如何实现一个高效的单向链表逆序输出

问题：如何实现一个高效的单向链表逆序输出？
出题人：阿里巴巴出题专家：昀龙/阿里云弹性人工智能负责人
参考答案：下面是其中一种写法，也可以有不同的写法，比如递归等。供参考。

typedef struct node{
    int           data;
    struct node*  next;
    node(int d):data(d), next(NULL){}}node;

void reverse(node* head)
{if(head == NULL){return;}

    node* pleft = NULL;
    node* pcurrent = head;
    node* pright = head->next;

    while(pright){
        pcurrent->next = pleft;
        node *ptemp = pright->next;
        pright->next = pcurrent;
        pleft = pcurrent;
        pcurrent = pright;
        pright = ptemp;
    }

    while(pcurrent != NULL){
        cout<< pcurrent->data << "\t";
        pcurrent = pcurrent->next;
    }
}

class Solution<T> {public void reverse(ListNode<T> head) {if (head == null || head.next == null) {return ;}
       ListNode<T> currentNode = head;
       Stack<ListNode<T>> stack = new Stack<>();
       while (currentNode != null) {stack.push(currentNode);
           ListNode<T> tempNode = currentNode.next;
           currentNode.next = null; // 断开连接
           currentNode = tempNode;
       }

       head = stack.pop();
       currentNode = head;

       while (!stack.isEmpty()) {currentNode.next = stack.pop();
           currentNode = currentNode.next;
       }
    }
}

class ListNode<T>{
    T val;
    public ListNode(T val) {this.val = val;}
    ListNode<T> next;
}