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‎en/1-1000/203-remove-linked-list-elements.md

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# 203. Remove Linked List Elements - Best Practices of LeetCode Solutions
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LeetCode link: [203. Remove Linked List Elements](https://leetcode.com/problems/remove-linked-list-elements),
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[203. 移除链表元素](https://leetcode.cn/problems/remove-linked-list-elements)
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LeetCode link: [203. Remove Linked List Elements](https://leetcode.com/problems/remove-linked-list-elements), difficulty: **Easy**.
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[中文题解](#中文题解)
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## LeetCode problem description
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## LeetCode description of "203. Remove Linked List Elements"
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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_.
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### [Example 1]
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- `0 <= val <= 50`
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## Intuition
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[中文题解](#中文题解)
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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`.
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To delete `d`, just set `p.next = p.next.next`.
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```
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// Welcome to create a PR to complete the code of this language, thanks!
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```
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## 问题描述
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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** 。
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### [Example 1]
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![](../../images/examples/203_1.jpg)
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**Input**: `head = [1,2,6,3,4,5,6], val = 6`
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**Output**: `[1,2,3,4,5]`
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## 中文题解
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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`。
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因为用到了`p.next.next`，所以循环条件应为`while (p.next != null)`，而不是`while (p != null)`。
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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。
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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。
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办法是引入`dummy`节点，`dummy.next = head`。

‎zh/1-1000/203-remove-linked-list-elements.md

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# 203. Remove Linked List Elements - Best Practices of LeetCode Solutions
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LeetCode link: [203. Remove Linked List Elements](https://leetcode.com/problems/remove-linked-list-elements),
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[203. 移除链表元素](https://leetcode.cn/problems/remove-linked-list-elements)
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# 203. 移除链表元素 - 力扣题解最佳实践
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力扣链接：[203. 移除链表元素](https://leetcode.cn/problems/remove-linked-list-elements), 难度: **简单**。
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[中文题解](#中文题解)
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## LeetCode problem description
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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_.
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## 力扣“203. 移除链表元素”问题描述
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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** 。
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### [Example 1]
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### [示例 1]
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![](../../images/examples/203_1.jpg)
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**Input**: `head = [1,2,6,3,4,5,6], val = 6`
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**输入**: `head = [1,2,6,3,4,5,6], val = 6`
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**Output**: `[1,2,3,4,5]`
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**输出**: `[1,2,3,4,5]`
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### [Example 2]
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**Input**: `head = [], val = 1`
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### [示例 2]
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**输入**: `head = [], val = 1`
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**Output**: `[]`
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**输出**: `[]`
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### [Example 3]
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**Input**: `head = [7,7,7,7], val = 7`
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### [示例 3]
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**输入**: `head = [7,7,7,7], val = 7`
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**Output**: `[]`
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**输出**: `[]`
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### [Constraints]
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- The number of nodes in the list is in the range `[0, 10000]`.
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### [约束]
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- 列表中的节点数目在范围 `[0, 10000]` 内
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- `1 <= Node.val <= 50`
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- `0 <= val <= 50`
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## Intuition
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[中文题解](#中文题解)
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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`.
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To delete `d`, just set `p.next = p.next.next`.
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## 思路
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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`。
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Because `p.next.next` is used, the loop condition should be `while (p.next != null)` instead of `while (p != null)`.
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因为用到了`p.next.next`，所以循环条件应为`while (p.next != null)`，而不是`while (p != null)`。
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But there is no node before the `head` node, which means that the `head` node needs to be treated specially.
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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。
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Is there a way to make the `head` node no longer special? In this way, there is no need to treat the `head` specially.
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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。
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The way is to introduce a `dummy` node, `dummy.next = head`.
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办法是引入`dummy`节点，`dummy.next = head`。
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## Complexity
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* Time: `O(n)`.
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* Space: `O(1)`.
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## 复杂度
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* 时间：`O(N)`。
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* 空间：`O(1)`。
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## Java
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```java
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```
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// Welcome to create a PR to complete the code of this language, thanks!
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```
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## 问题描述
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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** 。
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### [Example 1]
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![](../../images/examples/203_1.jpg)
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**Input**: `head = [1,2,6,3,4,5,6], val = 6`
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**Output**: `[1,2,3,4,5]`
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## 中文题解
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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`。
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因为用到了`p.next.next`，所以循环条件应为`while (p.next != null)`，而不是`while (p != null)`。
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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。
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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。
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办法是引入`dummy`节点，`dummy.next = head`。

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