• Difficulty Level : Easy
• Last Updated : 08 Dec, 2021

Given two linked lists, insert nodes of second list into first list at alternate positions of 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 second list should only be inserted when there are positions available. For example, if the first list is 1->2->3 and second list is 4->5->6->7->8, then first list should become 1->4->2->5->3->6 and 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. Expected time complexity is O(n) where n is number of nodes in first list.

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

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.



Output:

1 2 3