How do you solve Reverse Linked List on LeetCode?

Reverse Linked List (LeetCode #206) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Reversing a linked list can be done in place, which is more efficient than creating a new list.

What is the brute force solution for Reverse Linked List?

The brute force approach for Reverse Linked List is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves creating a new list and copying elements from the original list in reverse order. This is straightforward but inefficient due to extra space usage and multiple traversals. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Reverse Linked List?

Reverse Linked List uses the following concepts: Linked List, Recursion. The recommended patterns to study are: Two Pointers, In-Place Reversal.

What are the interview tips for Reverse Linked List?

Explain your thought process clearly as you code. Consider edge cases and mention them during your explanation. Practice pointer manipulation problems to build confidence.

What are common mistakes when solving Reverse Linked List?

Forgetting to update the next pointer before changing the current pointer. Not handling edge cases like an empty list or a single node.

#206

Reverse Linked List

Easy

Linked ListRecursionTwo PointersIn-Place Reversal

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

The optimal solution reverses the linked list in place using a three-pointer technique. This is efficient as it only requires a single pass through the list and uses constant space.

⚙️

Algorithm

3 steps

1Step 1: Initialize three pointers: prev as null, current as head, and next as null.
2Step 2: Iterate through the list, and for each node, store the next node, reverse the current node's pointer to point to prev, then move prev and current one step forward.
3Step 3: Once the end of the list is reached, return prev as the new head of the reversed list.

solution.py15 lines

1# Full working Python code
2class ListNode:
3    def __init__(self, val=0, next=None):
4        self.val = val
5        self.next = next
6
7def reverseList(head):
8    prev = None
9    current = head
10    while current:
11        next_node = current.next
12        current.next = prev
13        prev = current
14        current = next_node
15    return prev

ℹ

Complexity note: This approach runs in linear time because we traverse the list once, and it uses constant space since we only use a few pointers.

1Reversing a linked list can be done in place, which is more efficient than creating a new list.
2Understanding pointer manipulation is crucial for linked list problems.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.