Delete Node in a Linked List

最新推荐文章于 2023-01-23 13:18:59 发布
原创 最新推荐文章于 2023-01-23 13:18:59 发布 · 246 阅读 · 0 ·
CC 4.0 BY-SA版权
版权声明：本文为博主原创文章，遵循 CC 4.0 BY-SA 版权协议，转载请附上原文出处链接和本声明。

本文介绍如何在单链表中删除指定节点，不修改整个链表结构，仅通过给定节点指针进行操作。

Delete Node in a Linked List

Write a function to delete a node (except the tail) in a singly linked list, given only access to that node.

Supposed the linked list is 1 -> 2 -> 3 -> 4 and you are given the third node with value 3, the linked list should become 1 -> 2 -> 4 after calling your function.

结果是要删除的那个元素，而不是整个链表


思路

其实是把给出的待删除节点赋值为下一个节点，然后删除下一个节点

临时变量指向->next，然后赋值，最后free


代码

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
void deleteNode(struct ListNode* node) {
    struct ListNode* tmp = node->next;
    node->val = tmp ->val;
    node->next = tmp->next;
    free(tmp);  
}
Leetcode 237. Delete Node in a Linked List-删除链表的指定节点，不给链表的头节点
Top5软件工程硕士，先后在京东、字节从事多年Java后端开发、实时和离线大数据开发
04-05 1022
Write a function todelete a nodein a singly-linked list. You willnotbe given access to theheadof the list, instead you will be given access tothe node to be deleteddirectly. It isguaranteedthat the node to be deleted isnot a tail nodein the li...
Easy 237题 Delete Node in a Linked List
brenda_zhao的博客
10-26 276
Question: Write a function to delete a node (except the tail) in a singly linked list, given only access to that node. Supposed the linked list is 1 -> 2 -> 3 -> 4 and you are given the third
372.Delete Node in a Linked List
vestlee的博客
06-02 204
描述给定一个单链表中的一个等待被删除的节点(非表头或表尾)。请在在O(1)时间复杂度删除该链表节点。您在真实的面试中是否遇到过这个题？  是样例Linked list is 1->2->3->4, and given node 3, delete the node in place 1->2->4第一眼审题有点懵不知道这是啥意思，只给了要删除的节点。所以这道题的意思是...
【简单】Lintcode 372:Delete Node in a Linked List
majichen95的博客
05-18 256
  Implement an algorithm to delete a node in the middle of a singly linked list, given only access to that node. Example Linked list is 1->2->3->4, and given node 3, delete the node in pla...
237. Delete Node in a Linked List
擦肩的阳光的专栏
03-03 394
//本文内容来自StarSight，欢迎访问。 Write a function to delete a node (except the tail) in a singly linked list, given only access to that node. Supposed the linked list is 1 -> 2 -> 3 -> 4 and you ar
Leetcode_237_Delete Node in a Linked List
皮斯特劳沃
08-07 1877
本文是在学习中的总结，欢迎转载但请注明出处：http://blog.youkuaiyun.com/pistolove/article/details/47334649 Write a function to delete a node (except the tail) in a singly linked list, given only access to that node. Supposed the linked list is 1 -> 2 -> 3 -> 4 and you are
Delete Node in a Linked List Java LeetCode
Json随笔
07-31 1227
Write a function to delete a node (except the tail) in a singly linked list, given only access to that node.Supposed the linked list is 1 -> 2 -> 3 -> 4 and you are given the third node with value 3, t
leetcode 237: Delete Node in a Linked List
热门推荐
西施豆腐渣
07-15 1万+
leetcode 237: Delete Node in a Linked List PYTHON JAVA C++
【LeetCode】237. Delete Node in a Linked List 删除链表中的节点（Easy）（JAVA）
qq_16927853的博客
12-07 203
【LeetCode】237. Delete Node in a Linked List 删除链表中的节点（Easy）（JAVA） 题目地址: https://leetcode.com/problems/delete-node-in-a-linked-list/ 题目描述： Write a function to delete a node in a singly-linked list. You will not be given access to the head of the list, instea
leetcode 237 Delete Node in a Linked List（链表删除）
祝如愿的博客
01-23 118
leetcode 237 Delete Node in a Linked List题解
python-leetcode题解之237-Delete-Node-in-a-Linked-List.py
09-12
针对leetcode题库中的“Delete Node in a Linked List”题目，我们主要探讨如何在不访问链表头节点的情况下，删除给定节点。按照标准的链表删除操作，通常我们会要求知道前一个节点以便将其next指针指向空，表示删除...
(LeetCode)Delete Node in a Linked List --- 删除链表中结点
杨鑫newlife的专栏
08-24 1176
(LeetCode)Delete Node in a Linked List --- 删除链表中结点
[LeetCode]--237. Delete Node in a Linked List
杜鲁门的博客
10-14 671
Write a function to delete a node (except the tail) in a singly linked list, given only access to that node.Supposed the linked list is 1 -> 2 -> 3 -> 4 and you are given the third node with value 3, t
snapshot
github_30830155的博客
09-21 2731
环境 文件说明 流程图 代码说明 成员变量 成员函数 检查有效性合法性 setget属性值 单元测试说明 github链接 改进之处环境c++ boost库(split和正则） 单元测试google test工具文件说明 SnapShot.h : 头文件 SnapShot.cpp : 实现函数 SnapShotTest.cpp : 使用gtest工具单元测试代码 run_main.cpp :
删除链表奇数元素
github_30830155的博客
07-03 2550
/* 功能： 删除队列中的奇数元素 思路： 队列是先进先出，前面删除后面插入的了，所以呢 我们就从前面开始，删除，偶数的话就插到队尾，奇数就不管咯 结束条件呢，就是原来的队尾都删除了吧 */ // 如果要用链队好了 #include #include typedef struct qnode { int data; struct qnode *n
Print Book
github_30830155的博客
07-29 1371
#include #include #include #include using namespace std; class Book { public: Book(string format) { this->book_format = format; setPageStyle(book_format); this->FirstLineChar = '='; th
BigInt
github_30830155的博客
09-14 1211
题目用string模拟int大数运算，记得网易的一个笔试题最后就是的调试错误解决 #include <string>依旧报错string未定义 因为木有using namespace std; 转换到 coff 期间失败 项目属性 — 配置属性 — 清单工具 — 点击输入输出 将右侧出现的嵌入清单的属性值修改为否，点击应用，点击确定 参考程序地址git
删除链队奇数元素
github_30830155的博客
07-03 784
/* 功能： 删除队列中的奇数元素 思路： 队列是先进先出，前面删除后面插入的了，所以呢 我们就从前面开始，删除，偶数的话就插到队尾，奇数就不管咯 结束条件呢，就是原来的队尾都删除了吧 */ // 如果要用链队好了 #include #include typedef struct qnode { int data; struct qnode *n
#include "linked_list.h" linked_list_ptr LINKED_LIST_CREATE () { linked_list_ptr linked_list_ptr_H = NULL; if (NULL == (linked_list_ptr_H = (linked_list_ptr)malloc (sizeof (linked_list_node)))) { return linked_list_ptr_H; } memset (&(linked_list_ptr_H->data), 'H', sizeof (linked_list_ptr_H->data)); linked_list_ptr_H->next = NULL; return linked_list_ptr_H; } int LINKED_LIST_FREE (linked_list_ptr linked_list_ptr_H) { if (NULL == linked_list_ptr_H) { return -1; } int num = -1; linked_list_ptr linked_list_ptr_node = linked_list_ptr_H; while (NULL != (linked_list_ptr_node = linked_list_ptr_node->next)) { free (linked_list_ptr_H); num ++; linked_list_ptr_H = linked_list_ptr_node; } free (linked_list_ptr_H); num ++; return num; } int LINKED_LIST_COUNT (linked_list_ptr linked_list_ptr_H) { int num = -1; while (NULL != linked_list_ptr_H) { linked_list_ptr_H = linked_list_ptr_H->next; num ++; } return num; } int LINKED_LIST_INSERT (linked_list_ptr linked_list_ptr_H, data_t data, int pos) { if (NULL == linked_list_ptr_H) { return -1; } int num = 0; if (-1 == pos) { while (NULL != linked_list_ptr_H->next) { linked_list_ptr_H = linked_list_ptr_H->next; num ++; } if (NULL == (linked_list_ptr_H->next = (linked_list_ptr)malloc (sizeof (linked_list_node)))) { return -2; } linked_list_ptr_H = linked_list_ptr_H->next; linked_list_ptr_H->data = data; linked_list_ptr_H->next = NULL; } else { while (NULL != linked_list_ptr_H->next && 0 != pos --) { linked_list_ptr_H = linked_list_ptr_H->next; num ++; } if (NULL == linked_list_ptr_H->next) { if (NULL == (linked_list_ptr_H->next = (linked_list_ptr)malloc (sizeof (linked_list_node)))) { return -2; } linked_list_ptr_H = linked_list_ptr_H->next; linked_list_ptr_H->data = data; linked_list_ptr_H->next = NULL; } else { linked_list_ptr linked_list_ptr_node = linked_list_ptr_H->next; if (NULL == (linked_list_ptr_H->next = (linked_list_ptr)malloc (sizeof (linked_list_node)))) { return -2; } linked_list_ptr_H = linked_list_ptr_H->next; linked_list_ptr_H->data = data; linked_list_ptr_H->next = linked_list_ptr_node; } } return num; } int LINKED_LIST_DELETE (linked_list_ptr linked_list_ptr_H, int pos) { if (NULL == linked_list_ptr_H) { return -1; } if (NULL == linked_list_ptr_H->next) { return -2; } int num = 0; if (-1 == pos) { while (NULL != linked_list_ptr_H->next->next) { linked_list_ptr_H = linked_list_ptr_H->next; num ++; } free (linked_list_ptr_H->next); linked_list_ptr_H->next = NULL; } else { while (NULL != linked_list_ptr_H->next->next && 0 != pos --) { linked_list_ptr_H = linked_list_ptr_H->next; num ++; } if (NULL == linked_list_ptr_H->next->next) { free (linked_list_ptr_H->next); linked_list_ptr_H->next = NULL; } else { linked_list_ptr linked_list_ptr_node = linked_list_ptr_H->next->next; free (linked_list_ptr_H->next); linked_list_ptr_H->next = linked_list_ptr_node; } } return num; } int LINKED_LIST_PURGE (linked_list_ptr linked_list_ptr_H) { if (NULL == linked_list_ptr_H) { return -1; } if (NULL == linked_list_ptr_H->next) { return 0; } if (NULL == linked_list_ptr_H->next->next) { return 1; } int num = 2; linked_list_ptr linked_list_ptr_node1 = NULL, linked_list_ptr_node2 = NULL; while (NULL != (linked_list_ptr_H = linked_list_ptr_H->next)->next->next) { linked_list_ptr_node1 = linked_list_ptr_H; while (NULL != linked_list_ptr_node1->next->next) { if (linked_list_ptr_H->data == linked_list_ptr_node1->next->data) { linked_list_ptr_node2 = linked_list_ptr_node1->next->next; free (linked_list_ptr_node1->next); linked_list_ptr_node1->next = linked_list_ptr_node2; continue; } linked_list_ptr_node1 = linked_list_ptr_node1->next; } if (linked_list_ptr_H->data == linked_list_ptr_node1->next->data) { free (linked_list_ptr_node1->next); linked_list_ptr_node1->next =NULL; } num ++; } return num; } linked_list_ptr LINKED_LIST_LOCATE_NUM2ADD (linked_list_ptr linked_list_ptr_H, int pos) { if (NULL == linked_list_ptr_H) { return NULL; } if (NULL == linked_list_ptr_H->next) { return NULL; } if (-1 == pos) { while (NULL != (linked_list_ptr_H = linked_list_ptr_H->next)->next); } else { while (NULL != (linked_list_ptr_H = linked_list_ptr_H->next) && 0 != pos --); } return linked_list_ptr_H; } int LINKED_LIST_LOCATE_ADD2NUM (linked_list_ptr linked_list_ptr_H, linked_list_ptr linked_list_ptr_node) { if (NULL == linked_list_ptr_H) { return -1; } if (NULL == linked_list_ptr_node) { return -2; } if (linked_list_ptr_H == linked_list_ptr_node) { return -3; } if (NULL == linked_list_ptr_H->next) { return -4; } int num = 0; while (NULL != (linked_list_ptr_H = linked_list_ptr_H->next) && linked_list_ptr_node != linked_list_ptr_H) { num ++; } if (linked_list_ptr_H != linked_list_ptr_node) { return -5; } return num; } int LINKED_LIST_REVERSE (linked_list_ptr linked_list_ptr_H) { if (NULL == linked_list_ptr_H) { return -1; } if (NULL == linked_list_ptr_H->next) { return 0; } if (NULL == linked_list_ptr_H->next->next) { return 1; } int num = 2; linked_list_ptr linked_list_ptr_node1 = linked_list_ptr_H->next, linked_list_ptr_node2 = linked_list_ptr_H->next->next, linked_list_ptr_node3 = linked_list_ptr_H->next->next->next; linked_list_ptr_node1->next = NULL; while (NULL != linked_list_ptr_node3) { linked_list_ptr_node2->next = linked_list_ptr_node1; linked_list_ptr_node1 = linked_list_ptr_node2; linked_list_ptr_node2 = linked_list_ptr_node3; linked_list_ptr_node3 = linked_list_ptr_node3->next; num ++; } linked_list_ptr_node2->next = linked_list_ptr_node1; linked_list_ptr_H->next = linked_list_ptr_node2; return num; } int LINKED_LIST_BUBBLE_SORT_MIN (linked_list_ptr linked_list_ptr_H) { int num = -1; linked_list_ptr linked_list_ptr_node = linked_list_ptr_H; while (NULL != linked_list_ptr_node) { linked_list_ptr_node = linked_list_ptr_node->next; num ++; } if (2 > num) { return num; } linked_list_ptr linked_list_ptr_node1 = linked_list_ptr_H->next, linked_list_ptr_node2 = linked_list_ptr_H->next->next; for (int i = 0; i < num - 1; i ++) { for (int j = 0; j < num - 1 - i; j ++) { if (linked_list_ptr_node1->data > linked_list_ptr_node2->data) { linked_list_ptr_node1->data ^= linked_list_ptr_node2->data; linked_list_ptr_node2->data ^= linked_list_ptr_node1->data; linked_list_ptr_node1->data ^= linked_list_ptr_node2->data; } linked_list_ptr_node1 = linked_list_ptr_node2; linked_list_ptr_node2 = linked_list_ptr_node2->next; } linked_list_ptr_node1 = linked_list_ptr_H->next; linked_list_ptr_node2 = linked_list_ptr_H->next->next; } return num; } int LINKED_LIST_BUBBLE_SORT_MAX (linked_list_ptr linked_list_ptr_H) { int num = -1; linked_list_ptr linked_list_ptr_node = linked_list_ptr_H; while (NULL != linked_list_ptr_node) { linked_list_ptr_node = linked_list_ptr_node->next; num ++; } if (2 > num) { return num; } linked_list_ptr linked_list_ptr_node1 = linked_list_ptr_H->next, linked_list_ptr_node2 = linked_list_ptr_H->next->next; for (int i = 0; i < num - 1; i ++) { for (int j = 0; j < num - 1 - i; j ++) { if (linked_list_ptr_node1->data < linked_list_ptr_node2->data) { linked_list_ptr_node1->data ^= linked_list_ptr_node2->data; linked_list_ptr_node2->data ^= linked_list_ptr_node1->data; linked_list_ptr_node1->data ^= linked_list_ptr_node2->data; } linked_list_ptr_node1 = linked_list_ptr_node2; linked_list_ptr_node2 = linked_list_ptr_node2->next; } linked_list_ptr_node1 = linked_list_ptr_H->next; linked_list_ptr_node2 = linked_list_ptr_H->next->next; } return num; } int LINKED_LIST_PRINTF (linked_list_ptr linked_list_ptr_H) { if (NULL == linked_list_ptr_H) { return -1; } int num = 0; linked_list_ptr linked_list_ptr_node = linked_list_ptr_H; while (NULL != (linked_list_ptr_node = linked_list_ptr_node->next)) { printf ("%d ", linked_list_ptr_node->data); num ++; } if (NULL != linked_list_ptr_H->next) { putchar ('\n'); } return num; } 寻找这段代码的错误
最新发布
12-06
linked_list_ptr newNode = (linked_list_ptr)malloc(sizeof(linked_list_node)); if (!newNode) return -1; newNode->data = data; // 定位到插入位置的前驱节点（即要插入在哪个节点之后） linked_list_ptr ...
评论
成就一亿技术人!
拼手气红包6.0元
还能输入1000个字符
 
红包 添加红包
表情包 插入表情
表情包 代码片
 条评论被折叠 查看
添加红包

请填写红包祝福语或标题

个

红包个数最小为10个

元

红包金额最低5元

当前余额3.43元 前往充值 >
需支付：10.00元
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付元
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明：

1.余额是钱包充值的虚拟货币，按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载，可以购买VIP、付费专栏及课程。

余额充值