# Reverse a linked list

### Problem Statement :

```Given the pointer to the head node of a linked list, change the next pointers of the nodes so that their order is reversed. The head pointer given may be null meaning that the initial list is empty.

Example:
head references the list 1->2->3->Null.

Manipulate the next pointers of each node in place and return head, now referencing the head of the list 3->2->1->Null.

Function Description:

Complete the reverse function in the editor below.

reverse has the following parameter:
SinglyLinkedListNode pointer head: a reference to the head of a list
Returns
SinglyLinkedListNode pointer: a reference to the head of the reversed list

Input Format:

The first line contains an integer t, the number of test cases.
Each test case has the following format:
1. The first line contains an integer n, the number of elements in the linked list.
2. Each of the next n lines contains an integer, the data values of the elements in the linked list.

Constraints:
1.   1<=t<=10
2.   1<=n<=1000
3.   1<=list[i]<=1000```

### Solution :

```                            ```Solution in C :

In C:

//the following function is all that is needed to complete the
//challenge in hackerrank platform.

}

}

In C++:

//the following function is all that is needed to complete the
//challenge in hackerrank platform.

{
// Complete this method
Node* prev   = NULL;
Node* current = head;
Node* next;

while (current != NULL)
{
next  = current->next;
current->next = prev;
prev = current;
current = next;
}

return prev;
}

In Java:

//the following method is all that is needed to complete the
//challenge in hackerrank platform.

SinglyLinkedListNode next = null;
SinglyLinkedListNode previous = null;
while(current!=null)
{
next = current.next;
current.next = previous;
previous = current;
current = next;
}
}

In Python 3:

//the following method is all that is needed to complete the
//challenge in hackerrank platform.

prev = None
next = None
while current is not None:
next = current.next
current.next = prev;
prev = current;
current = next;

return prev```
```

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