https://leetcode.com/problems/remove-linked-list-elements/description/
Remove all elements from a linked list of integers that have value val.
Example:
Input: 1->2->6->3->4->5->6, val = 6
Output: 1->2->3->4->5
这个一个链表基本操作的题目,思路就不多说了。
- 链表的基本操作(删除指定节点)
- 虚拟节点dummy 简化操作
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? 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。
- 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。 eg:
// 只有下个节点不是要删除的节点才更新current
if (!next || next.val !== val) {
current = next;
}
/*
* @lc app=leetcode id=203 lang=javascript
*
* [203] Remove Linked List Elements
*
* https://leetcode.com/problems/remove-linked-list-elements/description/
*
* algorithms
* Easy (35.32%)
* Total Accepted: 211.9K
* Total Submissions: 598.6K
* Testcase Example: '[1,2,6,3,4,5,6]\n6'
*
* Remove all elements from a linked list of integers that have value val.
*
* Example:
*
*
* Input: 1->2->6->3->4->5->6, val = 6
* Output: 1->2->3->4->5
*
*
*/
/**
* Definition for singly-linked list.
* function ListNode(val) {
* this.val = val;
* this.next = null;
* }
*/
/**
* @param {ListNode} head
* @param {number} val
* @return {ListNode}
*/
var removeElements = function(head, val) {
const dummy = {
next: head
}
let current = dummy;
while(current && current.next) {
let next = current.next;
if (next.val === val) {
current.next = next.next;
next = next.next;
}
if (!next || next.val !== val) {
current = next;
}
}
return dummy.next;
};