C Program For Finding The Middle Element Of A Given Linked List - GeeksforGeeks

08 Dec, 2021
Given a singly linked list, find the middle of the linked list. For example, if the given linked list is 1->2->3->4->5 then the output should be 3. If there are even nodes, then there would be two middle nodes, we need to print the second middle element. For example, if given linked list is 1->2->3->4->5->6 then the output should be 4. 
Method 1: Traverse the whole linked list and count the no. of nodes. Now traverse the list again till count/2 and return the node at count/2. 
Method 2: Traverse linked list using two pointers. Move one pointer by one and the other pointers by two. When the fast pointer reaches the end slow pointer will reach the middle of the linked list.
Below image shows how printMiddle function works in the code :
C
// C program to find middle of linked list#include<stdio.h> #include<stdlib.h>   // Link list nodestruct Node {     int data;     struct Node* next; };   // Function to get the middle of // the linked listvoid printMiddle(struct Node *head) {     struct Node *slow_ptr = head;     struct Node *fast_ptr = head;       if (head!=NULL)     {         while (fast_ptr != NULL &&                fast_ptr->next != NULL)         {             fast_ptr = fast_ptr->next->next;             slow_ptr = slow_ptr->next;         }         printf("The middle element is [%d]",                 slow_ptr->data);     } }   void push(struct Node** head_ref,           int new_data) {     // Allocate node     struct Node* new_node =            (struct Node*) malloc(sizeof(struct Node));       // Put in the data    new_node->data = new_data;       // Link the old list off the new node    new_node->next = (*head_ref);       // Move the head to point to the new node     (*head_ref) = new_node; }   // A utility function to print a given//  linked list void printList(struct Node *ptr) {     while (ptr != NULL)     {         printf("%d->", ptr->data);         ptr = ptr->next;     }     printf("NULL"); }   // Driver codeint main() {     // Start with the empty list     struct Node* head = NULL;     int i;       for (i = 5; i > 0; i--)     {         push(&head, i);         printList(head);         printMiddle(head);     }     return 0; } 
Output:
5->NULL
The middle element is [5]

4->5->NULL
The middle element is [5]

3->4->5->NULL
The middle element is [4]

2->3->4->5->NULL
The middle element is [4]

1->2->3->4->5->NULL
The middle element is [3]
Method 3: Initialize mid element as head and initialize a counter as 0. Traverse the list from head, while traversing increment the counter and change mid to mid->next whenever the counter is odd. So the mid will move only half of the total length of the list. Thanks to Narendra Kangralkar for suggesting this method. 
C
// C program to implement the // above approach#include <stdio.h>#include <stdlib.h>  // Link list nodestruct node {    int data;    struct node* next;};  // Function to get the middle of // the linked listvoid printMiddle(struct node* head){    int count = 0;    struct node* mid = head;      while (head != NULL)    {        // Update mid, when 'count'         // is odd number        if (count & 1)            mid = mid->next;          ++count;        head = head->next;    }      // If empty list is provided    if (mid != NULL)        printf("The middle element is [%d]",                 mid->data);}  void push(struct node** head_ref,           int new_data){    // Allocate node     struct node* new_node =            (struct node*)malloc(sizeof(struct node));      // Put in the data      new_node->data = new_data;      // Link the old list off the new node     new_node->next = (*head_ref);      // Move the head to point to the new node    (*head_ref) = new_node;}  // A utility function to print a // given linked listvoid printList(struct node* ptr){    while (ptr != NULL)     {        printf("%d->", ptr->data);        ptr = ptr->next;    }    printf("NULL");}  // Driver codeint main(){    // Start with the empty list     struct node* head = NULL;    int i;      for (i = 5; i > 0; i--)     {        push(&head, i);        printList(head);        printMiddle(head);    }    return 0;}
Output:
5->NULL
The middle element is [5]

4->5->NULL
The middle element is [5]

3->4->5->NULL
The middle element is [4]

2->3->4->5->NULL
The middle element is [4]

1->2->3->4->5->NULL
The middle element is [3]
Please refer complete article on Find the middle of a given linked list for more details!
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