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? 

考察：单链表的原地逆置，可以采用递归或者迭代的解法。

Method 1. 递归解法

/**
 * 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) {
        if (head == NULL || head->next == NULL)
            return head;
        
        ListNode *p = head;
        
        head = reverseList(p->next);
        p->next->next = p;
        p->next = NULL;
        return head;
        
        
    }
};

Method 2. 迭代解法 

/**
 * 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) {
        if (head == NULL)
            return head;
        
        ListNode *dummy = new ListNode(-1);
        dummy->next = head;
        ListNode *p = head;
        
        while(p->next) {
            ListNode *t = p->next;
            p->next = t->next;
            
            t->next = dummy->next;
            dummy->next = t;
        }
        
        return dummy->next;
    }
};

 Method 3. 头插法来逆置单链表

/**
 * 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) {
        if (head == NULL || head->next == NULL)
            return head;
        
        ListNode *dummy = new ListNode(-1);
        dummy->next = NULL;
        
        while (head) {
            ListNode *p = head;
            head = head->next;
            
            p->next = dummy->next;
            dummy->next = p;
        }
        return dummy->next;
    }
};

完，