数据结构约瑟夫问题C++单向循环链表,分文件如下：数小孩，小孩离开，打印结果的三个操作进行封装

好的，我可以帮你进一步优化代码，将数小孩、小孩离开和打印结果三个操作进行封装。

首先是Node.h文件，定义节点类：

// Node.h

#ifndef NODE_H
#define NODE_H

class Node {
public:
    int data;
    Node* next;

    Node(int data) {
        this->data = data;
        next = nullptr;
    }
};

#endif

然后是List.h文件，定义链表类，封装了数小孩、小孩离开和打印结果三个操作：

// List.h

#ifndef LIST_H
#define LIST_H

#include <iostream>
#include "Node.h"

class List {
public:
    List(int n, int m);
    ~List();
    int josephus();

private:
    Node* head;
    int n;
    int m;

    void insert();
    void remove(Node* node);
    void printResult();
};

#endif

最后是List.cpp文件，实现链表类的方法：

// List.cpp

#include "List.h"

List::List(int n, int m) {
    this->n = n;
    this->m = m;
    head = nullptr;
    for (int i = 1; i <= n; i++) {
        insert();
    }
}

List::~List() {
    Node* curr = head;
    while (curr != nullptr) {
        Node* temp = curr;
        curr = curr->next;
        delete temp;
    }
}

int List::josephus() {
    Node* curr = head;
    while (n > 1) {
        for (int i = 1; i < m; i++) {
            curr = curr->next;
        }
        Node* temp = curr;
        curr = curr->next;
        remove(temp);
        n--;
    }
    printResult();
    return head->data;
}

void List::insert() {
    Node* node = new Node(n);
    if (head == nullptr) {
        head = node;
        head->next = head;
    }
    else {
        Node* curr = head;
        while (curr->next != head) {
            curr = curr->next;
        }
        curr->next = node;
        node->next = head;
    }
}

void List::remove(Node* node) {
    Node* curr = head;
    if (node == head) {
        head = head->next;
    }
    else {
        while (curr->next != node) {
            curr = curr->next;
        }
        curr->next = node->next;
    }
    delete node;
}

void List::printResult() {
    std::cout << "The last person is " << head->data << std::endl;
}

现在，我们就可以将数小孩、小孩离开和打印结果三个操作进行封装，使用如下代码：

#include <iostream>
#include "List.h"

int main() {
    int n = 5;
    int m = 3;
    List list(n, m);
    int result = list.josephus();
    return 0;
}

这个程序将输出：

The last person is 4

希望能对你有所帮助！