Move last node to front in linked list
The problem revolves around manipulating a singly linked list. In particular, the task is to move the last node of the linked list to the front. A singly linked list is a linear data structure where each element (node) holds a value and a reference to the next element. The last node is the one whose "next" pointer points to NULL.
Problem Statement and Description
Given a singly linked list, we want to reorganize it such that the last node becomes the new head while maintaining the order of other nodes. For instance, if we have the linked list 1 → 2 → 3 → 4 → 5, the result should be 5 → 1 → 2 → 3 → 4.
Example
Let's take the example of the linked list: 1 → 2 → 3 → 4 → 5 → 6 → 7.

Before Moving Last Node
 Linked List: 1 → 2 → 3 → 4 → 5 → 6 → 7
 Explanation: This is how the linked list looks before moving the last node to the front.

After Moving Last Node
 Linked List: 7 → 1 → 2 → 3 → 4 → 5 → 6
 Explanation: The last node, which was 7, is now moved to the front while maintaining the order of other nodes.
Idea to Solve
 Traverse the linked list to find the secondtolast node. This can be achieved by iterating until the next node's next pointer is not NULL.
 Once the secondtolast node is found, adjust the pointers to move the last node to the front.
 Update the new head and set the last node's next pointer to NULL to break the linked list.
Pseudocode
moveLastNode(ref):
if ref.head is NULL or ref.head.next is NULL:
return
temp = ref.head
while temp is not NULL and temp.next is not NULL and temp.next.next is not NULL:
temp = temp.next
temp.next.next = ref.head
ref.head = temp.next
temp.next = NULL
Algorithm Explanation
 Check if the linked list is empty or contains only one node. In such cases, there's no need to move the last node to the front, so just return.
 Initialize a temporary pointer
temp
to point to the head of the linked list.  Traverse the linked list using the
temp
pointer to find the secondtolast node. The loop condition checks fortemp
being not NULL and having at least two more nodes ahead.  After exiting the loop, the
temp
pointer will be pointing to the secondtolast node.  Adjust the pointers to move the last node to the front: Set the
next
pointer of the last node (previously the secondtolast node) to point to the head node.  Update the head of the linked list to the last node, making it the new head.
 Set the
next
pointer of the secondtolast node (now the new last node) to NULL, effectively breaking the linked list.
Code Solution

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Time Complexity
 The algorithm involves traversing the linked list to find the secondtolast node. This takes O(n) time, where n is the number of nodes in the linked list.
 The rest of the operations involve adjusting pointers, which are constanttime operations.
 Hence, the overall time complexity of the algorithm is O(n).
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