# 203. 移除链表元素 - LeetCode Python/Java/C++/JS/C#/Go/Ruby 题解 访问原文链接：[203. 移除链表元素 - LeetCode Python/Java/C++/JS/C#/Go/Ruby 题解](https://leetcode.to/zh/leetcode/203-remove-linked-list-elements)，体验更佳！ 力扣链接：[203. 移除链表元素](https://leetcode.cn/problems/remove-linked-list-elements), 难度等级：**简单**。 ## LeetCode “203. 移除链表元素”问题描述 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** 。 ### [示例 1] ![](../../images/examples/203_1.jpg) **输入**: `head = [1,2,6,3,4,5,6], val = 6` **输出**: `[1,2,3,4,5]` ### [示例 2] **输入**: `head = [], val = 1` **输出**: `[]` ### [示例 3] **输入**: `head = [7,7,7,7], val = 7` **输出**: `[]` ### [约束] - 列表中的节点数目在范围 `[0, 10000]` 内 - `1 <= Node.val <= 50` - `0 <= val <= 50` ## 思路 - 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`。 - 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)`。 - 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。
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办法是引入`dummy`节点，`dummy.next = head`。

## 复杂度 - 时间复杂度: `O(N)`. - 空间复杂度: `O(1)`. ## Java ```java /** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode() {} * ListNode(int val) { this.val = val; } * ListNode(int val, ListNode next) { this.val = val; this.next = next; } * } */ class Solution { public ListNode removeElements(ListNode head, int val) { var dummyHead = new ListNode(); dummyHead.next = head; var node = dummyHead; while (node.next != null) { if (node.next.val == val) { node.next = node.next.next; } else { node = node.next; } } return dummyHead.next; } } ``` ## Python ```python # Definition for singly-linked list. # class ListNode: # def __init__(self, val=0, next=None): # self.val = val # self.next = next 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 ``` ## C++ ```cpp /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode() : val(0), next(nullptr) {} * ListNode(int x) : val(x), next(nullptr) {} * ListNode(int x, ListNode *next) : val(x), next(next) {} * }; */ class Solution { public: ListNode* removeElements(ListNode* head, int val) { auto dummyHead = new ListNode(); dummyHead->next = head; auto node = dummyHead; while (node->next != nullptr) { if (node->next->val == val) { auto next_node = node->next; node->next = node->next->next; delete next_node; } else { node = node->next; } } return dummyHead->next; } }; ``` ## JavaScript ```javascript /** * Definition for singly-linked list. * function ListNode(val, next) { * this.val = (val===undefined ? 0 : val) * this.next = (next===undefined ? null : next) * } */ var removeElements = function (head, val) { const dummyHead = new ListNode() dummyHead.next = head let node = dummyHead while (node.next != null) { if (node.next.val == val) { node.next = node.next.next } else { node = node.next } } return dummyHead.next }; ``` ## C# ```csharp /** * Definition for singly-linked list. * public class ListNode { * public int val; * public ListNode next; * public ListNode(int val=0, ListNode next=null) { * this.val = val; * this.next = next; * } * } */ public class Solution { public ListNode RemoveElements(ListNode head, int val) { var dummyHead = new ListNode(); dummyHead.next = head; var node = dummyHead; while (node.next != null) { if (node.next.val == val) { node.next = node.next.next; } else { node = node.next; } } return dummyHead.next; } } ``` ## Go ```go /** * Definition for singly-linked list. * type ListNode struct { * Val int * Next *ListNode * } */ func removeElements(head *ListNode, val int) *ListNode { dummyHead := &ListNode{} dummyHead.Next = head node := dummyHead for node.Next != nil { if node.Next.Val == val { node.Next = node.Next.Next } else { node = node.Next } } return dummyHead.Next } ``` ## Ruby ```ruby # Definition for singly-linked list. # class ListNode # attr_accessor :val, :next # def initialize(val = 0, _next = nil) # @val = val # @next = _next # end # end def remove_elements(head, val) dummy_head = ListNode.new dummy_head.next = head node = dummy_head until node.next.nil? if node.next.val == val node.next = node.next.next else node = node.next end end dummy_head.next end ``` ## Other languages ```java // Welcome to create a PR to complete the code of this language, thanks! ``` 亲爱的力扣人，为了您更好的刷题体验，请访问 [LeetCode.to](https://leetcode.to/zh)。 本站敢称力扣题解最佳实践，终将省你大量刷题时间！ 原文链接：[203. 移除链表元素 - LeetCode Python/Java/C++/JS/C#/Go/Ruby 题解](https://leetcode.to/zh/leetcode/203-remove-linked-list-elements). GitHub 仓库: [leetcode-python-java](https://github.com/leetcode-python-java/leetcode-python-java).