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 (Using Dummy Nodes):
Here is an 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.
Java
// Java program to implement// the above approachstatic void AlternatingSplit(Node source, Node aRef, Node bRef) { Node aDummy = new Node(); // Points to the last node in 'a' Node aTail = aDummy; Node bDummy = new Node(); // 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), current); // Advance the 'a' tail aTail = aTail.next; if (current != null) { MoveNode((bTail.next), current); bTail = bTail.next; } } aRef = aDummy.next; bRef = bDummy.next;}// This code is contributed by rutvik_56 |
Time Complexity: O(n) where n is number of node in the given linked list.
Space Complexity: O(1),No extra space is used other than the three references to the nodes.
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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