# 移除链表元素

给你一个链表的头节点 <code>head</code> 和一个整数 <code>val</code> ，请你删除链表中所有满足 <code>Node.val == val</code> 的节点，并返回 <strong>新的头节点</strong> 。
<p> </p>

<p><strong>示例 1：</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/03/06/removelinked-list.jpg" style="width: 500px; height: 142px;" />
<pre>
<strong>输入：</strong>head = [1,2,6,3,4,5,6], val = 6
<strong>输出：</strong>[1,2,3,4,5]
</pre>

<p><strong>示例 2：</strong></p>

<pre>
<strong>输入：</strong>head = [], val = 1
<strong>输出：</strong>[]
</pre>

<p><strong>示例 3：</strong></p>

<pre>
<strong>输入：</strong>head = [7,7,7,7], val = 7
<strong>输出：</strong>[]
</pre>

<p> </p>

<p><strong>提示：</strong></p>

<ul>
	<li>列表中的节点数目在范围 <code>[0, 10<sup>4</sup>]</code> 内</li>
	<li><code>1 <= Node.val <= 50</code></li>
	<li><code>0 <= val <= 50</code></li>
</ul>


## template

```python

class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None


class Solution(object):
    def removeElements(self, head, val):
        """
        :type head: ListNode
        :type val: int
        :rtype: ListNode
        """
        p = ListNode(0)
        p.next = head
        while p.next:
            if p.next.val == val:
                if p.next == head:
                    head = head.next
                    p.next = head
                else:
                    p.next = p.next.next
            else:
                p = p.next
        return head
```

## 答案

```python

```

## 选项

### A

```python

```

### B

```python

```

### C

```python

```