91 lines
1.9 KiB
Markdown
91 lines
1.9 KiB
Markdown
## 题目地址
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https://leetcode.com/problems/remove-linked-list-elements/description/
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## 题目描述
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```
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Remove all elements from a linked list of integers that have value val.
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Example:
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Input: 1->2->6->3->4->5->6, val = 6
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Output: 1->2->3->4->5
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```
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## 思路
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这个一个链表基本操作的题目，思路就不多说了。
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## 关键点解析
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- 链表的基本操作（删除指定节点）
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- 虚拟节点 dummy 简化操作
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> 其实设置 dummy 节点就是为了处理特殊位置（头节点），这这道题就是如果头节点是给定的需要删除的节点呢？
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为了保证代码逻辑的一致性，即不需要为头节点特殊定制逻辑，才采用的虚拟节点。
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- 如果连续两个节点都是要删除的节点，这个情况容易被忽略。
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eg:
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```js
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// 只有下个节点不是要删除的节点才更新 current
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if (!next || next.val !== val) {
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current = next;
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}
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```
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## 代码
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* 语言支持：JS，Python
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Javascript Code:
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```js
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/**
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* @param {ListNode} head
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* @param {number} val
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* @return {ListNode}
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*/
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var removeElements = function(head, val) {
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const dummy = {
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next: head
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}
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let current = dummy;
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while(current && current.next) {
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let next = current.next;
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if (next.val === val) {
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current.next = next.next;
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next = next.next;
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}
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if (!next || next.val !== val) {
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current = next;
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}
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}
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return dummy.next;
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};
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```
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Python Code:
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```python
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# Definition for singly-linked list.
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# class ListNode:
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# def __init__(self, x):
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# self.val = x
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# self.next = None
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class Solution:
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def removeElements(self, head: ListNode, val: int) -> ListNode:
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prev = ListNode(0)
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prev.next = head
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cur = prev
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while cur.next:
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if cur.next.val == val:
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cur.next = cur.next.next
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else:
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cur = cur.next
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return prev.next
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```
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