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Move last element to front of a given Linked List in C++
Given a linked list, we have to move the last element to the front. Let's see an example.
Input
1 -> 2 -> 3 -> 4 -> 5 -> NULL
Output
5 -> 1 -> 2 -> 3 -> 4 -> NULL
Algorithm
Initialise the linked list.
- Return if the linked list is empty or it has single node.
Find the last node and second last nodes of the linked list.
Make the last node as new head.
Update the link of second last node.
Implementation
Following is the implementation of the above algorithm in C++
#include <bits/stdc++.h>
using namespace std;
struct Node {
int data;
struct Node* next;
};
void moveFirstNodeToEnd(struct Node** head) {
if (*head == NULL || (*head)->next == NULL) {
return;
}
struct Node* secondLastNode = *head;
struct Node* lastNode = *head;
while (lastNode->next != NULL) {
secondLastNode = lastNode;
lastNode = lastNode->next;
}
secondLastNode->next = NULL;
lastNode->next = *head;
*head = lastNode;
}
void addNewNode(struct Node** head, int new_data) {
struct Node* newNode = new Node;
newNode->data = new_data;
newNode->next = *head;
*head = newNode;
}
void printLinkedList(struct Node* node) {
while (node != NULL) {
cout << node->data << "->";
node = node->next;
}
cout << "NULL" << endl;
}
int main() {
struct Node* head = NULL;
addNewNode(&head, 1);
addNewNode(&head, 2);
addNewNode(&head, 3);
addNewNode(&head, 4);
addNewNode(&head, 5);
addNewNode(&head, 6);
addNewNode(&head, 7);
addNewNode(&head, 8);
addNewNode(&head, 9);
moveFirstNodeToEnd(&head);
printLinkedList(head);
return 0;
}Output
If you run the above code, then you will get the following result.
1->9->8->7->6->5->4->3->2->NULL
Updated on: 25-Oct-2021
676 Views
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