You are given the pointer to the head node of a linked list and an integer to add to the list. Create a new node with the given integer. Insert this node at the tail of the linked list and return the head node of the linked list formed after inserting this new node. The given head pointer may be null, meaning that the initial list is empty.

**Input Format**

You have to complete the

method. It takes two arguments: the head of the linked list and the integer to insert at tail. You should **SinglyLinkedListNode insertAtTail(SinglyLinkedListNode head, int data)****not** read any input from the stdin/console.

The input is handled by code editor and is as follows:

The first line contains an integer * n*, denoting the elements of the linked list.

The next

*lines contain an integer each, denoting the element that needs to be inserted at tail.*

**n****Constraints**

**1 <= n <= 1000****1 <= list’i <= 1000**

**Output Format**

Insert the new node at the tail and just

the head of the updated linked list. Do **return****not** print anything to stdout/console.

The output is handled by code in the editor and is as follows:

Print the elements of the linked list from head to tail, each in a new line.

**Sample Input**

```
5
141
302
164
530
474
```

**Sample Output**

```
141
302
164
530
474
```

**Explanation**

First the linked list is NULL. After inserting 141, the list is* 141 -> NULL*.

After inserting

*, the list is*

**302***.*

**141 -> 302 -> NULL**After inserting

*, the list is*

**164***.*

**141 -> 302 -> 164 -> NULL**After inserting

*, the list is*

**530**

**141 -> 302 -> 164 -> 530 ->***After inserting*

**NULL.***, the list is*

**474***, which is the final list.*

**141 -> 302 -> 164 -> 530 -> 474 -> NULL****Code:**

```
static SinglyLinkedListNode insertNodeAtTail(SinglyLinkedListNode head, int data) {
SinglyLinkedListNode temp = head;
SinglyLinkedListNode insertNode = new SinglyLinkedListNode(data);
if(head == null) {
return insertNode;
}
while(temp.next != null) {
temp = temp.next;
}
temp.next = insertNode;
insertNode.next = null;
return head;
}
```

Time complexity of insertat tail is * O(n)* where

*is the number of nodes in linked list. Since there is a loop from*

**n***to*

**head***, the function does*

**end***work.*

**O(n)**This method can also be optimized to work in

**by keeping an extra pointer to tail of linked list.**

*O(1)*Hope you guys understand the simple solution of inserting a node at tail in linked list in java.

In next blog we will see other operations on linked list till then Bye Bye..!

**Thanks for reading…!**

**Happy Coding..!**

**Thank You…!**

Categories: HackerRank Questions, JAVA

why return head?

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