C++ C++;排序双链接列表:在列表中间插入问题

C++ C++;排序双链接列表:在列表中间插入问题,c++,linked-list,C++,Linked List,下面是我当前正在进行的代码,它将单链接列表转换为双链接列表。我还没有触及删除功能。我已经在空列表中插入了insert,列表的末尾和开头都工作了 但是,在中间插入的节点似乎无法创建与前一个节点的链接。我插入的调试行似乎同时显示了N-> NEXT和N-> PREV以及正确的内存地址,但是当我回到ReavePress时,中间插入的任何节点都会丢失,链接也就消失了。在这方面我错在哪里 代码如下: #include <iostream> #include <string> usin

下面是我当前正在进行的代码,它将单链接列表转换为双链接列表。我还没有触及删除功能。我已经在空列表中插入了insert,列表的末尾和开头都工作了

但是,在中间插入的节点似乎无法创建与前一个节点的链接。我插入的调试行似乎同时显示了N-> NEXT和N-> PREV以及正确的内存地址,但是当我回到ReavePress时,中间插入的任何节点都会丢失,链接也就消失了。在这方面我错在哪里

代码如下:

#include <iostream>
#include <string>
using namespace std;

// define a node for storage and linking
class node {
public:
    string name;
    node *next;
    node *prev; 
};

class linkedList {
public:
    linkedList() :top(NULL) {}
    bool empty() { return top == NULL; }
    node *getTop() { return top; }
    node *getEnd() { return end; }
    void setTop(node *n) { top = n; }
    void setEnd(node *p) { end = p; }
    void add(string);
    int menu();
    void remove(string);
    ~linkedList();
     void reversePrint(); 
    friend ostream& operator << (ostream&, const linkedList&); // default output is in-order print.
private:
    node *top;
    node *end; 
};

void main() {
    linkedList l;
    cout << l.empty() << endl;
    int option = 0;
    string s;
    bool go = true;
    while (go) {
        option = l.menu();
        switch (option) {
        case 1: cout << "enter a name: "; cin >> s; l.add(s); break;
        case 2: cout << "enter name to be deleted: "; cin >> s; l.remove(s); break;
        case 3: cout << l; break;
        //case 4: cout << "can not be done with a singly linked list" << endl;
        case 4: l.reversePrint(); break;
        case 5: cout << "exiting" << endl; go = false; break;
        }
    }


    system("pause");
}


void linkedList::remove(string s) {
    bool found = false;
    node *curr = getTop(), *prev = NULL;
    while (curr != NULL) {

        // match found, delete
        if (curr->name == s) {
            found = true;

            // found at top
            if (prev == NULL) {
                node *temp = getTop();
                setTop(curr->next);
                delete(temp);

                // found in list - not top
            }
            else {
                prev->next = curr->next;
                delete(curr);
            }
        }

        // not found, advance pointers
        if (!found) {
            prev = curr;
            curr = curr->next;
        }

        // found, exit loop
        else curr = NULL;
    }
    if (found)cout << "Deleted " << s << endl;
    else cout << s << " Not Found " << endl;
}

void linkedList::add(string s) {
    node *n = new node();
    n->name = s;
    n->next = NULL;
    n->prev = NULL;

    // take care of empty list case
    if (empty()) {
        top = n;
        end = n;

        // take care of node belongs at beginning case
    }
    else if (getTop()->name > s) {


        n->next = getTop();
        n->prev = NULL;
        setTop(n);

        node *temp;
        temp = n->next;
        temp->prev = n;

        // take care of inorder and end insert
    }
    else {

        // insert in order case
        node *curr = getTop(), *prev = curr;
        while (curr != NULL) {
            if (curr->name > s)break;
            prev = curr;
            curr = curr->next;
        }
        if (curr != NULL) { // search found insert point
            n->next = curr;
            cout << n->name << " " << n << " prev " << prev << " " << prev->name << endl;
            n->prev = prev;

            prev->next = n;
            cout << "n->prev is: " << n->prev << " " << n->prev->name << endl;
            cout << "n->next is: " << n->next << " " << n->next->name << endl;
        }

        // take care of end of list insertion
        else if (curr == NULL) {// search did not find insert point
            prev->next = n;
            n->prev = prev;
            cout << "n->prev is: " << n->prev << " " << n->prev->name << endl;
            setEnd(n);
        }
    }
}

ostream& operator << (ostream& os, const linkedList& ll) {
    //linkedList x = ll; // put this in and the code blows up - why?
    node *n = ll.top;
    if (n == NULL)cout << "List is empty." << endl;
    else
        while (n != NULL) {
            os << n->name << endl;
            os << n << endl;
            if (n->next != NULL) {
                os << "next is " << n->next << endl;
            }

            n = n->next;
        }
    return os;
}

void linkedList::reversePrint() {
    node *n = end;


    if (n == NULL)cout << "List is empty." << endl;
    else
        while (n != NULL) {
            //cout << n->name << endl;
            cout << "memory address of " << n->name << " is " << n << endl;

            if (n->prev != NULL) {
                cout << "prev is " << n->prev << endl;
            }
            n = n->prev;
        }
    return;
}
// return memory to heap
linkedList::~linkedList() {
    cout << "~linkedList called." << endl;
    node *curr = getTop(), *del;
    while (curr != NULL) {
        del = curr;
        curr = curr->next;
        delete(del);
    }
}

int linkedList::menu() {
    int choice = 0;
    while (choice < 1 || choice > 5) {
        cout << "\nEnter your choice" << endl;
        cout << " 1. Add a name." << endl;
        cout << " 2. Delete a name." << endl;
        cout << " 3. Show list." << endl;
        cout << " 4. Show reverse list. " << endl; 
        cout << " 5. EXIT " << endl;
        cin >> choice;
    }
    return choice;
}
#包括
#包括
使用名称空间std;
//定义用于存储和链接的节点
类节点{
公众:
字符串名;
节点*下一步;
节点*prev;
};
类链接列表{
公众:
linkedList():top(NULL){}
bool empty(){return top==NULL;}
node*getTop(){return top;}
node*getEnd(){return end;}
void setTop(node*n){top=n;}
void setEnd(node*p){end=p;}
无效添加(字符串);
int菜单();
脱空(串);
~linkedList();
void reversePrint();
friend-ostream&运算符prev=NULL;
setTop(n);
节点*温度;
temp=n->next;
温度->上一个=n;
//注意顺序和末端插入
}
否则{
//在订单箱中插入
节点*curr=getTop(),*prev=curr;
while(curr!=NULL){
如果(curr->name>s)中断;
上一次=当前;
当前=当前->下一步;
}
如果(curr!=NULL){//搜索找到插入点
n->next=当前;

cout name插入到中间时,您没有设置当前的prev,只需执行以下操作:

n->next = curr;
curr->prev = n;  // <-- this
n->next=curr;

curr->prev=n;//不相关:
main
必须始终具有返回类型
int
,只需在其中添加那一行就行了,其他什么都没有。现在我必须找出原因。。。。