# C++ Program to Find the middle of a given linked list

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#### In this Program, you’ll learn how to Find the middle of a given linked list.

To nicely understand this example to Find the middle of a given linked list, you should have the knowledge of  C++ programming.

Given a singly linked list, find the middle of the linked list. For example, if given linked list is 1->2->3->4->5 then output should be 3.

If there are even nodes, then there would be two middle nodes, we need to print second middle element. For example, if given linked list is 1->2->3->4->5->6 then output should be 4.

#### Q. Write a Program to find the middle element of the linked list in C++

• 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 other pointers by two. When the fast pointer reaches end slow pointer will reach the middle of the linked list.

```#include<stdio.h>
#include<stdlib.h>

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

/* Function to get the middle of the linked list*/
{

{
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]\n\n", 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 */

/* move the head to point to the 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\n");
}

/* Drier program to test above function*/
int main()
{
int i;

for (i=5; i>0; i--)
{
}

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 the 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.

```#include<stdio.h>
#include<stdlib.h>

struct node
{
int data;
struct node* next;
};

/* Function to get the middle of the linked list*/
{
int count = 0;

{
/* update mid, when 'count' is odd number */
if (count & 1)
mid = mid->next;

++count;
}

/* if empty list is provided */
if (mid != NULL)
printf("The middle element is [%d]\n\n", 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 */

/* move the head to point to the 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\n");
}

/* Drier program to test above function*/
int main()
{
int i;

for (i=5; i>0; i--)
{
}

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]```

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