```cpp #include <iostream> using namespace std; struct node { int data; node* link; node(int x) : data(x), link(null) {} }; // 查找最大节点及其前一个节点 void findmaxandprev(node* list, node*& maxnode, node*& prevmax) { node* curr = list; maxnode = list; prevmax = null; node* prev = null; while (curr!= null) { if (curr->data > maxnode->data) { maxnode = curr; prevmax = prev; } prev = curr; curr = curr->link; } } // 将最大节点移到链表末尾 void movemaxtoend(node*& list) { node* maxnode = null; node* prevmax = null;
以下是继续完善这个 C++ 代码的步骤: ```cpp #include <iostream> using namespace std; struct node { int data; node* link; node(int x) : data(x), link(nullptr) {} }; // 查找最大节点及其前一个节点 void findmaxandprev(node* list, node*& maxnode, node*& prevmax) { node* curr = list; maxnode = list; prevmax = nullptr; node* prev = nullptr; while (curr!= nullptr) { if (curr->data > maxnode->data) { maxnode = curr; prevmax = prev; } prev = curr; curr = curr->link; } } // 将最大节点移到链表末尾 void movemaxtoend(node*& list) { node* maxnode = nullptr; node* prevmax = nullptr; findmaxandprev(list, maxnode, prevmax); if (prevmax!= nullptr) { prevmax->link = maxnode->link; } else { list = list->link; } node* curr = list; while (curr->link!= nullptr) { curr = curr->link; } curr->link = maxnode; maxnode->link = nullptr; } // 打印链表 void printList(node* list) { node* curr = list; while (curr!= nullptr) { std::cout << curr->data << " "; curr = curr->link; } std::cout << std::endl; } int main() { node* list = new node(10); list->link = new node(20); list->link->link = new node(5); list->link->link->link = new node(30); std::cout << "Original List: "; printList(list); movemaxtoend(list); std::cout << "Modified List: "; printList(list); return 0; } ``` 在上述代码中,我们添加了 `movemaxtoend` 函数的完整实现,用于将最大节点移到链表末尾。添加了 `printList` 函数用于打印链表,以及 `main` 函数进行测试。 [2024-10-14 22:55:13 | AI写代码神器 | 549点数解答]
- [字节豆包] ```cpp #include <iostream> using namespace std; struct node { int data; node* link; node(int x) : data(x), link(null) {} }; // 查找最大节点及其前一个节点 void findmaxandprev(node* list, node*& maxnode, node*& prevmax) { node* curr = list; maxnode = list; prevmax = null; node* prev = null; while (curr!= null) { if (curr->data > maxnode->data) { maxnode = curr; prevmax = prev; } prev = curr; curr = curr->link; } } // 将最大节点移到链表末尾 void movemaxtoend(node*& list) { node* maxnode = null; node* prevmax = null;(549点数解答 | 2024-10-14 22:55:13)294
- [字节豆包] 7-1 单向链表的创建与输出 分数 100 作者 王群芳 单位 合肥师范学院 本题目要求补充两个函数,实现如下功能: 输入若干个正整数,以-1结束,采取向链表中添加节点的方式来建立一个单链表,并输出这个单链表。 向链表尾部添加节点函数: Link AppendNode(Link head,int data); 其中Link结构定义如下: typedef struct link { int data; struct link *next; }*Link; head:链表头指针,如果head为NULL,则会新建头节点 data:要添加的节点数据值 函数返回值:添加节点后的链表头指针 输出链表函数: void DisplyNode(Link head); head :链表头指针 主函数样例: #include <stdio.h> #include <stdlib.h> typedef struct link { int data; struct link *next; }*Link; Link AppendNode(Link head,int d(351点数解答 | 2025-12-19 20:53:08)30
- [字节豆包] 7-1 单向链表的创建与输出 分数 100 作者 王群芳 单位 合肥师范学院 本题目要求补充两个函数,实现如下功能: 输入若干个正整数,以-1结束,采取向链表中添加节点的方式来建立一个单链表,并输出这个单链表。 向链表尾部添加节点函数: Link AppendNode(Link head,int data); 其中Link结构定义如下: typedef struct link { int data; struct link *next; }*Link; head:链表头指针,如果head为NULL,则会新建头节点 data:要添加的节点数据值 函数返回值:添加节点后的链表头指针 输出链表函数: void DisplyNode(Link head); head :链表头指针 主函数样例: #include <stdio.h> #include <stdlib.h> typedef struct link { int data; struct link *next; }*Link; Link AppendNode(Link head,int d(354点数解答 | 2025-12-19 22:00:13)52
- [字节豆包] #include<stdio.h> #include<stdlib.h> #include<time.h> int producerand(int remainder); void initprocess(); void chosedisplace(); struct linknode* fifo(struct linknode* head, int randcount); void optimal(struct linknode* head, int randprocess); struct linknode* lru(struct linknode* head, int randprocess); struct linknode* initlink(); void choicestey(); int allotment(struct linknode* head); int checkfifooptimal(struct linknode* head, int checkpage); void recover(struct linknode* head, int randproc(60点数解答 | 2024-12-13 20:02:21)280
- [阿里通义] #include<stdio.h> #include<stdlib.h> #include<time.h> int producerand(int remainder); void initprocess(); void chosedisplace(); struct linknode* fifo(struct linknode* head, int randcount); void optimal(struct linknode* head, int randprocess); struct linknode* lru(struct linknode* head, int randprocess); struct linknode* initlink(); void choicestey(); int allotment(struct linknode* head); int checkfifooptimal(struct linknode* head, int checkpage); void recover(struct linknode* head, int randproc(858点数解答 | 2024-12-13 20:03:47)308
- [字节豆包] #include <stdio.h> #include <stdlib.h> #include <string.h> typedef struct sqlist //单链表结构体 { int data; struct sqlist *next; //指针域 }sqlist; void initlist (sqlist *l, sqlist *m, sqlist *n, sqlist *h) { //todo list yes l->next = null; m->next = null; n->next = null; h->next = null; } void emp(sqlist *l) { if (l->next==null) //todo list)yes printf("链表为空表。\n"); else printf("链表为非空表。\n"); } void length(sqlist *l) { sqlist *p; int length = 0; p=l->next; while(p!=null) { length++; p=p->next; } //todo list(345点数解答 | 2024-10-23 23:11:22)246
- [字节豆包] #include <iostream> #include <sstream> #include <string> #include <cctype> #include <vector> // 将字符串转换为小写 std::string toLowerCase(const std::string& str) { std::string result = str; for (char& c : result) { c = std::tolower(c); } return result; } int main() { std::string target; std::getline(std::cin, target); target = toLowerCase(target); std::string article; std::getline(std::cin, article); std::stringstream ss(article); std::vector<std::(1381点数解答 | 2025-12-15 17:32:28)31
- [字节豆包] #include <stdio.h> #include <stdlib.h> #include <string.h> typedef struct sqlist //单链表结构体 { int data; struct sqlist *next; //指针域 }sqlist; void initlist (sqlist *l, sqlist *m, sqlist *n, sqlist *h) { //todo list yes l->next = null; m->next = null; n->next = null; h->next = null; } void emp(sqlist *l) { if (l->next==null) //todo list)yes printf("链表为空表。\n"); else printf("链表为非空表。\n"); } void length(sqlist *l) { sqlist *p; int length = 0; p=l->next; while(p!=null)(428点数解答 | 2024-10-23 23:09:51)251
- [讯飞星火] 快速排序:# include< iostream>using namespace std;# include< stdlib. h># include< time. h>const int n =10;const int min= 10;const int max=99;int getrand( int min, int max){return ( rand()%( max-min+1)+ min);}void set( int r[], int n){int i;for(i=0;i<n;i++)r[i]=getrand(min, max);}void show( int r[], int n){int i;for(i=0;i<n;i++)cout << r[i]<<" ";cout << endl;}int partition( int r[], int i, int j){//由你完成}void quicksort( int r[], int i, int j){//由你完成}int main(){srand( time(0));int a[n];set(a, n);cout<<"(710点数解答 | 2024-11-14 10:44:25)331
- [阿里通义] 快速排序:# include< iostream>using namespace std;# include< stdlib. h># include< time. h>const int n =10;const int min= 10;const int max=99;int getrand( int min, int max){return ( rand()%( max-min+1)+ min);}void set( int r[], int n){int i;for(i=0;i<n;i++)r[i]=getrand(min, max);}void show( int r[], int n){int i;for(i=0;i<n;i++)cout << r[i]<<" ";cout << endl;}int partition( int r[], int i, int j){//由你完成}void quicksort( int r[], int i, int j){//由你完成}int main(){srand( time(0));int a[n];set(a, n);cout<<"(951点数解答 | 2024-11-14 10:46:29)314
- [字节豆包] 快速排序:# include< iostream>using namespace std;# include< stdlib. h># include< time. h>const int n =10;const int min= 10;const int max=99;int getrand( int min, int max){return ( rand()%( max-min+1)+ min);}void set( int r[], int n){int i;for(i=0;i<n;i++)r[i]=getrand(min, max);}void show( int r[], int n){int i;for(i=0;i<n;i++)cout << r[i]<<" ";cout << endl;}int partition( int r[], int i, int j){//由你完成}void quicksort( int r[], int i, int j){//由你完成}int main(){srand( time(0));int a[n];set(a, n);cout<<"(472点数解答 | 2024-11-14 10:51:44)307
- [DeepSeek] using communication_protocols; using FTS_F02391A; using NLog; using ScottPlot.Plottable; using Sunny.UI; using System; using System.Collections.Generic; using System.ComponentModel; using System.Data; using System.Drawing; using System.Linq; using System.Text; using System.Threading; using System.Threading.Tasks; using System.Windows.Forms; namespace 窗体界面设计 { public partial class FormHome : UIPage { private readonly DataLogger Log1;//波形图数据记录器 bool bViewSlide = false(562点数解答 | 2025-07-09 16:57:56)182