Name 1.a. Given a linked list, write a Simple (Java, etc) Function to print the alternate nodes of linked list. Be very Brief. Examples: Input: Output: 1 -> 8 -> 3 -> 10 -> 17 -> 22 -> 29 -> 42 1 -> 3 -> 17 -> 29 Alternate nodes: 1 -> 3 -> 17 -> 29 Input: Output: 10 -> 17 -> 33 -> 38 -> 73 10 -> 33 -> 73 Alternate nodes: 10 -> 33 -> 73
Added by Angela C.
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First, we need to define a class for the linked list node. Each node will have a value and a next pointer. ```python class Node: def __init__(self, value): self.value = value self.next = None ``` Show more…
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1. initialize three pointers: prev to None, curr to the head of the linked list, and next to None. 2. Iterate through the linked list using a loop. 3. Inside the loop, before changing the next pointer of curr, store the next node in the next variable. next_node = curr.next 4. Update the next pointer of curr to point to the previous node, i.e., curr.next = prev 5. Update the prev pointer to the current node and curr to the next node using the next variable. _______________ _______________
Akash M.
Starting from node "A", apply depth-first search traversing all remaining nodes. If multiple alternative nodes are available at any step, your choice should be based on alphabetical order. In your solution, provide: a) The output list showing the order in which every node is visited. b) A tree search that indicates the resulting hierarchy from depth-first search.
Aarya B.
Q2: Link to List Write a function "link_to_list" that takes in a linked list and returns the sequence as a Python list. You may assume that the input list is shallow; none of the elements are another linked list. Try to find both an iterative and recursive solution for this problem! def link_to_list(link): # code to convert linked list to list pass link = Link(1, Link(2, Link(3, Link(4)))) print(link_to_list(link)) # [1, 2, 3, 4] empty_link = Link.empty print(link_to_list(empty_link)) # []
Michael F.
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