# Mastering LeetCode's 'Middle of the Linked List': A Comprehensive Guide for Software Engineers

## Unlock the secrets to efficiently finding the middle node in a singly linked list with walkthrough and solutions in Python, TypeScript, and Java.

**Introduction to the Problem**

In software engineering interviews, demonstrating proficiency in data structures is pivotal. Today, I'll unpack a common problem that often perplexes many: finding the middle node of a singly linked list, as featured on LeetCode (LeetCode 876. Middle of Linked List).

This challenge tests your understanding of linked list traversal and two-pointer techniques. Consider a linked list where you're asked to identify the central element. For instance, given a list `[1,2,3,4,5]`

, the output should be `[3,4,5]`

, pinpointing node 3 as the midpoint. In scenarios with an even number of nodes, say `[1,2,3,4,5,6]`

, the expectation is to return the second middle node, resulting in `[4,5,6]`

.

Given the

`head`

of a singly linked list, returnthe middle node of the linked list.If there are two middle nodes, return

the second middlenode.

Example 1:`Input: head = [1,2,3,4,5] Output: [3,4,5] Explanation: The middle node of the list is node 3.`

Example 2:`Input: head = [1,2,3,4,5,6] Output: [4,5,6] Explanation: Since the list has two middle nodes with values 3 and 4, we return the second one.`

**Solving the Problem**

The crux of solving this problem lies in the two-pointer strategy: employing a slow pointer that moves one step at a time and a fast pointer advancing two steps per turn. This approach ensures that when the fast pointer reaches the end of the list, the slow pointer will be at the middle, elegantly sidestepping the need to count elements beforehand.

The beauty of this solution is its efficiency, boasting a time complexity of O(n) where n is the number of nodes in the list, and a space complexity of O(1), as it only utilizes two pointers regardless of the list's size.

### The Solution in Python

Let's dive into the Python solution, which embraces simplicity and efficiency.

```
def middleNode(head: Optional[ListNode]) -> Optional[ListNode]:
slow = head # Starts at the beginning
quick = head # Also starts at the beginning
while quick and quick.next: # Continues until the fast pointer reaches the end
slow = slow.next # Moves one step
quick = quick.next.next # Moves two steps
return slow # When fast pointer is at the end, slow is at the middle
```

### The Solution in TypeScript

TypeScript, often used for its strong typing features, offers a structured way to tackle this problem.

```
function middleNode(head: ListNode | null): ListNode | null {
let slow: ListNode | null = head; // Initialize slow pointer
let quick: ListNode | null = head; // Initialize fast pointer
while (quick !== null && quick.next !== null) {
slow = slow.next; // Move slow pointer one step
quick = quick.next.next; // Move fast pointer two steps
}
return slow; // Return the slow pointer as the middle node
}
```

**The Solution in Java**

Java offers a class-based approach to solving the problem, emphasizing readability and robustness.

```
public class Solution {
public ListNode middleNode(ListNode head) {
ListNode slow = head; // Starts at the beginning
ListNode quick = head; // Also starts at the beginning
while (quick != null && quick.next != null) {
slow = slow.next; // Moves one step
quick = quick.next.next; // Moves two steps
}
return slow; // Slow is at the middle when fast reaches the end
}
}
```

## Conclusion

Mastering this problem not only boosts your confidence in handling linked lists but also sharpens your problem-solving strategy with the two-pointer technique.

Whether you prefer Python, TypeScript, or Java, understanding the underlying concept remains key. As you practice, remember that the elegance of a solution often lies in its simplicity and efficiency.

Happy coding, and best of luck in your software engineering interviews!