# 206.Reverse Linked List

**206.Reverse Linked List**

难度:Easy

> 反转一个单链表。

```
示例:

输入: 1->2->3->4->5->NULL
输出: 5->4->3->2->1->NULL
```

进阶: 你可以迭代或递归地反转链表。你能否用两种方法解决这道题？

递归： 由于递归都是一层一层处理的，如果要递归，首先要处理最后一个节点，然后将剩余的继续反转，代码如下:

```
/**
 * 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) {
        if(!head) return NULL;
        if(!head->next) {head->next=NULL; return head;}
        ListNode* pt=head;
        while(pt->next->next)
        {
            pt=pt->next;
        }
        ListNode* t=pt->next;
        pt->next=NULL;
        t->next=reverseList(head);
        return t;
        
        
    }
};
```

迭代： 在遍历列表时，将当前节点的下一个节点之前前一个元素，实现反转：

```
/**
 * 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) return NULL;
        ListNode* prev=NULL;
        ListNode* curr=head;
        while(curr!=NULL)
        {
           ListNode* tmp=curr->next;
            curr->next=prev;
            prev=curr;
            curr=tmp;
        }

        

        return prev;
    }
    
};
```


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