639 lines
17 KiB
Markdown
639 lines
17 KiB
Markdown
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<p align="center">
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<a href="https://programmercarl.com/other/xunlianying.html" target="_blank">
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<img src="../pics/训练营.png" width="1000"/>
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</a>
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<p align="center"><strong><a href="https://mp.weixin.qq.com/s/tqCxrMEU-ajQumL1i8im9A">参与本项目</a>,贡献其他语言版本的代码,拥抱开源,让更多学习算法的小伙伴们收益!</strong></p>
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> 链表操作中,可以使用原链表来直接进行删除操作,也可以设置一个虚拟头结点再进行删除操作,接下来看一看哪种方式更方便。
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# 203.移除链表元素
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[力扣题目链接](https://leetcode.cn/problems/remove-linked-list-elements/)
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题意:删除链表中等于给定值 val 的所有节点。
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示例 1:
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输入:head = [1,2,6,3,4,5,6], val = 6
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输出:[1,2,3,4,5]
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示例 2:
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输入:head = [], val = 1
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输出:[]
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示例 3:
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输入:head = [7,7,7,7], val = 7
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输出:[]
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## 算法公开课
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**[《代码随想录》算法视频公开课](https://programmercarl.com/other/gongkaike.html):[链表基础操作| LeetCode:203.移除链表元素](https://www.bilibili.com/video/BV18B4y1s7R9),相信结合视频再看本篇题解,更有助于大家对本题的理解**。
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## 思路
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这里以链表 1 4 2 4 来举例,移除元素4。
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如果使用C,C++编程语言的话,不要忘了还要从内存中删除这两个移除的节点, 清理节点内存之后如图:
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**当然如果使用java ,python的话就不用手动管理内存了。**
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还要说明一下,就算使用C++来做leetcode,如果移除一个节点之后,没有手动在内存中删除这个节点,leetcode依然也是可以通过的,只不过,内存使用的空间大一些而已,但建议依然要养成手动清理内存的习惯。
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这种情况下的移除操作,就是让节点next指针直接指向下下一个节点就可以了,
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那么因为单链表的特殊性,只能指向下一个节点,刚刚删除的是链表的中第二个,和第四个节点,那么如果删除的是头结点又该怎么办呢?
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这里就涉及如下链表操作的两种方式:
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* **直接使用原来的链表来进行删除操作。**
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* **设置一个虚拟头结点在进行删除操作。**
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来看第一种操作:直接使用原来的链表来进行移除。
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移除头结点和移除其他节点的操作是不一样的,因为链表的其他节点都是通过前一个节点来移除当前节点,而头结点没有前一个节点。
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所以头结点如何移除呢,其实只要将头结点向后移动一位就可以,这样就从链表中移除了一个头结点。
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依然别忘将原头结点从内存中删掉。
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这样移除了一个头结点,是不是发现,在单链表中移除头结点 和 移除其他节点的操作方式是不一样,其实在写代码的时候也会发现,需要单独写一段逻辑来处理移除头结点的情况。
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那么可不可以 以一种统一的逻辑来移除 链表的节点呢。
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其实**可以设置一个虚拟头结点**,这样原链表的所有节点就都可以按照统一的方式进行移除了。
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来看看如何设置一个虚拟头。依然还是在这个链表中,移除元素1。
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这里来给链表添加一个虚拟头结点为新的头结点,此时要移除这个旧头结点元素1。
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这样是不是就可以使用和移除链表其他节点的方式统一了呢?
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来看一下,如何移除元素1 呢,还是熟悉的方式,然后从内存中删除元素1。
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最后呢在题目中,return 头结点的时候,别忘了 `return dummyNode->next;`, 这才是新的头结点
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**直接使用原来的链表来进行移除节点操作:**
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```CPP
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class Solution {
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public:
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ListNode* removeElements(ListNode* head, int val) {
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// 删除头结点
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while (head != NULL && head->val == val) { // 注意这里不是if
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ListNode* tmp = head;
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head = head->next;
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delete tmp;
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}
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// 删除非头结点
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ListNode* cur = head;
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while (cur != NULL && cur->next!= NULL) {
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if (cur->next->val == val) {
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ListNode* tmp = cur->next;
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cur->next = cur->next->next;
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delete tmp;
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} else {
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cur = cur->next;
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}
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}
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return head;
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}
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};
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```
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* 时间复杂度: O(n)
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* 空间复杂度: O(1)
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**设置一个虚拟头结点在进行移除节点操作:**
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```CPP
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class Solution {
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public:
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ListNode* removeElements(ListNode* head, int val) {
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ListNode* dummyHead = new ListNode(0); // 设置一个虚拟头结点
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dummyHead->next = head; // 将虚拟头结点指向head,这样方便后面做删除操作
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ListNode* cur = dummyHead;
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while (cur->next != NULL) {
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if(cur->next->val == val) {
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ListNode* tmp = cur->next;
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cur->next = cur->next->next;
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delete tmp;
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} else {
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cur = cur->next;
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}
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}
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head = dummyHead->next;
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delete dummyHead;
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return head;
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}
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};
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```
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* 时间复杂度: O(n)
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* 空间复杂度: O(1)
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## 其他语言版本
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### C:
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用原来的链表操作:
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```c
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struct ListNode* removeElements(struct ListNode* head, int val){
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struct ListNode* temp;
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// 当头结点存在并且头结点的值等于val时
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while(head && head->val == val) {
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temp = head;
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// 将新的头结点设置为head->next并删除原来的头结点
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head = head->next;
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free(temp);
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}
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struct ListNode *cur = head;
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// 当cur存在并且cur->next存在时
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// 此解法需要判断cur存在因为cur指向head。若head本身为NULL或者原链表中元素都为val的话,cur也会为NULL
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while(cur && (temp = cur->next)) {
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// 若cur->next的值等于val
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if(temp->val == val) {
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// 将cur->next设置为cur->next->next并删除cur->next
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cur->next = temp->next;
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free(temp);
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}
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// 若cur->next不等于val,则将cur后移一位
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else
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cur = cur->next;
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}
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// 返回头结点
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return head;
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}
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```
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设置一个虚拟头结点:
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```c
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/**
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* Definition for singly-linked list.
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* struct ListNode {
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* int val;
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* struct ListNode *next;
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* };
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*/
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struct ListNode* removeElements(struct ListNode* head, int val){
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typedef struct ListNode ListNode;
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ListNode *shead;
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shead = (ListNode *)malloc(sizeof(ListNode));
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shead->next = head;
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ListNode *cur = shead;
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while(cur->next != NULL){
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if (cur->next->val == val){
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ListNode *tmp = cur->next;
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cur->next = cur->next->next;
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free(tmp);
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}
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else{
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cur = cur->next;
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}
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}
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head = shead->next;
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free(shead);
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return head;
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}
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```
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### Java:
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```java
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/**
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* 添加虚节点方式
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* 时间复杂度 O(n)
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* 空间复杂度 O(1)
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* @param head
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* @param val
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* @return
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*/
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public ListNode removeElements(ListNode head, int val) {
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if (head == null) {
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return head;
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}
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// 因为删除可能涉及到头节点,所以设置dummy节点,统一操作
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ListNode dummy = new ListNode(-1, head);
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ListNode pre = dummy;
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ListNode cur = head;
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while (cur != null) {
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if (cur.val == val) {
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pre.next = cur.next;
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} else {
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pre = cur;
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}
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cur = cur.next;
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}
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return dummy.next;
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}
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/**
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* 不添加虚拟节点方式
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* 时间复杂度 O(n)
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* 空间复杂度 O(1)
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* @param head
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* @param val
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* @return
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*/
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public ListNode removeElements(ListNode head, int val) {
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while (head != null && head.val == val) {
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head = head.next;
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}
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// 已经为null,提前退出
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if (head == null) {
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return head;
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}
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// 已确定当前head.val != val
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ListNode pre = head;
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ListNode cur = head.next;
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while (cur != null) {
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if (cur.val == val) {
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pre.next = cur.next;
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} else {
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pre = cur;
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}
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cur = cur.next;
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}
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return head;
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}
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/**
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* 不添加虚拟节点and pre Node方式
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* 时间复杂度 O(n)
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* 空间复杂度 O(1)
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* @param head
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* @param val
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* @return
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*/
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public ListNode removeElements(ListNode head, int val) {
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while(head!=null && head.val==val){
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head = head.next;
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}
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ListNode curr = head;
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while(curr!=null){
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while(curr.next!=null && curr.next.val == val){
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curr.next = curr.next.next;
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}
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curr = curr.next;
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}
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return head;
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}
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```
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### Python:
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```python
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(版本一)虚拟头节点法
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# Definition for singly-linked list.
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# class ListNode:
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# def __init__(self, val=0, next=None):
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# self.val = val
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# self.next = next
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class Solution:
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def removeElements(self, head: Optional[ListNode], val: int) -> Optional[ListNode]:
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# 创建虚拟头部节点以简化删除过程
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dummy_head = ListNode(next = head)
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# 遍历列表并删除值为val的节点
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current = dummy_head
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while current.next:
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if current.next.val == val:
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current.next = current.next.next
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else:
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current = current.next
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return dummy_head.next
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```
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### Go:
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```go
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/**
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* Definition for singly-linked list.
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* type ListNode struct {
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* Val int
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* Next *ListNode
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* }
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*/
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func removeElements(head *ListNode, val int) *ListNode {
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dummyHead := &ListNode{}
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dummyHead.Next = head
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cur := dummyHead
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for cur != nil && cur.Next != nil {
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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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}
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}
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return dummyHead.Next
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}
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```
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### JavaScript:
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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 ret = new ListNode(0, head);
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let cur = ret;
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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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continue;
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}
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cur = cur.next;
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}
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return ret.next;
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};
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```
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### TypeScript:
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版本一(在原链表上直接删除):
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```typescript
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/**
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* Definition for singly-linked list.
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* class ListNode {
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* val: number
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* next: ListNode | null
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* constructor(val?: number, next?: ListNode | null) {
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* this.val = (val===undefined ? 0 : val)
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* this.next = (next===undefined ? null : next)
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* }
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* }
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*/
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function removeElements(head: ListNode | null, val: number): ListNode | null {
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// 删除头部节点
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while (head !== null && head.val === val) {
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head = head.next;
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}
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if (head === null) return head;
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let pre: ListNode = head, cur: ListNode | null = head.next;
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// 删除非头部节点
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while (cur) {
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if (cur.val === val) {
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pre.next = cur.next;
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} else {
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//此处不加类型断言时:编译器会认为pre类型为ListNode, pre.next类型为ListNode | null
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pre = pre.next as ListNode;
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}
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cur = cur.next;
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}
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return head;
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};
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```
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版本二(虚拟头节点):
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```typescript
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function removeElements(head: ListNode | null, val: number): ListNode | null {
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// 添加虚拟节点
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const data = new ListNode(0, head);
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let pre = data, cur = data.next;
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while (cur) {
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if (cur.val === val) {
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pre.next = cur.next
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} else {
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pre = cur;
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}
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cur = cur.next;
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}
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return data.next;
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};
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```
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### Swift:
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```swift
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/**
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* Definition for singly-linked list.
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* public class ListNode {
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* public var val: Int
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* public var next: ListNode?
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* public init() { self.val = 0; self.next = nil; }
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* public init(_ val: Int) { self.val = val; self.next = nil; }
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* public init(_ val: Int, _ next: ListNode?) { self.val = val; self.next = next; }
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* }
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*/
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func removeElements(_ head: ListNode?, _ val: Int) -> ListNode? {
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let dummyNode = ListNode()
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dummyNode.next = head
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var currentNode = dummyNode
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while let curNext = currentNode.next {
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if curNext.val == val {
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currentNode.next = curNext.next
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} else {
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currentNode = curNext
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}
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}
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return dummyNode.next
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}
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```
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|
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### PHP:
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|
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```php
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/**
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* Definition for singly-linked list.
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* type ListNode struct {
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* Val int
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* Next *ListNode
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* }
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*/
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// 虚拟头+双指针
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func removeElements(head *ListNode, val int) *ListNode {
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dummyHead := &ListNode{}
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dummyHead.Next = head
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pred := dummyHead
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cur := head
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for cur != nil {
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if cur.Val == val {
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pred.Next = cur.Next
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} else {
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pred = cur
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}
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cur = cur.Next
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}
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return dummyHead.Next
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}
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```
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### Rust:
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|
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```rust
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// Definition for singly-linked list.
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// #[derive(PartialEq, Eq, Clone, Debug)]
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// pub struct ListNode {
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// pub val: i32,
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// pub next: Option<Box<ListNode>>
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// }
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//
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// impl ListNode {
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// #[inline]
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// fn new(val: i32) -> Self {
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// ListNode {
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// next: None,
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// val
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// }
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// }
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// }
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impl Solution {
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pub fn remove_elements(head: Option<Box<ListNode>>, val: i32) -> Option<Box<ListNode>> {
|
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let mut dummyHead = Box::new(ListNode::new(0));
|
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dummyHead.next = head;
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let mut cur = dummyHead.as_mut();
|
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// 使用take()替换std::men::replace(&mut node.next, None)达到相同的效果,并且更普遍易读
|
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while let Some(nxt) = cur.next.take() {
|
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if nxt.val == val {
|
||
cur.next = nxt.next;
|
||
} else {
|
||
cur.next = Some(nxt);
|
||
cur = cur.next.as_mut().unwrap();
|
||
}
|
||
}
|
||
dummyHead.next
|
||
}
|
||
}
|
||
```
|
||
|
||
### Scala:
|
||
|
||
```scala
|
||
/**
|
||
* Definition for singly-linked list.
|
||
* class ListNode(_x: Int = 0, _next: ListNode = null) {
|
||
* var next: ListNode = _next
|
||
* var x: Int = _x
|
||
* }
|
||
*/
|
||
object Solution {
|
||
def removeElements(head: ListNode, `val`: Int): ListNode = {
|
||
if (head == null) return head
|
||
var dummy = new ListNode(-1, head) // 定义虚拟头节点
|
||
var cur = head // cur 表示当前节点
|
||
var pre = dummy // pre 表示cur前一个节点
|
||
while (cur != null) {
|
||
if (cur.x == `val`) {
|
||
// 相等,就删除那么cur的前一个节点pre执行cur的下一个
|
||
pre.next = cur.next
|
||
} else {
|
||
// 不相等,pre就等于当前cur节点
|
||
pre = cur
|
||
}
|
||
// 向下迭代
|
||
cur = cur.next
|
||
}
|
||
// 最终返回dummy的下一个,就是链表的头
|
||
dummy.next
|
||
}
|
||
}
|
||
```
|
||
|
||
### Kotlin:
|
||
|
||
```kotlin
|
||
/**
|
||
* Example:
|
||
* var li = ListNode(5)
|
||
* var v = li.`val`
|
||
* Definition for singly-linked list.
|
||
* class ListNode(var `val`: Int) {
|
||
* var next: ListNode? = null
|
||
* }
|
||
*/
|
||
class Solution {
|
||
fun removeElements(head: ListNode?, `val`: Int): ListNode? {
|
||
// 使用虚拟节点,令该节点指向head
|
||
var dummyNode = ListNode(-1)
|
||
dummyNode.next = head
|
||
// 使用cur遍历链表各个节点
|
||
var cur = dummyNode
|
||
// 判断下个节点是否为空
|
||
while (cur.next != null) {
|
||
// 符合条件,移除节点
|
||
if (cur.next.`val` == `val`) {
|
||
cur.next = cur.next.next
|
||
}
|
||
// 不符合条件,遍历下一节点
|
||
else {
|
||
cur = cur.next
|
||
}
|
||
}
|
||
// 注意:返回的不是虚拟节点
|
||
return dummyNode.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)
|
||
{
|
||
ListNode dummyHead = new ListNode(0,head);
|
||
ListNode temp = dummyHead;
|
||
while(temp.next != null)
|
||
{
|
||
if(temp.next.val == val)
|
||
{
|
||
temp.next = temp.next.next;
|
||
}
|
||
else
|
||
{
|
||
temp = temp.next;
|
||
}
|
||
}
|
||
return dummyHead.next;
|
||
}
|
||
}
|
||
```
|
||
|
||
<p align="center">
|
||
<a href="https://programmercarl.com/other/kstar.html" target="_blank">
|
||
<img src="../pics/网站星球宣传海报.jpg" width="1000"/>
|
||
</a>
|
||
|