【问题标题】:C++ Sorted Doubly Linked List: Problems inserting in middle of listC++ 排序双向链表:在列表中间插入问题
【发布时间】:2018-11-17 05:50:46
【问题描述】:

下面是我当前正在进行的代码,将单链表转换为双链表。我还没有接触过删除功能。我已经在空列表中插入,列表末尾和列表开头显然有效。

然而,插入中间的节点似乎无法创建到前一个节点的链接。我插入的调试行似乎同时显示了具有正确内存地址的 n->next 和 n-> prev,但是当我进行反向打印时,插入中间的任何节点都被遗漏并且链接消失了。我在这方面哪里出错了?

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;
}

【问题讨论】:

  • 不相关:main 必须始终具有返回类型 int

标签: c++ linked-list


【解决方案1】:

您没有将当前插入的 prev 设置到中间,只需这样做:

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

【讨论】:

  • 这只是在其中添加该行而已。现在我必须弄清楚为什么......
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