# Java Algorithm – Find the middle of a given linked list

Java Algorithm - Find the middle of a given linked list - Linked List - Given a singly linked list, find middle of the linked list

Given a singly linked list, find 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

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

Java Programming:

``````// Java program to find middle of linked list
{

class Node
{
int data;
Node next;
Node(int d)
{
data = d;
next = null;
}
}

/* Function to print middle of linked list */
void printMiddle()
{
{
while (fast_ptr != null && fast_ptr.next != null)
{
fast_ptr = fast_ptr.next.next;
slow_ptr = slow_ptr.next;
}
System.out.println("The middle element is [" +
slow_ptr.data + "] \n");
}
}

/* Inserts a new Node at front of the list. */
public void push(int new_data)
{
/* 1 & 2: Allocate the Node &
Put in the data*/
Node new_node = new Node(new_data);

/* 3. Make next of new Node as head */

/* 4. Move the head to point to new Node */
}

/* This function prints contents of linked list
starting from  the given node */
public void printList()
{
while (tnode != null)
{
System.out.print(tnode.data+"->");
tnode = tnode.next;
}
System.out.println("NULL");
}

public static void main(String [] args)
{
for (int i=5; i>0; --i)
{
llist.push(i);
llist.printList();
llist.printMiddle();
}
}
}
// This code is contributed by Rajat Mishra``````

Output:

```5->NULL
The middle element is 

4->5->NULL
The middle element is 

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

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

1->2->3->4->5->NULL
The middle element is ```

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.

READ  C Algorithm - Find n’th node from the end of a Linked List

C Programming:

``````#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 

4->5->NULL
The middle element is 

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

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

1->2->3->4->5->NULL
The middle element is ``` 