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 (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.
Python
# Python program to alternatively split # a linked list into two halves # Node classclass Node: def __init__(self, data, next = None): self.data = data self.next = Noneclass LinkedList: def __init__(self): self.head = None # 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. def AlternatingSplit(self, a, b): first = self.head second = first.next while (first is not None and second is not None and first.next is not None): # Move a node to list 'a' self.MoveNode(a, first) # Move a node to list 'b' self.MoveNode(b, second) first = first.next.next if first is None: break second = first.next # Pull off the front node of the # source and put it in dest def MoveNode(self, dest, node): # Make the new node new_node = Node(node.data) if dest.head is None: dest.head = new_node else: # Link the old dest off the # new node new_node.next = dest.head # Move dest to point to the # new node dest.head = new_node # Utility Functions # Function to insert a node at # the beginning of the linked list def push(self, data): # 1 & 2 allocate the Node & # put the data new_node = Node(data) # Make the next of new Node as head new_node.next = self.head # Move the head to point to # new Node self.head = new_node # Function to print nodes in a given # linked list def printList(self): temp = self.head while temp: print temp.data, temp = temp.next print("")# Driver Codeif __name__ == "__main__": # Start with empty list llist = LinkedList() a = LinkedList() b = LinkedList() # Created linked list will be # 0->1->2->3->4->5 llist.push(5) llist.push(4) llist.push(3) llist.push(2) llist.push(1) llist.push(0) llist.AlternatingSplit(a, b) print "Original Linked List: ", llist.printList() print "Resultant Linked List 'a' : ", a.printList() print "Resultant Linked List 'b' : ", b.printList() # This code is contributed by kevalshah5 |
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 time complexity for this algorithm is O(n) as we are traversing through the linked list only once.
Please refer complete article on Alternating split of a given Singly Linked List | Set 1 for more details!
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