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

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution(object):def removeElements(self, head, val):""":type head: ListNode:type val: int:rtype: ListNode"""dummy_head = ListNode(next = head)current = dummy_headwhile current.next:if current.next.val == val:current.next = current.next.nextelse:current = current.nextreturn dummy_head.next