Skip to content
Related Articles

Related Articles

Improve Article
Save Article
Like Article

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

  • 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++




/* C++ Program to alternatively split
   a linked list into two halves */
#include <bits/stdc++.h>
using namespace std;
  
// Link list node 
class Node 
    public:
    int data; 
    Node* next; 
}; 
  
/* Pull off the front node of
   the source and put it in dest */
void MoveNode(Node** destRef, 
              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(Node* source, 
                      Node** aRef, 
                      Node** bRef) 
    /* Split the nodes of source 
       to these 'a' and 'b' lists */
    Node* a = NULL; 
    Node* b = NULL; 
          
    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 empty.     
   Before calling MoveNode(): 
   source == {1, 2, 3} 
   dest == {1, 2, 3} 
   After calling MoveNode(): 
   source == {2, 3}     
   dest == {1, 1, 2, 3} */
void MoveNode(Node** destRef, 
              Node** sourceRef) 
    // The front source node 
    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(Node** head_ref, 
          int new_data) 
    // Allocate node 
    Node* new_node = new 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(Node *node) 
    while(node != NULL) 
    
    cout << node->data << " "
    node = node->next; 
    
  
// Driver code
int main() 
    // Start with the empty list 
    Node* head = NULL; 
    Node* a = NULL; 
    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); 
      
    cout << "Original linked List: "
    printList(head); 
      
    // Remove duplicates from linked list 
    AlternatingSplit(head, &a, &b); 
      
    cout << "Resultant Linked List 'a' : "
    printList(a);         
      
    cout << "Resultant Linked List 'b' : "
    printList(b);         
      
    return 0; 
// This code is contributed by rathbhupendra

Output: 
 

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.

C++




void AlternatingSplit(Node* source, 
                      Node** aRef, 
                      Node** bRef) 
    Node aDummy; 
      
    // Points to the last node in 'a' 
    Node* aTail = &aDummy; 
    Node bDummy; 
      
    // Points to the last node in 'b' 
    Node* bTail = &bDummy; 
    Node* current = source; 
    aDummy.next = NULL; 
    bDummy.next = 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 = aDummy.next; 
    *bRef = bDummy.next; 
// This code is contributed by rathbhupendra

Time Complexity: O(n) where n is number of node in the given linked list.
Source: http://cslibrary.stanford.edu/105/LinkedListProblems.pdf
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!