# 反转链表

给你单链表的头节点 <code>head</code> ，请你反转链表，并返回反转后的链表。
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<p> </p>

<p><strong>示例 1：</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/02/19/rev1ex1.jpg" style="width: 542px; height: 222px;" />
<pre>
<strong>输入：</strong>head = [1,2,3,4,5]
<strong>输出：</strong>[5,4,3,2,1]
</pre>

<p><strong>示例 2：</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/02/19/rev1ex2.jpg" style="width: 182px; height: 222px;" />
<pre>
<strong>输入：</strong>head = [1,2]
<strong>输出：</strong>[2,1]
</pre>

<p><strong>示例 3：</strong></p>

<pre>
<strong>输入：</strong>head = []
<strong>输出：</strong>[]
</pre>

<p> </p>

<p><strong>提示：</strong></p>

<ul>
	<li>链表中节点的数目范围是 <code>[0, 5000]</code></li>
	<li><code>-5000 <= Node.val <= 5000</code></li>
</ul>

<p> </p>

<p><strong>进阶：</strong>链表可以选用迭代或递归方式完成反转。你能否用两种方法解决这道题？</p>
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## template

```cpp
#include <bits/stdc++.h>
using namespace std;

struct ListNode
{
    int val;
    ListNode *next;
    ListNode(int x) : val(x), next(NULL) {}
};

class Solution
{
public:
    ListNode *reverseList(ListNode *head)
    {
        if (head == NULL)
            return NULL;
        ListNode *node = NULL;
        ListNode *temp = head;
        ListNode *temp1 = head;
        while (true)
        {
            if (temp->next == NULL)
            {
                temp1 = temp;
                temp1->next = node;
                break;
            }
            temp1 = temp;
            temp = temp->next;
            temp1->next = node;
            node = temp1;
        }
        return temp1;
    }
};


```

## 答案

```cpp

```

## 选项

### A

```cpp

```

### B

```cpp

```

### C

```cpp

```