Commit 275a673

LeetCode problem link: [203. Remove Linked List Elements](https://leetcode.com/problems/remove-linked-list-elements)

Given the `head` of a linked list and an integer `val`, remove all the nodes of the linked list that has `Node.val == val`, and return _the new head_.

**Input**: `head = [1,2,6,3,4,5,6], val = 6`

**Output**: `[1,2,3,4,5]`

**Input**: `head = [], val = 1`

**Output**: `[]`

**Input**: `head = [7,7,7,7], val = 7`

**Output**: `[]`

- The number of nodes in the list is in the range `[0, 10000]`.

- `1 <= Node.val <= 50`

- `0 <= val <= 50`

Assume that the node to be deleted in the linked list is `d`, and the previous node of `d` is `p`, so `p.next` is `d`.

To delete `d`, just set `p.next = p.next.next`.

But there is no node before the `head` node, which means that the `head` node needs to be treated specially.

Is there a way to make the `head` node no longer special? In this way, there is no need to treat the `head` specially.

The way is to introduce a `dummy` node, `dummy.next = head`.

* Time: `O(n)`.

* Space: `O(1)`.

class Solution:

def removeElements(self, head: Optional[ListNode], val: int) -> Optional[ListNode]:

dummy_head = ListNode()

dummy_head.next = head

node = dummy_head

while node.next:

if node.next.val == val:

node.next = node.next.next

else:

node = node.next

return dummy_head.next

假设链表中待删除的节点是`d`，`d`的前一个节点是`p`，所以`p.next`就是`d`。 删除`d`，只需要把`p.next = p.next.next`。

但`head`节点前面没有节点，这就意味着需要对`head`节点进行特殊处理。

是否有方法能够让`head`节点的不再特殊呢？

这样就不需要特殊处理`head`了。

办法是引入`dummy`节点，`dummy.next = head`。

* Cannot delete an item. Actually, it is overwrite. Delete a item of array will call the latter items move 1 to left. So it is `O(n)` time complexity.