The linked list data structure is a crucial topic to prepare for technical rounds. You can see a variety of ticklish and straightforward questions on this topic. This article will introduce you to the basic problem of a linked list.
What is a linked list?
A linked list is a linear data structure where every element acts as a node with two parts data part(which stores the actual data of the element) and the address part(which stores the address of the location of the next node). Linked lists are dynamic, which eases the insertion and deletion of the elements.
Implementation of Linked List
class Node: # constructor for defining the data and next def __init__(self, data = None, next=None): self.data = data self.next = next # A Linked List class with a single head node class LinkedList: def __init__(self): self.head = None # insertion method def insert(self, data): newNode = Node(data) if(self.head): current = self.head while(current.next): current = current.next current.next = newNode else: self.head = newNode # print function def printLL(self): current = self.head while(current): print(current.data) current = current.next # Singly Linked List with insertion and print methods LL = LinkedList() LL.insert(3) LL.insert(4) LL.insert(5) LL.printLL()
Reversing a Linked List
We are given a singly linked list, and our task is to reverse the linked list by changing the links between the nodes.
Here as you can see the list has been reversed.
Now 4->3, 3->2, 2->1.
Now 4 has become the head of the list as it is the starting node of this linked list after performing the reversal action.
Here the only change we can observe is the change in the links. So we have to think of an approach that can change the links between the nodes.
Let’s move to our different approaches to solving this problem.
Recommended: Do try to attempt the Problem yourself before moving on to the solution.
- Create two nodes-> current and next.
- Initially, current(node) will point to the first element(head), and prev will point to null.
- Store the next the current element in next.
- Traverse through the list till the current become null.
- In the end, the prev will be our head.
The current element will now point to that element stored in the 'prev' variable.
class Node: # Constructor def __init__(self, data): self.data = data self.next = None class LinkedList: # head function def __init__(self): self.head = None # To reverse the list def reverse(self): prev = None current = self.head while(current is not None): next = current.next current.next = prev prev = current current = next self.head = prev def push(self, new_data): new_node = Node(new_data) new_node.next = self.head self.head = new_node # To print the list def printL(self): temp = self.head while(temp): print temp.data, temp = temp.next # Main function llist = LinkedList() llist.push(1) llist.push(2) llist.push(3) llist.push(4) print "Given Linked List" llist.printL() llist.reverse() print "\nReversed Linked List" llist.printL()
Given Linked list 1 2 3 4 Reversed linked list 4 3 2 1
Time Complexity - O(n).
Space Complexity - O(1).
- when the curr node is the last node starts with it and also runs after the previous curr.
- Save the following curr to next.
- Next, create from the current point to the previous point.
- Repeat with curr as curr and curr as prev.
Pic referenced from prepbytes.
class Node: # Constructor def __init__(self, data): self.data = data self.next = None class LinkedList: # head function def __init__(self): self.head = None def reverse_Util(self, curr, prev): # If last node mark it head if curr.next is None: self.head = curr curr.next = prev return # Save curr.next node for recursive call next = curr.next # update next curr.next = prev self.reverse_Util(next, curr) # This function calls reverse_Util() # with previous as Null def reverse(self): if self.head is None: return self.reverse_Util(self.head, None) # To insert the new node at the beginningof the list def push(self, new_data): new_node = Node(new_data) new_node.next = self.head self.head = new_node # To print the list def printL(self): temp = self.head while(temp): print temp.data, temp = temp.next # Main function llist = LinkedList() llist.push(1) llist.push(2) llist.push(3) llist.push(4) print "Given linked list" llist.printL() llist.reverse() print "\nReverse linked list" llist.printL()
Given linked list 4 3 2 1 Reverse linked list 1 2 3 4
Time Complexity: O(n)
Space Complexity: O(n)
To read about Fibonacci Series in Python click here.
Must Read C Program to Reverse a Number
Frequently Asked Questions
What do you mean by reverse a linked list?
You will be given a linked list, and you have to print the reverse linked list by changing the links between the nodes.
What are the approaches we used to find the reverse of a linked list?
We have used two efficient approaches to solve this problem, i.e., Iterative and Recursive.
To print the reverse of a linked list is one of the most straightforward problems of a linked list topic. We have seen the implementation of a singly linked list and two approaches to finding the reverse of the list in Python language.
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- Doubly Linked List
- Reverse a Stack using Recursion
- Java Program to Reverse a Linked List
- Reverse a Linked List in Groups
- Reversing a Linked List
- Reverse a Doubly Linked List
- Sort a Linked List
- Reverse a Linked List recursively
- Problems on Linked Lists
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