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.
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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