课程设计｜C++实现两个链表的合并

Posted 2022-06-18 海轰Pro

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1. 实现两个链表的合并

基本功能与要求

（1）建立两个链表A和B，链表元素个数分别为m和n个。

（2）假设元素分别为(x1,x2,…,xm)，和(y1,y2,…,yn)。把它们合并成一个线性表C：
当m>=n时，C=x1,y1,x2,y2,…,xn,yn,…,xm
当n>m时， C=y1,x1,y2,x2,…,ym,xm,…,yn；

（3）输出线性表C。

（4）用直接插入排序法对C进行升序排序，生成链表D，并输出链表D。

测试数据

（1） A表（30，41，15，12，56，80）
B表（23，56，78，23，12，33，79，90，55）

（2） A表（30，41，15，12，56，80，23，12，34）
B表（23，56，78，23，12）

代码

#include <iostream>
using namespace std;

template <class t>
class node

public:
    t data;
    node<t> *next;
    node(node<t> *ptr = NULL)  next = ptr; 
    node(t a, node<t> *ptr = NULL)
    
        data = a;
        next = ptr;
    
;

template <class t>
class linklist

public:
    void merge(linklist<t> &l1, linklist<t> &l2)
    
        int m = l1.length();
        int n = l2.length();
        node<t> *r;
        node<t> *s;
        r = first;
        node<t> *p = l1.first->next;
        cout << p->data << endl;
        node<t> *q = l2.first->next;
        if (m >= n)
        
            while (p != NULL && q != NULL)
            
                s = new node<t>(p->data);
                r->next = s;
                r = s;
                s = new node<t>(q->data);
                r->next = s;
                r = s;
                p = p->next;
                q = q->next;
            
            r->next = p;
        
        else
        
            while (p != NULL && q != NULL)
            
                s = new node<t>(q->data);
                r->next = s;
                r = s;
                s = new node<t>(p->data);
                r->next = s;
                r = s;
                p = p->next;
                q = q->next;
            
            r->next = q;
        
    
    void sort()
    
        node<t> *L = first->next;
        node<t> *p, *q, *pre;
        p = L->next->next;
        L->next->next = NULL;
        while (p)
        
            q = p->next;
            pre = L;
            while (pre->next != NULL && pre->next->data < p->data)
                pre = pre->next;
            p->next = pre->next;
            pre->next = p;
            p = q;
        
        // first = L
        // node<t> *r;
        // node<t> *p = l.first->next;
        // r = first;
        // node<t> *newNode = new node<t>(p->data);
        // r->next = newNode;
        // r = newNode;
        // p = p->next;
        // while(p != NULL) 
        //     t num = r->data;
        //     newNode = new node<t>(p->data);
        //     if(p->data > num) 
        //         // newNode = new node<t>(p->data);
        //         r->next = newNode;
        //         r = newNode;
        //      else 

        //     
        //     p = p->next;
        // 
    
    void rel()
    
        node<t> *p;
        node<t> *s;
        p = first->next;
        s = p->next;
        first->next = NULL;
        while (p != NULL)

        
            p->next = first->next;
            first->next = p;
            p = s;
            // cout<<"1"<<endl;
            if (p != NULL)

            
                s = p->next;
            
        
    
    linklist()  first = new node<t>; 
    linklist(t x)  first = new node<t>(x); 
    void creat(t end)
    
        node<t> *newnode;
        t val;
        cin >> val;
        while (val != end)
        
            newnode = new node<t>(val);
            newnode->next = first->next;
            first->next = newnode;
            cin >> val;
        
    

    void creat1(t end)
    
        node<t> *r;
        node<t> *s;
        r = first;
        t val;
        cin >> val;
        while (val != end)
        

            s = new node<t>(val);
            r->next = s;
            r = s;
            cin >> val;
        
        // first=new node<t>
        r->next = NULL;
    
    void print()
    

        node<t> *p = first->next;
        while (p != NULL)
        
            cout << p->data << "--->>";
            p = p->next;
        
        cout << "NULL" << endl;
    

    int length()
    
        node<t> *p;
        p = first->next;
        int count = 0;
        while (p != NULL)
        
            p = p->next;
            count++;
        
        return count;
    
    t get(int i)
    

        node<t> *p;
        int count = 1;
        p = first->next;

        while (p != NULL && count < i)
        
            p = p->next;
            count++;
        
        if (p == NULL)
            throw "位置";
        else
            return p->data;
    

    void insert(t x, int i)
    
        node<t> *p;
        node<t> *s;
        p = first;
        int count = 0;
        while (p != NULL && count < i - 1)
        
            p = p->next;
            count++;
        
        if (p == NULL)
            throw "位置";
        else
        
            s = new node<t>(x);
            s->next = p->next;
            p->next = s;
        
    

    int locate(t x)
    
        node<t> *p;
        p = first->next;
        int count = 1;
        while (p != NULL)
        
            if (p->data == x)
                return count;
            p = p->next;
            count++;
        
        return 0;
    

    void del(int i)
    
        node<t> *p;
        node<t> *q;
        p = first;
        int count = 0;
        while (p != NULL && count < i - 1)
        
            p = p->next;
            count++;
        
        if (p == NULL || p->next == NULL)
            throw "位置";
        else
        
            q = p->next;
            p->next = q->next;
            delete q;
        
    

    ~linklist()
    
        node<t> *p;
        while (first != NULL)
        

            p = first;
            first = first->next;
            delete p;
        
    

    // private:
    node<t> *first;
;

linklist<int> merge(linklist<int> &l1, linklist<int> &l2)

    int m = l1.length();
    int n = l2.length();
    cout << "m =" << m << endl;
    linklist<int> ans;
    // linklist<int> ans;
    // cout << "((" << endl;
    node<int> *cur = ans.first;
    cout << "((" << endl;
    // cout << ans.first->data;
    node<int> *p = l1.first;
    node<int> *q = l2.first;
    if (m >= n)
    
        while (p != NULL && q != NULL)
        
            cur->next = p;
            cur = cur->next;
            cur->next = q;
            cur = cur->next;
            p = p->next;
            q = q->next;
        
        cur->next = p;
    
    else
    
        while (p != NULL && q != NULL)
        
            cur->next = q;
            cur = cur->next;
            cur->next = p;
            cur = cur->next;
            p = p->next;
            q = q->next;
        
        cur->next = q;
    
    return ans;

int main()


    linklist<int> L;
    linklist<int> A;
    linklist<int> B;
    linklist<int> C;
    linklist<int> D;
    int key = 1;
    int choice;
    int choice1;
    int choice2;
    while (key == 1)
    
        cout << "+++++++++++++++++++++++++++++" << endl;
        cout << "+       单链表              +" << endl;
        cout << "+  1.创建单链表（头部插入） +" << endl;
        cout << "+  2. 创建单链表（尾部插入）+" << endl;
        cout << "+  3.查找第i个结点的值      +" << endl;
        cout << "+  4.查找元素x的结点位置    +" << endl;
        cout << "+  5.在第i个位置插入元素x   +" << endl;
        cout << "+  6.删除第i个位置的值      +" << endl;
        cout << "+  7.输出单链表各元素       +" << endl;
        cout << "+  8.旋转链表               +" << endl;
        cout << "+  9.新建两个链表               +" << endl;
        cout << "+  10.直接插入排序              +" << endl;
        cout << "输入你的选择：（1---10）" << endl;
        cin >> choice;
        switch (choice)
        
        case 1:
        
            cout << "输入需要存储的数据 以0结尾 例如 1 2 3 4 0" << endl;
            L.creat(0);
            cout << "单链表创建完成！" << endl;
        
        break;
        case 2:
        
            cout << "输入需要存储的数据 以0结尾 例如 1 2 3 4 0" << endl;
            L.creat1(0);
            cout << "单链表创建完成！" << endl;
        
        break;
        case 3:
        

            cout << "输入查找元素的位置i（第i个结点）：";
            cin >> choice1;
            cout << "第" << choice1 << "个结点的值是：";
            cout << L.get(choice1) << endl;
        
        break;
        case 4:
        
            cout << "输入查找元素的值：";
            cin >> choice1;

            cout << "该元素在第" << L.locate(choice1) << "个位置！" << endl;
        
        break;
        case 5:
        
            cout << "输入插入元素的位置（第i个） ：";
            cin >> choice1;
            cout << "输入插入元素的值：";
            cin >> choice2;
            L.insert(choice2, choice1);
            cout << "插入完成！" << endl;
        
        break;
        case 6:
        
            cout << "输入删除元素的位置（第i个）：" << endl;
            cin >> choice1;
            L.del(choice1);
            cout << "删除完成！" << endl;
        
        break;
        case 7:
        
            cout << "该链表元素的值是：" << endl;
            L.print();
        
        break;
        case 8:
        

            L.rel();
            // cout<<"1"<<endl;
        
        case 9:
        
            cout << "输入需要存储的数据 以0结尾 例如 1 2 3 4 0" << endl;
            A.creat1(0);
            cout << "单链表创建完成！" << endl;

            cout << "输入需要存储的数据 以0结尾 例如 1 2 3 4 0" << endl;
            B.creat1(0);
            cout << "单链表创建完成！" << endl;

            C.merge(A, B);

            C.print();
            // break;
        
        break;
                case 10:
        
            C.sort();
            C.print();
        
        break;
        
        cout << "是否继续？（1：继续 0：退出）" << endl;
        cin >> key;
        system("cls");
    
    return 0;