How do you solve Middle of the Linked List on LeetCode?

Middle of the Linked List (LeetCode #876) 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: Using two pointers can significantly reduce time complexity.

What is the brute force solution for Middle of the Linked List?

The brute force approach for Middle of the Linked List is Brute Force, with O(n²) time complexity and O(1) space complexity. In this approach, we traverse the linked list to count the total number of nodes. Then, we traverse the list again to reach the middle node based on the count. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Middle of the Linked List?

Middle of the Linked List uses the following concepts: Linked List, Two Pointers. The recommended patterns to study are: Two Pointers, Linked List.

What are the interview tips for Middle of the Linked List?

Explain your thought process clearly during the interview. Consider edge cases before jumping into coding. Practice the two-pointer technique with other problems.

What are common mistakes when solving Middle of the Linked List?

Not handling edge cases like a single node list. Confusing the middle node with the middle index.

#876

Middle of the Linked List

Easy

Linked ListTwo PointersTwo PointersLinked List

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)

Using the two-pointer technique, we can find the middle node in a single pass through the list. One pointer moves twice as fast as the other, allowing us to find the middle efficiently.

⚙️

Algorithm

4 steps

1Step 1: Initialize two pointers, slow and fast, both pointing to the head.
2Step 2: Move slow by one step and fast by two steps until fast reaches the end of the list.
3Step 3: When fast is null or fast.next is null, slow will be at the middle node.
4Step 4: Return the slow pointer.

solution.py13 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 middleNode(head):
8    slow = head
9    fast = head
10    while fast and fast.next:
11        slow = slow.next
12        fast = fast.next.next
13    return slow

ℹ

Complexity note: The time complexity is O(n) because we traverse the list only once. The space complexity is O(1) since we only use a fixed amount of space for the pointers.

1Using two pointers can significantly reduce time complexity.
2Understanding the structure of linked lists is crucial for solving related problems.

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