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Commit 17fb14c

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add 剑指 Offer 18. 删除链表的节点
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![](https://imgconvert.csdnimg.cn/aHR0cHM6Ly9jZG4uanNkZWxpdnIubmV0L2doL2Nob2NvbGF0ZTE5OTkvY2RuL2ltZy8yMDIwMDgyODE0NTUyMS5qcGc?x-oss-process=image/format,png)
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>仰望星空的人,不应该被嘲笑
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## 题目描述
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给定单向链表的头指针和一个要删除的节点的值,定义一个函数删除该节点。
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返回删除后的链表的头节点。
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注意:此题对比原题有改动
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示例 1:
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```javascript
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输入: head = [4,5,1,9], val = 5
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输出: [4,1,9]
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解释: 给定你链表中值为 5 的第二个节点,那么在调用了你的函数之后,该链表应变为 4 -> 1 -> 9.
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```
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示例 2:
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```javascript
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输入: head = [4,5,1,9], val = 1
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输出: [4,5,9]
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解释: 给定你链表中值为 1 的第三个节点,那么在调用了你的函数之后,该链表应变为 4 -> 5 -> 9.
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```
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说明:
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题目保证链表中节点的值互不相同
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若使用 C 或 C++ 语言,你不需要 free 或 delete 被删除的节点
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来源:力扣(LeetCode)
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链接:https://leetcode-cn.com/problems/shan-chu-lian-biao-de-jie-dian-lcof
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著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
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## 解题思路
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创建一个新链表,遇到相同值的情况,将当前节点的next指向下一个节点的next,否则继续遍历。
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```javascript
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/**
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* Definition for singly-linked list.
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* function ListNode(val) {
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* this.val = val;
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* this.next = null;
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* }
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*/
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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 deleteNode = function(head, val) {
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let dummyHead = new ListNode();
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dummyHead.next = head;
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let p = dummyHead;
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while(p.next){
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if(p.next.val === val){
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p.next = p.next.next;
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}else{
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p = p.next;
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}
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}
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return dummyHead.next;
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};
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```
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## 最后
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文章产出不易,还望各位小伙伴们支持一波!
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往期精选:
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<a href="https://github.com/Chocolate1999/Front-end-learning-to-organize-notes">小狮子前端の笔记仓库</a>
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<a href="https://github.com/Chocolate1999/leetcode-javascript">leetcode-javascript:LeetCode 力扣的 JavaScript 解题仓库,前端刷题路线(思维导图)</a>
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小伙伴们可以在Issues中提交自己的解题代码,🤝 欢迎Contributing,可打卡刷题,Give a ⭐️ if this project helped you!
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<a href="https://yangchaoyi.vip/">访问超逸の博客</a>,方便小伙伴阅读玩耍~
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![](https://img-blog.csdnimg.cn/2020090211491121.png#pic_center)
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```javascript
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学如逆水行舟,不进则退
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```
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