Objective: Reverse the given linked list.

Example:

```Input : ->30->25->20->15->10->5

Reversed : ->5->10->15->20->25->30```

NOTE : Click Reverse a Linked List – Part 2 to see the another implementation of this problem.

Approach:

Iterative:

• Create 3 nodes, currNode, PrevNode and nextNode.
• Initialize them as currNode = head; nextNode = null;prevNode = null;
• Now keep reversing the pointers one by one till currNode!=null.

while

`(currNode!=`

null

```){
nextNode = currNode.next;
currNode.next = prevNode;
prevNode = currNode;
currNode = nextNode;
}
```
• At the end set head = prevNode;
• See Example:

Note: If Image above is not clear, either zoom your browser or right click on image and open in a new tab.

• Recursive Approach:
• Take 3 nodes as Node ptrOne,Node ptrTwo, Node prevNode
• Reverse the ptrOne and ptrTwo
• Make a recursive call for reverseRecursion(ptrOne.next,ptrTwo.next,null)

Code:

```->30->25->20->15->10->5
Reverse Through Iteration
->5->10->15->20->25->30
___________________
Original Link List 2 : ->36->35->34->33->32->31
Reverse Through Recursion
->31->32->33->34->35->36

```