In this series of posts, I will discuss coding questions on the `LinkedList`

Data structure.

The posts in this series will be organized in the following way,

- Question Link ❓
- Possible Explanation 📝
- Documented C++ Code 🧹
- Time and Space Complexity Analysis ⌛🌌

## The Question

Given a non-empty, singly linked list with head node `head`

, return a middle node of the linked list.

If there are two middle nodes, return the second middle node.

https://leetcode.com/problems/middle-of-the-linked-list/

💡 Give yourself at least 15-20 mins to figure out the solution :)

## Explanation

The naive approach is very simple but it is important to understand *what* & *why* we are doing this to fully understand the optimized approach.

- Find the length of LinkedList →
`L`

- Find the position (from start) of the node to be deleted:
`floor(L/2) + 1`

→`K`

- eg: for
`L = 5`

, required node is`3 = floor(5/2) + 1`

- eg: for
`L = 6`

, required node is`4 = floor(6/2) + 1`

- eg: for
- Reach the
`Kth`

node in**K-1**iterations and return it.

## C++ Code

### Definition of Linked List

```
//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) {}
};
```

### Solution

```
ListNode *middleNode(ListNode *head)
{
ListNode *temp = head;
int count = 0;
//- finding length of LL : O(n)
while (temp)
{
count++;
temp = temp->next;
}
//pointing temp to head again
temp = head;
int nodeNo = count / 2 + 1; //doesn't matter if count is odd or even
//-Time: O(k-1)
//If I want to reach kth node, I need to go k-1 deep
for (int i = 1; i <= nodeNo - 1; i++)
{
temp = temp->next;
}
head = temp;
return head;
}
```

## Complexity Analysis

### Time Complexity: O(n)

O(n) for finding length of List + O(n/2) for reaching the required node = O(n)

### Space Complexity: O(1)

We didn't use any extra space

## Discussion (0)