct Node** sourceRef) 
  // The front source node  
  struct Node* newNode = *sourceRef; 
  assert(newNode != NULL);
  // Advance the source pointer 
  *sourceRef = newNode->next;
  // Link the old dest off the 
  // new node 
  newNode->next = *destRef; 
  // Move dest to point to the 
  // new node 
  *destRef = newNode; 
// Utility Functions
/* Function to insert a node at the 
   beginning of the linked list */
void push(struct node** head_ref, 
          int new_data)
  // Allocate node 
  struct Node* new_node =
         (struct Node*) malloc(sizeof(struct Node));
  // Put in the data  
  new_node->data = new_data;
  // Link the old list off the 
  // new node 
  new_node->next = (*head_ref);     
  // Move the head to point to 
  // the new node 
  (*head_ref) = new_node;
/* Function to print nodes in a 
   given linked list */
void printList(struct Node *node)
  while(node != NULL)
   printf("%d ", node->data);
   node = node->next;
// Driver code
int main()
  // Start with the empty list 
  struct Node* head = NULL;
  struct Node* a = NULL;
  struct Node* b = NULL;  
  /* Let us create a sorted linked list 
     to test the functions
     Created linked list will be 
     0->1->2->3->4->5 */
  push(&head, 5);
  push(&head, 4);
  push(&head, 3);
  push(&head, 2);
  push(&head, 1);                                    
  push(&head, 0);  
  Original linked List:  ");
  // Remove duplicates from linked list 
  AlternatingSplit(head, &a, &b); 
  Resultant Linked List 'a' ");
  Resultant Linked List 'b' ");
  return 0;


Original linked List: 0 1 2 3 4 5 
Resultant Linked List 'a' : 4 2 0 
Resultant Linked List 'b' : 5 3 1

Time Complexity: O(n) where n is a number of nodes in the given linked list.

Method 2(Using Dummy Nodes):
Here is an alternative approach that builds the sub-lists in the same order as the source list. The code uses temporary dummy header nodes for the ‘a’ and ‘b’ lists as they are being built. Each sublist has a “tail” pointer that points to its current last node — that way new nodes can be appended to the end of each list easily. The dummy nodes give the tail pointers something to point to initially. The dummy nodes are efficient in this case because they are temporary and allocated in the stack. Alternately, local “reference pointers” (which always point to the last pointer in the list instead of to the last node) could be used to avoid Dummy nodes. 


void AlternatingSplit(struct Node* source, 
                      struct Node** aRef, 
                      struct Node** bRef) 
  struct Node aDummy;
  // Points to the last node in 'a' 
  struct Node* aTail = &aDummy; 
  struct Node bDummy;
  // Points to the last node in 'b' 
  struct Node* bTail = &bDummy; 
  struct Node* current = source; = NULL; = NULL;
  while (current != NULL) 
    // Add at 'a' tail 
    MoveNode(&(aTail->next), &t);
    // Advance the 'a' tail 
    aTail = aTail->next; 
    if (current != NULL) 
      MoveNode(&(bTail->next), &t);
      bTail = bTail->next;
  *aRef =;
  *bRef =;

Time Complexity: O(n) where n is number of node in the given linked list.
Please refer complete article on Alternating split of a given Singly Linked List | Set 1 for more details!

My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!