Understanding and Solving the " Reverse Linked List" Coding Problem.

206. Reverse Linked List

Given the head of a singly linked list, reverse the list, and return the reversed list.

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Example 1:

Input: head = [1,2,3,4,5]
Output: [5,4,3,2,1]
        

Example 2:

Input: head = [1,2]
Output: [2,1]
        

Example 3:

Input: head = []
Output: []        

Solution Steps

1. Initialize Three Pointers: The key to reversing a linked list is to change the direction of the pointers. We start by initializing three pointers: prev as null, curr as the head of the list, and next as a temporary pointer.
2. Traverse the List: We iterate through the list. In each iteration, we temporarily store the next node (next = curr.next), move curr.next to point to prev, shift prev to curr, and finally move curr to next.
3. Termination and Return: The iteration continues until curr becomes null, meaning we've reached the end of the original list. At this point, prev will be pointing to the new head of the reversed list.

function reverseList(head) {
    let prev = null;
    let curr = head;
    while (curr) {
        let nextTemp = curr.next;
        curr.next = prev;
        prev = curr;
        curr = nextTemp;
    }
    return prev;
}        

Time Complexity

The time complexity of this algorithm is O(n), where n is the number of nodes in the linked list. This is because we traverse the list once, and each node's pointer is changed exactly once.

Space Complexity

The space complexity is O(1), indicating constant space usage. This efficiency is due to not allocating any additional data structures; we're only using a few pointer variables regardless of the size of the input list.

Conclusion

Reversing a linked list is a fundamental problem that demonstrates the power of pointer manipulation. It's a great example of how a seemingly complex task can be achieved with a simple and efficient algorithm. Understanding this problem enhances one's grasp of linked lists, a vital concept in computer science and software engineering.

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