C++ Program For Moving Last Element To Front Of A Given Linked List
Write a function that moves the last element to the front in a given Singly Linked List. For example, if the given Linked List is 1->2->3->4->5, then the function should change the list to 5->1->2->3->4. Algorithm: Traverse the list till the last node. Use two pointers: one to store the address of the last node and the other for the address of the second last node. After the end of the loop do the following operations.
- Make second last as last (secLast->next = NULL).
- Set next of last as head (last->next = *head_ref).
- Make last as head ( *head_ref = last).
C++
#include <bits/stdc++.h>
using namespace std;
class Node
{
public :
int data;
Node *next;
};
void moveToFront(Node **head_ref)
{
if (*head_ref == NULL ||
(*head_ref)->next == NULL)
return ;
Node *secLast = NULL;
Node *last = *head_ref;
while (last->next != NULL)
{
secLast = last;
last = last->next;
}
secLast->next = NULL;
last->next = *head_ref;
*head_ref = last;
}
void push(Node** head_ref,
int new_data)
{
Node* new_node = new Node();
new_node->data = new_data;
new_node->next = (*head_ref);
(*head_ref) = new_node;
}
void printList(Node *node)
{
while (node != NULL)
{
cout << node->data << " " ;
node = node->next;
}
}
int main()
{
Node *start = NULL;
push(&start, 5);
push(&start, 4);
push(&start, 3);
push(&start, 2);
push(&start, 1);
cout <<
"Linked list before moving last to front" ;
printList(start);
moveToFront(&start);
cout <<
"Linked list after removing last to front" ;
printList(start);
return 0;
}
|
Output:
Linked list before moving last to front
1 2 3 4 5
Linked list after removing last to front
5 1 2 3 4
Time Complexity: O(n) where n is the number of nodes in the given Linked List.
Auxiliary Space: O(1) because using constant variables
Please refer complete article on Move last element to front of a given Linked List for more details!
Last Updated :
03 Aug, 2022
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