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

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: []

Constraints:

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

Explanation

Alright, imagine you have a train of toy cars, all linked together, and you want to reverse the order of the train.

Here’s how you’d do it in simple steps, like a recipe:

🚓Starting Point:

• You have a train of cars, with a “head” car at the front.
• You have an empty space called “previous car” (it’s empty at first).

🚓Take the head car:

• Grab the car at the front of the train. Let’s call it the “current car”.

🚓Remember the next car:

• Before you change anything, remember the car that’s linked to the “current car”. We’ll need it later. Let’s call this the “next car”.

🚓Unlink and Relink:

• Take the “current car” and unlink it from the rest of the train.
• Link the “current car” to the “previous car” space. (At first, this will link to nothing, but that’s okay!)

🚓Move the “previous car” marker:

• Now, the “current car” becomes the “previous car”.

🚓Move to the next car:

• The “next car” that you remembered becomes the “current car”.

🚓Repeat:

• Go back to step 2 and keep doing this until you run out of cars in the original train.

🚓The new head:

• When you run out of cars, the “previous car” is now the new head of the reversed train.

Example:

Imagine your train is: Car 1 -> Car 2 -> Car 3

Start: “previous car” is empty, “current car” is Car 1.

Remember: “next car” is Car 2.

Relink: Car 1 now links to nothing.

Move: “previous car” is Car 1, “current car” is Car 2.

Repeat:

“next car” is Car 3.

Car 2 links to Car 1.

“previous car” is car 2, “current car” is car 3.

Repeat:

next car is nothing.

Car 3 links to car 2.

“previous car” is car 3, “current car” is nothing.

Done! The new train is Car 3 -> Car 2 -> Car 1.

Implementation

Python

# # Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next

class Solution:
    def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:
        """
        Reverses a singly-linked list.

        Args:
            head: The head of the linked list.

        Returns:
            The head of the reversed linked list.
        """
        prev_node = None  # Initialize the previous node to None
        curr_node = head  # Initialize the current node to the head of the list

        # Iterate through the linked list
        while curr_node:
            # Store the next node in a temporary variable
            temp = curr_node.next

            # Reverse the current node's pointer to point to the previous node
            curr_node.next = prev_node

            # Move the previous node pointer to the current node
            prev_node = curr_node

            # Move the current node pointer to the next node (stored in temp)
            curr_node = temp

        # After the loop, prev_node will be the new head of the reversed list
        return prev_node

Typescript

// /**
//  * Definition for singly-linked list.
//  */
// class ListNode {
//     val: number;
//     next: ListNode | null;
//     constructor(val?: number, next?: ListNode | null) {
//         this.val = (val === undefined ? 0 : val);
//         this.next = (next === undefined ? null : next);
//     }
// }

function reverseList(head: ListNode | null): ListNode | null {
    /**
     * Reverses a singly-linked list.
     *
     * @param head - The head of the linked list.
     * @returns The head of the reversed linked list.
     */
    let prevNode: ListNode | null = null; // Initialize the previous node to null
    let currNode: ListNode | null = head; // Initialize the current node to the head of the list

    // Iterate through the linked list
    while (currNode) {
        // Store the next node in a temporary variable
        const temp: ListNode | null = currNode.next;

        // Reverse the current node's pointer to point to the previous node
        currNode.next = prevNode;

        // Move the previous node pointer to the current node
        prevNode = currNode;

        // Move the current node pointer to the next node (stored in temp)
        currNode = temp;
    }

    // After the loop, prevNode will be the new head of the reversed list
    return prevNode;
}

Go

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 * Val int
 * Next *ListNode
 * }
 */
func reverseList(head *ListNode) *ListNode {
    /**
     * Reverses a singly-linked list.
     *
     * @param head - The head of the linked list.
     * @returns The head of the reversed linked list.
     */
    var prevNode *ListNode = nil // Initialize the previous node to nil
    currNode := head // Initialize the current node to the head of the list

    // Iterate through the linked list
    for currNode != nil {
        // Store the next node in a temporary variable
        temp := currNode.Next

        // Reverse the current node's pointer to point to the previous node
        currNode.Next = prevNode

        // Move the previous node pointer to the current node
        prevNode = currNode

        // Move the current node pointer to the next node (stored in temp)
        currNode = temp
    }

    // After the loop, prevNode will be the new head of the reversed list
    return prevNode
}

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