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86. Partition ListGiven the head of a linked list and a value x, partition it such that all nodes less than x come before nodes greater than or equal to x. You should preserve the original relative order of the nodes in each of the two partitions. Example 1: 12Input: head = [1,4,3,2,5,2], x = 3Output: [1,2,2,4,3,5] Example 2: 12Input: head = [2,1], x = 2Output: [1,2] 题目大意给定链表的头节点 head 和一个值 x，要求将链表分隔成两部分：所有值小于 x 的节点排在所有值大于或等于 x...

82. Remove Duplicates from Sorted List IIGiven the head of a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list. Return the linked list sorted as well. Example 1: 12Input: head = [1,2,3,3,4,4,5]Output: [1,2,5] Example 2: 12Input: head = [1,1,1,2,3]Output: [2,3] 题目大意给定一个已排序的链表，要求删除所有存在重复的节点（即重复的节点一个都不保留），仅保留原链表中只出现过一次的节点，最终返回排序后的新链表头节点。 核心解题思路由于链表已排序，重复节点必然相邻，因此可通过「遍历链表 + 跳过重复节点」的思路解决，关键是： 虚拟头节点：避免删除头节点时的特殊处理（如示例 2 中头节点...

一、事务与并发控制基础1.1 事务ACID特性在分布式系统中，事务需要满足以下核心特性： 原子性：事务操作不可分割，全部完成或全部失败 一致性：事务执行前后数据库状态保持一致 隔离性：事务之间相互隔离，避免并发执行的干扰 持久性：事务提交后永久生效，即使系统崩溃也不会丢失 对于InnoDB存储引擎，ACID特性通过日志系统（Redo Log）和多版本并发控制（MVCC）实现，MySQL 8.0版本后已完全支持事务性存储引擎。 1.2 并发问题分类体系四个经典的并发问题及其表现形式： 问题类型 定义说明 典型场景 脏写 事务A写入数据后未提交，事务B又覆盖该数据，导致A的写入被B无效 两个事务对同一行数据进行并发修改，典型例子为数据库主从复制中的write-ahead...

CD\LS命令12345678910111213141516171819202122232425262728293031323334353637383940414243444546474849505152#include "Server.h"// CD命令实现int CmdCd(char *path, const int depth, FileBuf_t *msg, Node_t *p){ DIR *dir =opendir(path); if(dir == NULL){ sprintf(msg->buf, "(error: %s)", strerror(errno)); return -1; }else{ p->depth = depth; char *help; char *pathblock = strtok_r(path, " /", &help); ...

服务端信息处理123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657585960616263646566676869707172737475767778798081828384858687888990919293949596979899100101102103104105106107108109110111#include "Server.h"// 接收并处理命令int RecvCommand(Node_t *p){ int netfd = p->netfd; //接收指令 SignalSend_t *cmd = (SignalSend_t *)calloc(1, sizeof(SignalSend_t)); if (cmd == NULL){ printf("Error calloc\n"); ...

287. Find the Duplicate NumberGiven an array of integers nums containing n + 1 integers where each integer is in the range [1, n] inclusive. There is only one repeated number in nums, return this repeated number. You must solve the problem without modifying the array nums and using only constant extra space. Example 1: 12Input: nums = [1,3,4,2,2]Output: 2 Example 2: 12Input: nums = [3,1,3,4,2]Output: 3 Example 3: 12Input: nums = [3,3,3,3,3]Output: 3 题目大意给定一个包含 n + 1 个整数的数组 nums，其中每个整数都在...

151. Reverse Words in a StringGiven an input string s, reverse the order of the words. A word is defined as a sequence of non-space characters. The words in s will be separated by at least one space. Return a string of the words in reverse order concatenated by a single space. Note that s may contain leading or trailing spaces or multiple spaces between two words. The returned string should only have a single space separating the words. Do not include any extra spaces. Example 1: 12Input: s...

28. Find the Index of the First Occurrence in a StringGiven two strings needle and haystack, return the index of the first occurrence of needle in haystack, or -1 if needle is not part of haystack. Example 1: 1234Input: haystack = "sadbutsad", needle = "sad"Output: 0Explanation: "sad" occurs at index 0 and 6.The first occurrence is at index 0, so we return 0. Example 2: 123Input: haystack = "leetcode", needle = "leeto"Output: -1Explanation:...

服务端登录验证123456789101112131415161718192021222324252627282930313233343536373839#include "Server.h"//#include <shadow.h>//#include <crypt.h>//#include <syslog.h>int IsLoading(int netfd){ if(send(netfd, "请输入用户名和密码", sizeof("请输入用户名和密码"), 0) <= 0){ printf("error recv loadmsg\n"); return -1; } char load[2][PATHLEN]; char *username = load[0]; if(recv(netfd, load, 2 * PATHLEN, 0) <= 0){ ...

服务端头文件123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657585960616263646566676869707172737475767778798081828384#ifndef SERVER_H#define SERVER_H#include "my_header.h"#include <arpa/inet.h>#include <fcntl.h>#include <dirent.h>#include <sys/mman.h>#include <unistd.h>#include <pthread.h>#include <netinet/in.h>#include <sys/epoll.h>#include <sys/types.h>#include...

收发文件循环接收12345678910#include "Client.h"int recvn(int sockfd, void *buf, int size){ int total = 0; char *p = (char *)buf; while(total < size){ ssize_t sret = recv(sockfd, p+total, size-total, 0); total += sret; } return total;} 收文件根据对方发送的文件名，在当下实现文件的接收 123456789101112131415161718192021222324252627282930313233343536int RecvFile(int sockfd){ FileBuf train; char filename[128] = {0}; recv(sockfd, &train.size,...

导言最近学习完成了数据库和多人聊天室，想基于这些东西，设计一个老掉牙的网盘项目（才不是收到齐哥刺激）。目前是只有空闲时间能做，大概还是跟之前更新一样，大概一周一更慢慢写。 头文件1234567891011121314151617181920212223242526272829303132333435/* Usage: 实现收发功能，收消息没有限制，发消息需要信号，Switch判断发送，结构体发送指令，接收消息，接收文件*/#ifndef CLIENT_H#define CLIENT_H#include "my_header.h"#include <arpa/inet.h>#include <fcntl.h>#include <unistd.h>#include <pthread.h>#include <netinet/in.h>#include <sys/epoll.h>#include <sys/types.h>#include...

一、数据库设计1234567891011121314151617181920212223242526272829303132333435-- 创建游戏管理系统数据库CREATE DATABASE game_management_db;USE game_management_db;-- 创建游戏表（主表）CREATE TABLE games ( game_id INT PRIMARY KEY, -- 游戏编号，主键 game_name VARCHAR(100) NOT NULL, -- 游戏名称，非空约束 developer VARCHAR(50) NOT NULL, -- 开发商，非空约束 release_year YEAR, -- 发布年份 genre VARCHAR(30), -- 游戏类型 copies_in_stock INT DEFAULT 0 -- 库存数量，默认值0);-- 创建玩家表（主表）CREATE TABLE players ( player_id INT PRIMARY KEY, -- 玩家编号，主键 player_name...

3002. Maximum Size of a Set After RemovalsYou are given two 0-indexed integer arrays nums1 and nums2 of even length n. You must remove n / 2 elements from nums1 and n / 2 elements from nums2. After the removals, you insert the remaining elements of nums1 and nums2 into a set s. Return the maximum possible size of the set s. Example 1: 1234Input: nums1 = [1,2,1,2], nums2 = [1,1,1,1]Output: 2Explanation: We remove two occurences of 1 from nums1 and nums2. After the removals, the arrays become...

454. 4Sum IIGiven four integer arrays nums1, nums2, nums3, and nums4 all of length n, return the number of tuples (i, j, k, l) such that: 0 <= i, j, k, l < n nums1[i] + nums2[j] + nums3[k] + nums4[l] == 0 Example 1: 123456Input: nums1 = [1,2], nums2 = [-2,-1], nums3 = [-1,2], nums4 = [0,2]Output: 2Explanation:The two tuples are:1. (0, 0, 0, 1) -> nums1[0] + nums2[0] + nums3[0] + nums4[1] = 1 + (-2) + (-1) + 2 = 02. (1, 1, 0, 0) -> nums1[1] + nums2[1] + nums3[0] + nums4[0] = 2...

349. Intersection of Two ArraysGiven two integer arrays nums1 and nums2, return an array of their intersection. Each element in the result must be unique and you may return the result in any order. Example 1: 12Input: nums1 = [1,2,2,1], nums2 = [2,2]Output: [2] Example 2: 123Input: nums1 = [4,9,5], nums2 = [9,4,9,8,4]Output: [9,4]Explanation: [4,9] is also accepted. 题目大意给定两个整数数组 nums1 和...