Reverse 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++
// C++ program to reverse alternate // k nodes in a linked list #include <bits/stdc++.h> using namespace std; /* Link list node */class Node { public: int data; Node* next; }; /* Reverses alternate k nodes and returns the pointer to the new head node */Node *kAltReverse(Node *head, int k) { Node* current = head; Node* next; Node* prev = NULL; int count = 0; /*1) reverse first k nodes of the linked list */ while (current != NULL && count < k) { next = current->next; current->next = prev; prev = current; current = next; count++; } /* 2) Now head points to the kth node. So change next of head to (k+1)th node*/ if(head != NULL) head->next = current; /* 3) We do not want to reverse next k nodes. So move the current pointer to skip next k nodes */ count = 0; while(count < k-1 && current != NULL ) { current = current->next; count++; } /* 4) Recursively call for the list starting from current->next. And make rest of the list as next of first node */ if(current != NULL) current->next = kAltReverse(current->next, k); /* 5) prev is new head of the input list */ return prev; } /* UTILITY FUNCTIONS *//* Function to push a node */void push(Node** head_ref, int new_data) { /* allocate node */ Node* new_node = new Node(); /* put in the data */ new_node->data = new_data; /* link the old list off the new node */ new_node->next = (*head_ref); /* move the head to point to the new node */ (*head_ref) = new_node; } /* Function to print linked list */void printList(Node *node) { int count = 0; while(node != NULL) { cout<<node->data<<" "; node = node->next; count++; } } /* Driver code*/int main(void) { /* Start with the empty list */ Node* head = NULL; int i; // create a list 1->2->3->4->5...... ->20 for(i = 20; i > 0; i--) push(&head, i); cout<<"Given linked list \n"; printList(head); head = kAltReverse(head, 3); cout<<"\n Modified Linked list \n"; printList(head); return(0); } // This code is contributed by rathbhupendra |
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C
#include<stdio.h> #include<stdlib.h> /* Link list node */struct Node { int data; struct Node* next; }; /* Reverses alternate k nodes and returns the pointer to the new head node */struct Node *kAltReverse(struct Node *head, int k) { struct Node* current = head; struct Node* next; struct Node* prev = NULL; int count = 0; /*1) reverse first k nodes of the linked list */ while (current != NULL && count < k) { next = current->next; current->next = prev; prev = current; current = next; count++; } /* 2) Now head points to the kth node. So change next of head to (k+1)th node*/ if(head != NULL) head->next = current; /* 3) We do not want to reverse next k nodes. So move the current pointer to skip next k nodes */ count = 0; while(count < k-1 && current != NULL ) { current = current->next; count++; } /* 4) Recursively call for the list starting from current->next. And make rest of the list as next of first node */ if(current != NULL) current->next = kAltReverse(current->next, k); /* 5) prev is new head of the input list */ return prev; } /* UTILITY FUNCTIONS *//* Function to push a node */void push(struct Node** head_ref, int new_data) { /* allocate node */ struct Node* new_node = (struct Node*) malloc(sizeof(struct Node)); /* put in the data */ new_node->data = new_data; /* link the old list off the new node */ new_node->next = (*head_ref); /* move the head to point to the new node */ (*head_ref) = new_node; } /* Function to print linked list */void printList(struct Node *node) { int count = 0; while(node != NULL) { printf("%d ", node->data); node = node->next; count++; } } /* Driver program to test above function*/int main(void) { /* Start with the empty list */ struct Node* head = NULL; int i; // create a list 1->2->3->4->5...... ->20 for(i = 20; i > 0; i--) push(&head, i); printf("\n Given linked list \n"); printList(head); head = kAltReverse(head, 3); printf("\n Modified Linked list \n"); printList(head); getchar(); return(0); } |
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Java
// Java program to reverse alternate k nodes in a linked list class LinkedList { static Node head; class Node { int data; Node next; Node(int d) { data = d; next = null; } } /* Reverses alternate k nodes and returns the pointer to the new head node */ Node kAltReverse(Node node, int k) { Node current = node; Node next = null, prev = null; int count = 0; /*1) reverse first k nodes of the linked list */ while (current != null && count < k) { next = current.next; current.next = prev; prev = current; current = next; count++; } /* 2) Now head points to the kth node. So change next of head to (k+1)th node*/ if (node != null) { node.next = current; } /* 3) We do not want to reverse next k nodes. So move the current pointer to skip next k nodes */ count = 0; while (count < k - 1 && current != null) { current = current.next; count++; } /* 4) Recursively call for the list starting from current->next. And make rest of the list as next of first node */ if (current != null) { current.next = kAltReverse(current.next, k); } /* 5) prev is new head of the input list */ return prev; } void printList(Node node) { while (node != null) { System.out.print(node.data + " "); node = node.next; } } void push(int newdata) { Node mynode = new Node(newdata); mynode.next = head; head = mynode; } public static void main(String[] args) { LinkedList list = new LinkedList(); // Creating the linkedlist for (int i = 20; i > 0; i--) { list.push(i); } System.out.println("Given Linked List :"); list.printList(head); head = list.kAltReverse(head, 3); System.out.println(""); System.out.println("Modified Linked List :"); list.printList(head); } } // This code has been contributed by Mayank Jaiswal |
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Python3
# Python3 program to reverse alternate # k nodes in a linked list import math # Link list node class Node: def __init__(self, data): self.data = data self.next = None # Reverses alternate k nodes and #returns the poer to the new head node def kAltReverse(head, k) : current = head next = None prev = None count = 0 #1) reverse first k nodes of the linked list while (current != None and count < k) : next = current.next current.next = prev prev = current current = next count = count + 1; # 2) Now head pos to the kth node. # So change next of head to (k+1)th node if(head != None): head.next = current # 3) We do not want to reverse next k # nodes. So move the current # poer to skip next k nodes count = 0 while(count < k - 1 and current != None ): current = current.next count = count + 1 # 4) Recursively call for the list # starting from current.next. And make # rest of the list as next of first node if(current != None): current.next = kAltReverse(current.next, k) # 5) prev is new head of the input list return prev # UTILITY FUNCTIONS # Function to push a node def push(head_ref, new_data): # allocate node new_node = Node(new_data) # put in the data # new_node.data = new_data # link the old list off the new node new_node.next = head_ref # move the head to po to the new node head_ref = new_node return head_ref # Function to print linked list def prList(node): count = 0 while(node != None): print(node.data, end = " ") node = node.next count = count + 1 # Driver code if __name__=='__main__': # Start with the empty list head = None # create a list 1.2.3.4.5...... .20 for i in range(20, 0, -1): head = push(head, i) print("Given linked list ") prList(head) head = kAltReverse(head, 3) print("\nModified Linked list") prList(head) # This code is contributed by Srathore |
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C#
// C# program to reverse alternate // k nodes in a linked list using System; class LinkedList { static Node head; public class Node { public int data; public Node next; public Node(int d) { data = d; next = null; } } /* Reverses alternate k nodes and returns the pointer to the new head node */ Node kAltReverse(Node node, int k) { Node current = node; Node next = null, prev = null; int count = 0; /*1) reverse first k nodes of the linked list */ while (current != null && count < k) { next = current.next; current.next = prev; prev = current; current = next; count++; } /* 2) Now head points to the kth node. So change next of head to (k+1)th node*/ if (node != null) { node.next = current; } /* 3) We do not want to reverse next k nodes. So move the current pointer to skip next k nodes */ count = 0; while (count < k - 1 && current != null) { current = current.next; count++; } /* 4) Recursively call for the list starting from current->next. And make rest of the list as next of first node */ if (current != null) { current.next = kAltReverse(current.next, k); } /* 5) prev is new head of the input list */ return prev; } void printList(Node node) { while (node != null) { Console.Write(node.data + " "); node = node.next; } } void push(int newdata) { Node mynode = new Node(newdata); mynode.next = head; head = mynode; } // Driver code public static void Main(String []args) { LinkedList list = new LinkedList(); // Creating the linkedlist for (int i = 20; i > 0; i--) { list.push(i); } Console.WriteLine("Given Linked List :"); list.printList(head); head = list.kAltReverse(head, 3); Console.WriteLine(""); Console.WriteLine("Modified Linked List :"); list.printList(head); } } // This code has been contributed by Arnab Kundu |
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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)
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; /* Link list node */class node { public: int data; node* next; }; /* Helper function for kAltReverse() */node * _kAltReverse(node *node, int k, bool b); /* Alternatively reverses the given linked list in groups of given size k. */node *kAltReverse(node *head, int k) { return _kAltReverse(head, k, true); } /* Helper function for kAltReverse(). It reverses k nodes of the list only if the third parameter b is passed as true, otherwise moves the pointer k nodes ahead and recursively calls iteself */node * _kAltReverse(node *Node, int k, bool b) { if(Node == NULL) return NULL; int count = 1; node *prev = NULL; node *current = Node; node *next; /* The loop serves two purposes 1) If b is true, then it reverses the k nodes 2) If b is false, then it moves the current pointer */ while(current != NULL && count <= k) { next = current->next; /* Reverse the nodes only if b is true*/ if(b == true) current->next = prev; prev = current; current = next; count++; } /* 3) If b is true, then node is the kth node. So attach rest of the list after node. 4) After attaching, return the new head */ if(b == true) { Node->next = _kAltReverse(current, k, !b); return prev; } /* If b is not true, then attach rest of the list after prev. So attach rest of the list after prev */ else { prev->next = _kAltReverse(current, k, !b); return Node; } } /* UTILITY FUNCTIONS *//* Function to push a node */void push(node** head_ref, int new_data) { /* allocate node */ node* new_node = new node(); /* put in the data */ new_node->data = new_data; /* link the old list off the new node */ new_node->next = (*head_ref); /* move the head to point to the new node */ (*head_ref) = new_node; } /* Function to print linked list */void printList(node *node) { int count = 0; while(node != NULL) { cout << node->data << " "; node = node->next; count++; } } // Driver Code int main(void) { /* Start with the empty list */ node* head = NULL; int i; // create a list 1->2->3->4->5...... ->20 for(i = 20; i > 0; i--) push(&head, i); cout << "Given linked list \n"; printList(head); head = kAltReverse(head, 3); cout << "\nModified Linked list \n"; printList(head); return(0); } // This code is contributed by rathbhupendra |
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C
#include<stdio.h> #include<stdlib.h> /* Link list node */struct node { int data; struct node* next; }; /* Helper function for kAltReverse() */struct node * _kAltReverse(struct node *node, int k, bool b); /* Alternatively reverses the given linked list in groups of given size k. */struct node *kAltReverse(struct node *head, int k) { return _kAltReverse(head, k, true); } /* Helper function for kAltReverse(). It reverses k nodes of the list only if the third parameter b is passed as true, otherwise moves the pointer k nodes ahead and recursively calls iteself */ struct node * _kAltReverse(struct node *node, int k, bool b) { if(node == NULL) return NULL; int count = 1; struct node *prev = NULL; struct node *current = node; struct node *next; /* The loop serves two purposes 1) If b is true, then it reverses the k nodes 2) If b is false, then it moves the current pointer */ while(current != NULL && count <= k) { next = current->next; /* Reverse the nodes only if b is true*/ if(b == true) current->next = prev; prev = current; current = next; count++; } /* 3) If b is true, then node is the kth node. So attach rest of the list after node. 4) After attaching, return the new head */ if(b == true) { node->next = _kAltReverse(current,k,!b); return prev; } /* If b is not true, then attach rest of the list after prev. So attach rest of the list after prev */ else { prev->next = _kAltReverse(current, k, !b); return node; } } /* UTILITY FUNCTIONS *//* Function to push a node */void push(struct node** head_ref, int new_data) { /* allocate node */ struct node* new_node = (struct node*) malloc(sizeof(struct node)); /* put in the data */ new_node->data = new_data; /* link the old list off the new node */ new_node->next = (*head_ref); /* move the head to point to the new node */ (*head_ref) = new_node; } /* Function to print linked list */void printList(struct node *node) { int count = 0; while(node != NULL) { printf("%d ", node->data); node = node->next; count++; } } /* Driver program to test above function*/int main(void) { /* Start with the empty list */ struct node* head = NULL; int i; // create a list 1->2->3->4->5...... ->20 for(i = 20; i > 0; i--) push(&head, i); printf("\n Given linked list \n"); printList(head); head = kAltReverse(head, 3); printf("\n Modified Linked list \n"); printList(head); getchar(); return(0); } |
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Java
// Java program to reverse alternate k nodes in a linked list class LinkedList { static Node head; class Node { int data; Node next; Node(int d) { data = d; next = null; } } /* Alternatively reverses the given linked list in groups of given size k. */ Node kAltReverse(Node head, int k) { return _kAltReverse(head, k, true); } /* Helper function for kAltReverse(). It reverses k nodes of the list only if the third parameter b is passed as true, otherwise moves the pointer k nodes ahead and recursively calls iteself */ Node _kAltReverse(Node node, int k, boolean b) { if (node == null) { return null; } int count = 1; Node prev = null; Node current = node; Node next = null; /* The loop serves two purposes 1) If b is true, then it reverses the k nodes 2) If b is false, then it moves the current pointer */ while (current != null && count <= k) { next = current.next; /* Reverse the nodes only if b is true*/ if (b == true) { current.next = prev; } prev = current; current = next; count++; } /* 3) If b is true, then node is the kth node. So attach rest of the list after node. 4) After attaching, return the new head */ if (b == true) { node.next = _kAltReverse(current, k, !b); return prev; } /* If b is not true, then attach rest of the list after prev. So attach rest of the list after prev */ else { prev.next = _kAltReverse(current, k, !b); return node; } } void printList(Node node) { while (node != null) { System.out.print(node.data + " "); node = node.next; } } void push(int newdata) { Node mynode = new Node(newdata); mynode.next = head; head = mynode; } public static void main(String[] args) { LinkedList list = new LinkedList(); // Creating the linkedlist for (int i = 20; i > 0; i--) { list.push(i); } System.out.println("Given Linked List :"); list.printList(head); head = list.kAltReverse(head, 3); System.out.println(""); System.out.println("Modified Linked List :"); list.printList(head); } } // This code has been contributed by Mayank Jaiswal |
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Python
# Python code for above algorithm # Link list node class node: def __init__(self, data): self.data = data self.next = next # function to insert a node at the # beginning of the linked list def push(head_ref, new_data): # allocate node new_node = node(0) # put in the data new_node.data = new_data # link the old list to the new node new_node.next = (head_ref) # move the head to point to the new node (head_ref) = new_node return head_ref """ Alternatively reverses the given linked list in groups of given size k. """def kAltReverse(head, k) : return _kAltReverse(head, k, True) """ Helper function for kAltReverse(). It reverses k nodes of the list only if the third parameter b is passed as True, otherwise moves the pointer k nodes ahead and recursively calls iteself """def _kAltReverse(Node, k, b) : if(Node == None) : return None count = 1 prev = None current = Node next = None """ The loop serves two purposes 1) If b is True, then it reverses the k nodes 2) If b is false, then it moves the current pointer """ while(current != None and count <= k) : next = current.next """ Reverse the nodes only if b is True""" if(b == True) : current.next = prev prev = current current = next count = count + 1 """ 3) If b is True, then node is the kth node. So attach rest of the list after node. 4) After attaching, return the new head """ if(b == True) : Node.next = _kAltReverse(current, k, not b) return prev else : """ If b is not True, then attach rest of the list after prev. So attach rest of the list after prev """ prev.next = _kAltReverse(current, k, not b) return Node """ Function to print linked list """def printList(node) : count = 0 while(node != None) : print( node.data, end = " ") node = node.next count = count + 1 # Driver Code """ Start with the empty list """head = Nonei = 20 # create a list 1->2->3->4->5...... ->20 while(i > 0 ): head = push(head, i) i = i - 1 print( "Given linked list ") printList(head) head = kAltReverse(head, 3) print( "\nModified Linked list ") printList(head) # This code is contributed by Arnab Kundu |
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C#
// C# Program for converting // singly linked list into // circular linked list. using System; public class LinkedList { static Node head; public class Node { public int data; public Node next; public Node(int d) { data = d; next = null; } } /* Reverses alternate k nodes and returns the pointer to the new head node */ Node kAltReverse(Node node, int k) { Node current = node; Node next = null, prev = null; int count = 0; /*1) reverse first k nodes of the linked list */ while (current != null && count < k) { next = current.next; current.next = prev; prev = current; current = next; count++; } /* 2) Now head points to the kth node. So change next of head to (k+1)th node*/ if (node != null) { node.next = current; } /* 3) We do not want to reverse next k nodes. So move the current pointer to skip next k nodes */ count = 0; while (count < k - 1 && current != null) { current = current.next; count++; } /* 4) Recursively call for the list starting from current->next. And make rest of the list as next of first node */ if (current != null) { current.next = kAltReverse(current.next, k); } /* 5) prev is new head of the input list */ return prev; } void printList(Node node) { while (node != null) { Console.Write(node.data + " "); node = node.next; } } void push(int newdata) { Node mynode = new Node(newdata); mynode.next = head; head = mynode; } public static void Main(String[] args) { LinkedList list = new LinkedList(); // Creating the linkedlist for (int i = 20; i > 0; i--) { list.push(i); } Console.WriteLine("Given Linked List :"); list.printList(head); head = list.kAltReverse(head, 3); Console.WriteLine(""); Console.WriteLine("Modified Linked List :"); list.printList(head); } } // This code is contributed 29AjayKumar |
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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)
Please write comments if you find the above code/algorithm incorrect, or find other ways to solve the same problem.
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