Reverse a linked list

Problem Statement

This challenge is part of a tutorial track by MyCodeSchool and is accompanied by a video lesson.

You’re 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.

Input Format 
You have to complete the Node* Reverse(Node* head) method which takes one argument - the head of the linked list. You should NOT read any input from stdin/console.

Output Format 
Change the next pointers of the nodes that their order is reversed and return the head of the reversed linked list. Do NOT print anything to stdout/console.

Sample Input

NULL 
2 --> 3 --> NULL

Sample Output

NULL
3 --> 2 --> NULL

Explanation 
1. Empty list remains empty 

2. List is reversed from 2,3 to 3,2

/*
  Insert Node at the end of a linked list 
  head pointer input could be NULL as well for empty list
  Node is defined as 
  class Node {
     int data;
     Node next;
  }
*/
    // This is a "method-only" submission. 
    // You only need to complete this method. 

//三個指針，畫圖，q始終指向頭，head沒動，p在head後面
Node Reverse(Node head) {
    
    if(head==null||head.next==null)
        return head;

    Node p=new Node();
    p=head.next;
    Node q=new Node();
    q=head;
    while(p!=null){
        head.next=p.next;
        p.next=q;
        q=p;
        p=head.next;
    }
    return q;
}

//用棧，對鏈表的值替換
Node Reverse(Node head) {
    if(head==null||head.next==null)
        return head;

    Stack<Integer> s=new Stack<>();
    Node p=new Node();
    p=head;
    while(p!=null){
        s.push(p.data);
        p=p.next;
    }
    p=head;
    while(p!=null){
        p.data=s.pop();
        p=p.next;
    }
    return head;
}