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

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``#include``#include` `// 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``   ``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(``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); ` `  ``printf``("``  ``Original linked List:  ");``  ``printList(head);` `  ``// Remove duplicates from linked list``  ``AlternatingSplit(head, &a, &b);` `  ``printf``("``  ``Resultant Linked List ``'a'` `");``  ``printList(a);           ` `  ``printf``("``  ``Resultant Linked List ``'b'` `");``  ``printList(b);           ` `  ``getchar``();``  ``return` `0;``}`

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.

Space Complexity: O(1),The algorithm only uses two pointers, thus it has a constant space complexity of O(1).

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(``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;``  ``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;``}`

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

Space Complexity: O(1) ,we are not using any extra space.
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!

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