# Find the middle of a given linked list in Java

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.

```Input : list: 1->2->4->5
x = 3
Output : 1->2->3->4->5

Input : list: 5->10->4->32->16
x = 41
Output : 5->10->4->41->32->16

Method 1(Using length of the linked list):
Find the number of nodes or length of the linked using one traversal. Let it be len. Calculate c = (len/2), if len is even, else c = (len+1)/2, if len is odd. Traverse again the first c nodes and insert the new node after the cth node.```
 `// Java implementation to insert node` `// at the middle of the linked list` `import` `java.util.*;` `import` `java.lang.*;` `import` `java.io.*;` `class` `LinkedList` `{` `    ``static` `Node head; ``// head of list` `    ``/* Node Class */` `    ``static` `class` `Node {` `        ``int` `data;` `        ``Node next;` `        ` `        ``// Constructor to create a new node` `        ``Node(``int` `d) {` `            ``data = d;` `            ``next = ``null``;` `        ``}` `    ``}` `    ``// function to insert node at the ` `    ``// middle of the linked list` `    ``static` `void` `insertAtMid(``int` `x)` `    ``{` `        ``// if list is empty` `        ``if` `(head == ``null``)` `            ``head = ``new` `Node(x);` `        ``else` `{` `            ``// get a new node` `            ``Node newNode = ``new` `Node(x);` `            ``Node ptr = head;` `            ``int` `len = ``0``;` `            ``// calculate length of the linked list` `            ``//, i.e, the number of nodes` `            ``while` `(ptr != ``null``) {` `                ``len++;` `                ``ptr = ptr.next;` `            ``}` `            ``// 'count' the number of nodes after which` `            ``// the new node is to be inserted` `            ``int` `count = ((len % ``2``) == ``0``) ? (len / ``2``) :` `                                        ``(len + ``1``) / ``2``;` `            ``ptr = head;` `            ``// 'ptr' points to the node after which ` `            ``// the new node is to be inserted` `            ``while` `(count-- > ``1``)` `                ``ptr = ptr.next;` `            ``// insert the 'newNode' and adjust ` `            ``// the required links` `            ``newNode.next = ptr.next;` `            ``ptr.next = newNode;` `        ``}` `    ``}` `    ``// function to display the linked list` `    ``static` `void` `display()` `    ``{` `        ``Node temp = head;` `        ``while` `(temp != ``null``) ` `        ``{` `            ``System.out.print(temp.data + ``" "``);` `            ``temp = temp.next;` `        ``}` `    ``}` `    ``// Driver program to test above` `    ``public` `static` `void` `main (String[] args) ` `    ``{ ` `        ``// Creating the list 1.2.4.5` `        ``head = ``null``;` `        ``head = ``new` `Node(``1``);` `        ``head.next = ``new` `Node(``2``);` `        ``head.next.next = ``new` `Node(``4``);` `        ``head.next.next.next = ``new` `Node(``5``);` `        ` `        ``System.out.println(``"Linked list before "``+` `                           ``"insertion: "``);` `        ``display();` `        ``int` `x = ``3``;` `        ``insertAtMid(x);` `        ``System.out.println(``"\nLinked list after"``+` `                           ``" insertion: "``);` `        ``display();` `    ``} ` `}` `// This article is contributed by Chhavi`

Output:

```Linked list before insertion: 1 2 4 5
Linked list after insertion: 1 2 3 4 5
```

Time Complexity: O(n)

Method 2(Using two pointers):
Based on the tortoise and hare algorithm which uses two pointers, one known as slow and the other known as fast. This algorithm helps in finding the middle node of the linked list. It is explained in the front and black split procedure of this post. Now, you can insert the new node after the middle node obtained from the above process. This approach requires only a single traversal of the list.

 `// Java implementation to insert node ` `// at the middle of the linked list` `import` `java.util.*;` `import` `java.lang.*;` `import` `java.io.*;` `class` `LinkedList` `{` `    ``static` `Node head; ``// head of list` `    ``/* Node Class */` `    ``static` `class` `Node {` `        ``int` `data;` `        ``Node next;` `        ` `        ``// Constructor to create a new node` `        ``Node(``int` `d) {` `            ``data = d;` `            ``next = ``null``;` `        ``}` `    ``}` `    ``// function to insert node at the ` `    ``// middle of the linked list` `    ``static` `void` `insertAtMid(``int` `x)` `    ``{` `        ``// if list is empty` `        ``if` `(head == ``null``)` `        ``head = ``new` `Node(x);` `        ``else` `{` `            ``// get a new node` `            ``Node newNode = ``new` `Node(x);` `            ``// assign values to the slow ` `            ``// and fast pointers` `            ``Node slow = head;` `            ``Node fast = head.next;` `            ``while` `(fast != ``null` `&& fast.next ` `                                  ``!= ``null``) ` `            ``{` `                ``// move slow pointer to next node` `                ``slow = slow.next;` `                ``// move fast pointer two nodes ` `                ``// at a time` `                ``fast = fast.next.next;` `            ``}` `            ``// insert the 'newNode' and adjust ` `            ``// the required links` `            ``newNode.next = slow.next;` `            ``slow.next = newNode;` `        ``}` `    ``}` `    ``// function to display the linked list` `    ``static` `void` `display()` `    ``{` `        ``Node temp = head;` `        ``while` `(temp != ``null``) ` `        ``{` `            ``System.out.print(temp.data + ``" "``);` `            ``temp = temp.next;` `        ``}` `    ``}` `    ``// Driver program to test above` `    ``public` `static` `void` `main (String[] args) ` `    ``{ ` `        ``// Creating the list 1.2.4.5` `        ``head = ``null``;` `        ``head = ``new` `Node(``1``);` `        ``head.next = ``new` `Node(``2``);` `        ``head.next.next = ``new` `Node(``4``);` `        ``head.next.next.next = ``new` `Node(``5``);` `        ` `        ``System.out.println(``"Linked list before"``+` `                           ``" insertion: "``);` `        ``display();` `        ``int` `x = ``3``;` `        ``insertAtMid(x);` `        ``System.out.println(``"\nLinked list after"``+` `                           ``" insertion: "``);` `        ``display();` `    ``} ` `}` `// This article is contributed by Chhavi`

Output:

```Linked list before insertion: 1 2 4 5
Linked list after insertion: 1 2 3 4 5
```

Time Complexity: O(n)