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1 | | -#203. Remove Linked List Elements - Best Practices of LeetCode Solutions |
2 | | -LeetCode link:[203. Remove Linked List Elements](https://leetcode.com/problems/remove-linked-list-elements), |
3 | | -[203. 移除链表元素](https://leetcode.cn/problems/remove-linked-list-elements) |
| 1 | +#203. 移除链表元素 - 力扣题解最佳实践 |
| 2 | +力扣链接：[203. 移除链表元素](https://leetcode.cn/problems/remove-linked-list-elements), 难度:**简单**。 |
4 | 3 |
|
5 | | -[中文题解](#中文题解) |
6 | | - |
7 | | -##LeetCode problem description |
8 | | -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_. |
| 4 | +##力扣“203. 移除链表元素”问题描述 |
| 5 | +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** 。 |
9 | 6 |
|
10 | | -###[Example 1] |
| 7 | +###[示例 1] |
11 | 8 |  |
12 | 9 |
|
13 | | -**Input**:`head = [1,2,6,3,4,5,6], val = 6` |
| 10 | +**输入**:`head = [1,2,6,3,4,5,6], val = 6` |
14 | 11 |
|
15 | | -**Output**:`[1,2,3,4,5]` |
| 12 | +**输出**:`[1,2,3,4,5]` |
16 | 13 |
|
17 | | -###[Example 2] |
18 | | -**Input**:`head = [], val = 1` |
| 14 | +###[示例 2] |
| 15 | +**输入**:`head = [], val = 1` |
19 | 16 |
|
20 | | -**Output**:`[]` |
| 17 | +**输出**:`[]` |
21 | 18 |
|
22 | | -###[Example 3] |
23 | | -**Input**:`head = [7,7,7,7], val = 7` |
| 19 | +###[示例 3] |
| 20 | +**输入**:`head = [7,7,7,7], val = 7` |
24 | 21 |
|
25 | | -**Output**:`[]` |
| 22 | +**输出**:`[]` |
26 | 23 |
|
27 | | -###[Constraints] |
28 | | --The number of nodes in the list is in the range`[0, 10000]`. |
| 24 | +###[约束] |
| 25 | +-列表中的节点数目在范围`[0, 10000]` 内 |
29 | 26 | -`1 <= Node.val <= 50` |
30 | 27 | -`0 <= val <= 50` |
31 | 28 |
|
32 | | -##Intuition |
33 | | -[中文题解](#中文题解) |
34 | | - |
35 | | -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`. |
36 | | - |
37 | | -To delete`d`, just set`p.next = p.next.next`. |
| 29 | +##思路 |
| 30 | +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`。 |
38 | 31 |
|
39 | | -Because`p.next.next` is used, the loop condition should be`while (p.next != null)` instead of`while (p != null)`. |
| 32 | +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)`。 |
40 | 33 |
|
41 | | -But there is no node before the`head` node, which means that the`head` node needs to be treated specially. |
| 34 | +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。 |
42 | 35 |
|
43 | | -Is there a way to make the`head` node no longer special? In this way, there is no need to treat the`head` specially. |
| 36 | +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。 |
44 | 37 |
|
45 | | -The way is to introduce a`dummy` node,`dummy.next = head`. |
| 38 | +办法是引入`dummy`节点，`dummy.next = head`。 |
46 | 39 |
|
47 | | -##Complexity |
48 | | -*Time:`O(n)`. |
49 | | -*Space:`O(1)`. |
| 40 | +##复杂度 |
| 41 | +*时间：`O(N)`。 |
| 42 | +*空间：`O(1)`。 |
50 | 43 |
|
51 | 44 | ##Java |
52 | 45 | ```java |
|
272 | 265 | ``` |
273 | 266 | // Welcome to create a PR to complete the code of this language, thanks! |
274 | 267 | ``` |
275 | | - |
276 | | -##问题描述 |
277 | | -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** 。 |
278 | | - |
279 | | -###[Example 1] |
280 | | - |
281 | | - |
282 | | -**Input**:`head = [1,2,6,3,4,5,6], val = 6` |
283 | | - |
284 | | -**Output**:`[1,2,3,4,5]` |
285 | | - |
286 | | -##中文题解 |
287 | | -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`。 |
288 | | - |
289 | | -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)`。 |
290 | | - |
291 | | -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。 |
292 | | - |
293 | | -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。 |
294 | | - |
295 | | -办法是引入`dummy`节点，`dummy.next = head`。 |