Given a singly linked list, rotate the linked list counter-clockwise by k nodes. Where k is a given positive integer. For example, if the given linked list is 10->20->30->40->50->60 and k is 4, the list should be modified to 50->60->10->20->30->40. Assume that k is smaller than the count of nodes in a linked list.
Method 1:
To rotate the linked list, we need to change the next of kth node to NULL, the next of the last node to the previous head node, and finally, change the head to (k+1)th node. So we need to get hold of three nodes: kth node, (k+1)th node, and last node.
Traverse the list from the beginning and stop at kth node. Store pointer to kth node. We can get (k+1)th node using kthNode->next. Keep traversing till the end and store a pointer to the last node also. Finally, change pointers as stated above.
Below image shows how to rotate function works in the code :
Python
# Python program to rotate # a linked list # Node class class Node: # Constructor to initialize # the node object def __init__(self, data): self.data = data self.next = None class LinkedList: # Function to initialize head def __init__(self): self.head = None # Function to insert a new node # at the beginning def push(self, new_data): # allocate node and put the data new_node = Node(new_data) # Make next of new node as head new_node.next = self.head # Move the head to point to the # new Node self.head = new_node # Utility function to print it the # linked LinkedList def printList(self): temp = self.head while(temp): print temp.data, temp = temp.next # This function rotates a linked list # counter-clockwise and updates the # head. The function assumes that k # is smaller than size of linked list. # It doesn't modify the list if k is # greater than of equal to size def rotate(self, k): if k == 0: return # Let us understand the below code # for example k = 4 and list = # 10->20->30->40->50->60 current = self.head # current will either point to kth # or NULL after this loop current # will point to node 40 in the above # example count = 1 while(count <k and current is not None): current = current.next count += 1 # If current is None, k is greater # than or equal to count of nodes # in linked list. Don't change # the list in this case if current is None: return # current points to kth node. Store # it in a variable kth node points # to node 40 in the above example kthNode = current # current will point to last node # after this loop current will point # to node 60 in above example while(current.next is not None): current = current.next # Change next of last node to previous # head Next of 60 is now changed to # node 10 current.next = self.head # Change head to (k + 1)th node # head is not changed to node 50 self.head = kthNode.next # change next of kth node to NULL # next of 40 is not NULL kthNode.next = None # Driver code llist = LinkedList() # Create a list # 10->20->30->40->50->60 for i in range(60, 0, -10): llist.push(i) print "Given linked list"llist.printList() llist.rotate(4) print "Rotated Linked list"llist.printList() # This code is contributed by Nikhil Kumar Singh(nickzuck_007) |
p>Output:
Given linked list 10 20 30 40 50 60 Rotated Linked list 50 60 10 20 30 40
Time Complexity: O(n) where n is the number of nodes in Linked List. The code traverses the linked list only once.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
Method 2:
To rotate a linked list by k, we can first make the linked list circular and then moving k-1 steps forward from head node, making (k-1)th node’s next to null and make kth node as head.
Python3
# Python3 program to rotate a # linked list counter clock wise # Link list node class Node: def __init__(self): self.data = 0 self.next = None # This function rotates a linked list # counter-clockwise and updates the # head. The function assumes that k is # smaller than size of linked list. def rotate(head_ref, k): if (k == 0): return # Let us understand the below # code for example k = 4 and # list = 10.20.30.40.50.60. current = head_ref # Traverse till the end. while (current.next != None): current = current.next current.next = head_ref current = head_ref # Traverse the linked list to k-1 # position which will be last # element for rotated array. for i in range(k - 1): current = current.next # Update the head_ref and last # element pointer to None head_ref = current.next current.next = None return head_ref # UTILITY FUNCTIONS # Function to push a node def push(head_ref, new_data): # Allocate node new_node = 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 return head_ref # Function to print linked list def printList(node): while (node != None): print(node.data, end = ' ') node = node.next # Driver code if __name__=='__main__': # Start with the empty list head = None # Create a list # 10.20.30.40.50.60 for i in range(60, 0, -10): head = push(head, i) print("Given linked list ") printList(head) head = rotate(head, 4) print("\nRotated Linked list ") printList(head) # This code is contributed by rutvik_56 |
Output:
Given linked list 10 20 30 40 50 60 Rotated Linked list 50 60 10 20 30 40
Please refer complete article on Rotate a Linked List for more details!
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