Given two linked lists, insert nodes of the second list into the first list at alternate positions of the first list.
For example, if first list is 5->7->17->13->11 and second is 12->10->2->4->6, the first list should become 5->12->7->10->17->2->13->4->11->6 and second list should become empty. The nodes of the second list should only be inserted when there are positions available. For example, if the first list is 1->2->3 and the second list is 4->5->6->7->8, then the first list should become 1->4->2->5->3->6 and the second list to 7->8.
Use of extra space is not allowed (Not allowed to create additional nodes), i.e., insertion must be done in-place. The expected time complexity is O(n) where n is the number of nodes in the first list.
The idea is to run a loop while there are available positions in first loop and insert nodes of second list by changing pointers. Following are implementations of this approach.
Python3
# Python program to merge a linked list # into another at alternate positions class Node(object): def __init__(self, data:int): self.data = data self.next = None class LinkedList(object): def __init__(self): self.head = None def push(self, new_data:int): new_node = Node(new_data) new_node.next = self.head # 4. Move the head to point to # new Node self.head = new_node # Function to print linked list from # the Head def printList(self): temp = self.head while temp != None: print(temp.data) temp = temp.next # Main function that inserts nodes of linked # list q into p at alternate positions. # Since head of first list never changes # but head of second list/ may change, # we need single pointer for first list and # double pointer for second list. def merge(self, p, q): p_curr = p.head q_curr = q.head # swap their positions until one # finishes off while p_curr != None and q_curr != None: # Save next pointers p_next = p_curr.next q_next = q_curr.next # make q_curr as next of p_curr # change next pointer of q_curr q_curr.next = p_next # change next pointer of p_curr p_curr.next = q_curr # update current pointers for next # iteration p_curr = p_next q_curr = q_next q.head = q_curr # Driver code llist1 = LinkedList() llist2 = LinkedList() # Creating Linked lists # 1. llist1.push(3) llist1.push(2) llist1.push(1) llist1.push(0) # 2. for i in range(8, 3, -1): llist2.push(i) print("First Linked List:") llist1.printList() print("Second Linked List:") llist2.printList() # Merging the LLs llist1.merge(p=llist1, q=llist2) print("Modified first linked list:") llist1.printList() print("Modified second linked list:") llist2.printList() # This code is contributed by Deepanshu Mehta |
Output:
First Linked List: 1 2 3 Second Linked List: 4 5 6 7 8 Modified First Linked List: 1 4 2 5 3 6 Modified Second Linked List: 7 8
Time Complexity: O(N)
Auxiliary Space: O(1)
Please refer complete article on Merge a linked list into another linked list at alternate positions for more details!
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