# 大厂面试题：如何实现一个高效的单向链表逆序输出？

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

{
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;
}

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;
}
``````

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