. Day 9 - Challenge 2 - Reverse Linked List Skip to main content

Day 9 - Challenge 2 - Reverse Linked List

Reversing a Singly Linked List in JavaScript: An In-Place Approach

Introduction: 

Singly linked lists are fundamental data structures in computer science that consist of a sequence of nodes, each containing data and a reference to the next node in the list. Reversing a singly linked list is a classic problem that challenges programmers to manipulate pointers effectively to achieve the desired outcome. In this blog post, we'll explore the problem of reversing a singly linked list using an in-place approach and provide a step-by-step solution in JavaScript.

Problem Statement: 

Given the head of a singly linked list, our task is to reverse the list in-place and return its new head. In other words, we need to modify the pointers of the nodes in such a way that the direction of the linked list is reversed.

Solution Approach: 

To solve this problem, we will iterate through the linked list while maintaining three pointers: previous, current, and next. The previous pointer will initially be null, as the first node in the reversed list will become the new head. We will update the next pointer of the current node to point to the previous node, effectively reversing the direction of the link. After reversing the link, we move the previous and current pointers one step forward in the list.

Here's a step-by-step breakdown of the solution:

  1. Initialize three pointers: previous (null), current (head of the original list), and next (null).

  2. Iterate through the list while current is not null: a. Store the next node in the next pointer. b. Update the next pointer of the current node to point to the previous node (reverse the link). c. Move the previous pointer to the current node. d. Move the current pointer to the next node.

  3. After the iteration, the previous pointer will be pointing to the last node of the original list, which will become the new head of the reversed list.

  4. Return the previous pointer as the new head of the reversed list.

JavaScript Implementation: 

Here's the JavaScript code that implements the solution:

class ListNode {
    constructor(val, next = null) {
        this.val = val;
        this.next = next;
    }
}

function reverseLinkedList(head) {
    let previous = null;
    let current = head;

    while (current !== null) {
        let nextNode = current.next;
        current.next = previous;
        previous = current;
        current = nextNode;
    }

    return previous; // The new head of the reversed list
}

Conclusion: 

Reversing a singly linked list in-place is a fundamental problem that tests a programmer's understanding of pointers and data manipulation. The provided JavaScript solution offers an efficient approach to solve this problem by iteratively reversing the links between the nodes. Understanding this solution can serve as a valuable skill for tackling similar problems and mastering the intricacies of linked list manipulation.

Now write the same program in your favorite language in comment section. 

Other Challenges:

  1. Day 6 Challenges 
  2. Day 7 Challenges
  3. Day 8 Challenges
Labels:

Comments

Post a Comment

Popular posts from this blog

Day 9 - Challenge 1 - Product of Array Except Self

Solving the "Product of Array Except Self" Problem in JavaScript Are you ready to dive into a common coding challenge that not only tests your programming skills but also sharpens your problem-solving mindset? If you're up for the challenge, let's tackle the "Product of Array Except Self" problem together using JavaScript. This problem requires us to return an array where each element at index i is the product of all the elements in the original array except the one at index i . Understanding the Problem:   Imagine you're given an array of integers, let's call it nums . Your task is to create a new array where the value at index i in this new array is the product of all the elements in nums , except the one at index i . In other words, you're calculating the product of all the elements to the left of nums[i] and the product of all the elements to the right of nums[i] , and then multiplying these two products to get the final value at index i ...
Post a Comment
Read more

Day 8 - Challenge 2 - Move Zeroes to the End

Moving Zeroes to the End: A JavaScript Solution Introduction:   When working with arrays, there are often times when we need to manipulate their elements to achieve a specific goal. One common problem is moving all zeroes to the end of an array while keeping the order of non-zero elements unchanged. In this blog post, we will explore an elegant solution to this problem using JavaScript. We'll discuss the problem statement, the approach we'll take, and provide a step-by-step guide to implementing the solution. The Problem:   Given an array of integers, the task is to move all zeroes to the end of the array while maintaining the relative order of the non-zero elements. This means that after rearranging the array, all the zeroes should be at the end, and the order of the non-zero elements should remain the same. The Approach:   To solve this problem, we can utilize a two-pointer approach. We'll maintain two pointers, one for iterating through the array and another for keepin...
Post a Comment
Read more