题目描述如下

Reverse a singly linked list.

Example:

Input: 1->2->3->4->5->NULL
Output: 5->4->3->2->1->NULL
Follow up:

A linked list can be reversed either iteratively or recursively. Could you implement both?

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/reverse-linked-list

逆转链表，其实就是先进后出，所以可以用栈实现。当然简单的头插法也可以，这里就不写了。

/**
 * DeFinition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int X) : val(X),next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* reverseList(ListNode* head) {
        stack<ListNode *> nodeStack;
        if(head==NULL)
            return NULL;
        ListNode * pNode=head;
        while(pNode!= NULL)
        {
            nodeStack.push(pNodE);
            pNode=pNode->next;
        }
        ListNode * newHead=NULL;
        newHead=nodeStack.top();
        nodeStack.pop();
        ListNode * pre=newHead;
        while(!nodeStack.empty())
        {
            ListNode * p=nodeStack.top();
            nodeStack.pop();
            pre->next=p;
            pre=p;
        }
        pre->next=NULL;
        return newHead;
    }
};

递归本质也是一个栈结构，所以也可以用递归实现，这里记录一下：

/**
 * DeFinition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int X) : val(X),next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* reverseList(ListNode* head) {
        ListNode * newHead=NULL;
        if(head ==NULL)  //链表是空的就返回空
            return NULL;
        ListNode * nextNode = head->next;
        ListNode * res=reverseList(nextNodE);
        if(nextNode ==NULL)
            return head;
        nextNode->next = head;
        head->next=NULL;
        
        return res;
    }
};