Find Middle Element of Linked List in DSA C - Build from Scratch

Find Middle Element of Linked List

📖 Scenario: You are working with a simple linked list that stores numbers. You want to find the middle number in the list, just like finding the middle book on a shelf.

🎯 Goal: Build a program that creates a linked list with 5 nodes, then finds and prints the middle element's value.

📋 What You'll Learn

Create a linked list with exactly 5 nodes with values 10, 20, 30, 40, 50

Use a pointer to traverse the list

Find the middle element using the slow and fast pointer technique

Print the middle element's value

💡 Why This Matters

🌍 Real World

Finding the middle element is useful in many applications like splitting data, balancing trees, or optimizing searches.

💼 Career

Understanding linked lists and pointer techniques is essential for software developers working with low-level data structures and memory management.

1

Create the linked list with 5 nodes

Create a struct called Node with an int data and a Node* called next. Then create 5 nodes named node1 to node5 with data values 10, 20, 30, 40, and 50 respectively. Link them so node1.next = &node2, node2.next = &node3, and so on, ending with node5.next = NULL. Finally, create a pointer head pointing to &node1.

DSA C

// Define struct Node and create 5 linked nodes with values 10, 20, 30, 40, 50
// Link nodes and set head pointer

Hint

Remember to define the struct first, then create nodes with the given values, link them, and set the head pointer.

2

Create pointers for slow and fast traversal

Inside main, create two pointers called slow and fast and set both to point to head.

DSA C

#include 
#include 

typedef struct Node {
    int data;
    struct Node* next;
} Node;

int main() {
    Node node1 = {10, NULL};
    Node node2 = {20, NULL};
    Node node3 = {30, NULL};
    Node node4 = {40, NULL};
    Node node5 = {50, NULL};

    node1.next = &node2;
    node2.next = &node3;
    node3.next = &node4;
    node4.next = &node5;
    node5.next = NULL;

    Node* head = &node1;

    // Create pointers slow and fast and set to head
    // Your code here

    return 0;
}

Hint

Both slow and fast pointers start at the head of the list.

3

Find the middle element using slow and fast pointers

Use a while loop to move fast two steps and slow one step until fast is NULL or fast->next is NULL. This will make slow point to the middle node.

DSA C

#include 
#include 

typedef struct Node {
    int data;
    struct Node* next;
} Node;

int main() {
    Node node1 = {10, NULL};
    Node node2 = {20, NULL};
    Node node3 = {30, NULL};
    Node node4 = {40, NULL};
    Node node5 = {50, NULL};

    node1.next = &node2;
    node2.next = &node3;
    node3.next = &node4;
    node4.next = &node5;
    node5.next = NULL;

    Node* head = &node1;

    Node* slow = head;
    Node* fast = head;

    // Use while loop to move fast two steps and slow one step
    // Your code here

    return 0;
}

Hint

Move slow by one node and fast by two nodes each loop until fast reaches the end.

4

Print the middle element's value

Use printf to print the data of the node pointed to by slow in the format: "Middle element: %d\n".

DSA C

#include 
#include 

typedef struct Node {
    int data;
    struct Node* next;
} Node;

int main() {
    Node node1 = {10, NULL};
    Node node2 = {20, NULL};
    Node node3 = {30, NULL};
    Node node4 = {40, NULL};
    Node node5 = {50, NULL};

    node1.next = &node2;
    node2.next = &node3;
    node3.next = &node4;
    node4.next = &node5;
    node5.next = NULL;

    Node* head = &node1;

    Node* slow = head;
    Node* fast = head;

    while (fast != NULL && fast->next != NULL) {
        slow = slow->next;
        fast = fast->next->next;
    }

    // Print the middle element's data
    // Your code here

    return 0;
}

Hint

Use printf("Middle element: %d\n", slow->data); to print the middle node's value.