反转一个单链表。

进阶:

链表可以迭代或递归地反转。你能否两个都实现一遍？

详见：https://leetcode.com/problems/reverse-linked-list/description/

Java实现：

迭代实现：



/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode reverseList(ListNode head) {
        if(head==null){
            return null;
        }
        ListNode cur=head;
        ListNode pre=null;
        ListNode next=null;
        while(cur!=null){
            next=cur.next;
            cur.next=pre;
            pre=cur;
            cur=next;
        }
        return pre;
    }
}


递归实现：



/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
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;
    }
}


 C++实现:

方法一：非递归



/**
 * 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==nullptr)
        {
            return nullptr;
        }
        ListNode *cur=head;
        ListNode *pre=nullptr;
        ListNode *next=nullptr;
        while(cur)
        {
            next=cur->next;
            cur->next=pre;
            pre=cur;
            cur=next;
        }
        return pre;
    }
};


 方法二：递归



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