Two sorted linked lists can be merged. This problem is common in IT interviews, because it demonstrates the practical application of linked list data structures. The algorithm to merge two sorted linked lists would generally be based on the "merge" process used in merge sort.
One possible way to accomplish this is by using a iterative solution, as follows:
1. Create a new head node, ‘mergedList‘.
2. Keep track of the current node in the ‘mergedList‘.
3. Set two pointers (‘list1Pointer‘ and ‘list2Pointer‘) to the heads of the two lists.
4. Compare the values at ‘list1Pointer‘ and ‘list2Pointer‘.
5. Append the lesser value to ‘mergedList‘.
6. Repeat steps 4-5 until one of the pointers reaches the end of its respective list.
7. Append the remainder of the non-empty list to ‘mergedList‘.
Here’s the pseudocode for merging two linked lists:
function mergeLists(list1Pointer, list2Pointer):
create a new node, mergedList
set current to mergedList
while list1Pointer is not null and list2Pointer is not null do
if list1Pointer.value <= list2Pointer.value then
current.next = new Node(list1Pointer.value)
list1Pointer = list1Pointer.next
else
current.next = new Node(list2Pointer.value)
list2Pointer = list2Pointer.next
end if
current = current.next
end while
if list1Pointer is not null then
current.next = list1Pointer
end if
if list2Pointer is not null then
current.next = list2Pointer
end if
return mergedList.next
end function
This pseudocode assumes a singly linked list where ‘next‘ points to the next node in the list, and ‘null‘ signifies the end of the list.
The time complexity of this algorithm is linear, i.e., O(n + m), where ‘n‘ and ‘m‘ are the sizes of the two linked lists. This is because in the worst case, we need to traverse both linked lists fully to merge them. The space complexity is O(1), as we are only using a constant amount of space to store the pointers.
Of course, this is just one method of merging two sorted linked lists. There are other ways of doing this, depending on the specific requirements of the problem, such as doing it in a recursive manner instead of using an iterative approach.