C++ Program For Reversing Alternate K Nodes In A Singly Linked List
Given a linked list, write a function to reverse every alternate k nodes (where k is an input to the function) in an efficient way. Give the complexity of your algorithm.
Example:
Inputs: 1->2->3->4->5->6->7->8->9->NULL and k = 3
Output: 3->2->1->4->5->6->9->8->7->NULL.
Method 1 (Process 2k nodes and recursively call for rest of the list):
This method is basically an extension of the method discussed in this post.
kAltReverse(struct node *head, int k)
1) Reverse first k nodes.
2) In the modified list head points to the kth node. So change next
of head to (k+1)th node
3) Move the current pointer to skip next k nodes.
4) Call the kAltReverse() recursively for rest of the n - 2k nodes.
5) Return new head of the list.
C++
#include <bits/stdc++.h>
using namespace std;
class Node
{
public :
int data;
Node* next;
};
Node *kAltReverse(Node *head, int k)
{
Node* current = head;
Node* next;
Node* prev = NULL;
int count = 0;
while (current != NULL && count < k)
{
next = current->next;
current->next = prev;
prev = current;
current = next;
count++;
}
if (head != NULL)
head->next = current;
count = 0;
while (count < k-1 &&
current != NULL )
{
current = current->next;
count++;
}
if (current != NULL)
current->next = kAltReverse(current->next, k);
return prev;
}
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)
{
int count = 0;
while (node != NULL)
{
cout<<node->data<< " " ;
node = node->next;
count++;
}
}
int main( void )
{
Node* head = NULL;
int i;
for (i = 20; i > 0; i--)
push(&head, i);
cout << "Given linked list " ;
printList(head);
head = kAltReverse(head, 3);
cout << "Modified Linked list " ;
printList(head);
return (0);
}
|
Output:
Given linked list
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Modified Linked list
3 2 1 4 5 6 9 8 7 10 11 12 15 14 13 16 17 18 20 19
Time Complexity: O(n)
Auxiliary Space: O(1)
Method 2 (Process k nodes and recursively call for rest of the list):
The method 1 reverses the first k node and then moves the pointer to k nodes ahead. So method 1 uses two while loops and processes 2k nodes in one recursive call.
This method processes only k nodes in a recursive call. It uses a third bool parameter b which decides whether to reverse the k elements or simply move the pointer.
_kAltReverse(struct node *head, int k, bool b)
1) If b is true, then reverse first k nodes.
2) If b is false, then move the pointer k nodes ahead.
3) Call the kAltReverse() recursively for rest of the n - k nodes and link
rest of the modified list with end of first k nodes.
4) Return new head of the list.
C++
#include <bits/stdc++.h>
using namespace std;
class node
{
public :
int data;
node* next;
};
node * _kAltReverse(node *node,
int k, bool b);
node *kAltReverse(node *head, int k)
{
return _kAltReverse(head, k, true );
}
node * _kAltReverse(node *Node, int k, bool b)
{
if (Node == NULL)
return NULL;
int count = 1;
node *prev = NULL;
node *current = Node;
node *next;
while (current != NULL && count <= k)
{
next = current->next;
if (b == true )
current->next = prev;
prev = current;
current = next;
count++;
}
if (b == true )
{
Node->next = _kAltReverse(current, k, !b);
return prev;
}
else
{
prev->next =
_kAltReverse(current, k, !b);
return Node;
}
}
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)
{
int count = 0;
while (node != NULL)
{
cout << node->data << " " ;
node = node->next;
count++;
}
}
int main( void )
{
node* head = NULL;
int i;
for (i = 20; i > 0; i--)
push(&head, i);
cout << "Given linked list " ;
printList(head);
head = kAltReverse(head, 3);
cout << "Modified Linked list " ;
printList(head);
return (0);
}
|
Output:
Given linked list
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Modified Linked list
3 2 1 4 5 6 9 8 7 10 11 12 15 14 13 16 17 18 20 19
Time Complexity: O(n)
Auxiliary Space: O(n) for call stack because it is using recursion
Please refer complete article on Reverse alternate K nodes in a Singly Linked List for more details!
Last Updated :
17 Aug, 2023
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