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 :



title-img 


                            Solution in C :

In C:

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

SinglyLinkedListNode* reverse(SinglyLinkedListNode* head) {
    struct SinglyLinkedListNode* new_head = NULL;
    struct SinglyLinkedListNode* temp;
    
    while(head != NULL){
        temp = head->next;
        head->next = new_head;
        new_head = head;
        head = temp;
    }
    
    return new_head;

}









In C++:

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


Node* Reverse(Node *head)
{
  // 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.

  static SinglyLinkedListNode reverse(SinglyLinkedListNode head) {
        SinglyLinkedListNode next = null;
        SinglyLinkedListNode current = head;
        SinglyLinkedListNode previous = null;
        while(current!=null)
        {
            next = current.next;
            current.next = previous;
            previous = current;
            current = next;
        }
        SinglyLinkedListNode singlyLinkedListNode = previous;
        return singlyLinkedListNode;
    }








In Python 3:

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

def Reverse(head):
    current = head
    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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