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LESSON 296 OF 409 · LINKED LISTS

Reverse Linked List in O(n) Time and O(1) Space using C++

Practice O(n) Read the full explanation

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

Example:

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


Note:

Your algorithm should run in O(n) time and use O(1) extra space.


Hints:

This problem is very similar to "Reverse String" problem, so it's natural to think of adapting that solution for Linked Lists. Unfortunately, traversing with two pointers doesn't work here because the right pointer should move to the left, and we cannot do that in a Linked List. The trick is actually in modifying the links of our Linked List to obtain its reversed version. How should we modify them?

We just have to reverse each link in the list. For example: 1 -> 2 -> 3 -> 4 should become 1 <- 2 <- 3 <- 4. And the head of the reversed list is the tail of the original one. How can we reverse these links?

One idea is to traverse the list from left to right using a variable node and convert each link node -> node.next to node.next -> node. In plain steps: start with node at the head; while node is not NULL, set nextNode to the node after node, make that next node point back at node, then move node forward to nextNode. But there is a problem with this algorithm. Can you find out what that is? Take some small example and run it by hand.

The problem is that the moment we reverse a link, we lose access to the next nodes. For example, if the list is 1 -> 2 -> 3. When node is 1, we reverse the link and the list becomes 1 <- 2 3. We lost the link between 2 and 3. And so at the next step node becomes 2 but node.next is 1 instead of 3. The solution is still good, but we need to implement it in such a way that this problem is fixed.

We should traverse the linked list, while using two pointers:
  • prev - the previous node
  • node - the current node
We will start with the first two nodes in the list: prev is the head and node is the node after the head. Then, for as long as node is not NULL, we will reverse the link prev -> node and update the two variables such that they point to the next two nodes. Can you figure out how to code this? Take the example above and figure out how you should update the pointers after each link reversal.

Start with prev at the head and node at the node after it. While node is not NULL:
  • save the node after node in nextNode;
  • make node point back at prev;
  • move prev to node;
  • move node to nextNode.
Don't forget to cut the old head's link to the second node, so the reversed list ends there, and return the new head.

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    C++ · LINKED LISTS

    Reverse Linked List in C++, explained

    Problem Definition

    The task is to reverse a singly linked list. Given the head of the list, we need to reverse the list and return the new head.

    Input:

    A singly linked list represented by its head node.

    Output:

    The head of the reversed linked list.

    Constraints:

    • The number of nodes in the list is in the range [0, 5000].
    • -5000 <= Node.val <= 5000

    Example:

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

    Understanding the Problem

    The core challenge is to reverse the links between nodes in a singly linked list. This problem is significant in many applications, such as reversing data streams or undoing operations. A common pitfall is losing access to the next node when reversing the link, which can be avoided by careful pointer manipulation.

    Approach

    To solve this problem, we need to reverse the direction of the links between nodes. We can achieve this by iterating through the list and adjusting the pointers.

    Naive Solution

    A naive approach might involve using additional space to store the nodes in a different order, but this would not meet the O(1) space requirement.

    Optimized Solution

    We can use an iterative approach with three pointers: prev, current, and next. This method ensures we reverse the list in O(n) time and O(1) space.

    Steps:

    1. Initialize three pointers: prev to null, current to head, and next to null.
    2. Iterate through the list, reversing the links by adjusting the next pointer of the current node to point to prev.
    3. Move the prev and current pointers one step forward.
    4. Continue until all nodes are processed.
    5. Return prev as the new head of the reversed list.

    Algorithm

    Here is a step-by-step breakdown of the algorithm:

    1. Initialize prev to null and current to head.
    2. While current is not null:
      • Store current->next in next.
      • Set current->next to prev.
      • Move prev to current.
      • Move current to next.
    3. Return prev as the new head.

    Code Implementation

    // Definition for singly-linked list.
struct ListNode {
    int val;
    ListNode *next;
    ListNode() : val(0), next(nullptr) {}
    ListNode(int x) : val(x), next(nullptr) {}
    ListNode(int x, ListNode *next) : val(x), next(next) {}
};

class Solution {
public:
    ListNode* reverseList(ListNode* head) {
        ListNode* prev = nullptr; // Initialize previous node to null
        ListNode* current = head; // Start with the head node
        ListNode* next = nullptr; // Initialize next node to null
        
        while (current != nullptr) { // Traverse the list
            next = current->next; // Store the next node
            current->next = prev; // Reverse the current node's pointer
            prev = current; // Move prev to current node
            current = next; // Move to the next node
        }
        
        return prev; // prev will be the new head of the reversed list
    }
};

    Complexity Analysis

    The time complexity of this approach is O(n) because we traverse the list once. The space complexity is O(1) as we only use a constant amount of extra space for the pointers.

    Edge Cases

    Consider the following edge cases:

    • An empty list (head is null): The function should return null.
    • A list with one node: The function should return the same node.

    Testing

    To test the solution comprehensively, consider the following test cases:

    1. Input: head = []
   Output: []

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

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

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

    Thinking and Problem-Solving Tips

    When approaching such problems, consider the following tips:

    • Understand the structure of the data (linked list in this case).
    • Visualize the problem with diagrams.
    • Break down the problem into smaller steps.
    • Consider edge cases and test your solution thoroughly.

    Conclusion

    Reversing a linked list is a fundamental problem that helps in understanding pointer manipulation and linked list operations. Mastering this problem enhances your problem-solving skills and prepares you for more complex data structure challenges.

    Additional Resources

    For further reading and practice, consider the following resources:

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