c++ - 克隆一个链表，其中每个节点都有一个指向链表中任何其他节点的随机指针

Question

请帮助我找出我的代码有什么问题
(1)。
您将获得一个包含N个节点的单链接列表，其中每个节点下一个指向其下一个节点。您还将获得M个随机指针，其中将获得M个对，它们表示两个节点a和b，即a-> arb = b。
任务是完成函数copyList()，该函数将一个参数作为要克隆的链表的开头，并应返回已克隆的链表的头。
注意:如果它们是没有给出任意指针的任何节点，则默认情况下为null。
我试图为上述问题编写代码..但是没有用

// { Driver Code Starts
#include <bits/stdc++.h>
using namespace std;

struct Node {
    int data;
    Node *next;
    Node *arb;

    Node(int x) {
        data = x;
        next = NULL;
        arb = NULL;
    }
};

void print(Node *root) {
    Node *temp = root;
    while (temp != NULL) {
        int k;
        if (temp->arb == NULL)
            k = -1;
        else
            k = temp->arb->data;
        cout << temp->data << " " << k << " ";
        temp = temp->next;
    }
}

Node *copyList(Node *head);

void append(Node **head_ref, Node **tail_ref, int new_data) {

    Node *new_node = new Node(new_data);
    if (*head_ref == NULL) {
        *head_ref = new_node;
    } else
        (*tail_ref)->next = new_node;
    *tail_ref = new_node;
}

bool validation(Node *head, Node *res, Node *cloned_addr,
                Node *generated_addr) {

    if (generated_addr == cloned_addr) return false;

    Node *temp1 = head;
    Node *temp2 = res;

    int len1 = 0, len2 = 0;
    while (temp1 != NULL) {
        len1++;
        temp1 = temp1->next;
    }
    while (temp2 != NULL) {
        len2++;
        temp2 = temp2->next;
    }

    /*if lengths not equal */

    if (len1 != len2) return false;

    temp1 = head;
    temp2 = res;
    while (temp1 != NULL) {
        if (temp1->data != temp2->data) return false;
        if (temp1->arb != NULL and temp2->arb != NULL) {
            if (temp1->arb->data != temp2->arb->data) return false;
        } else if (temp1->arb != NULL and temp2->arb == NULL)
            return false;
          else if (temp1->arb == NULL and temp2->arb != NULL)
            return false;
        temp1 = temp1->next;
        temp2 = temp2->next;
    }
    return true;
}

/* Driver program to test above function*/
int main() {
    int T, i, n, l, k;
    Node *generated_addr = NULL;
    /*reading input stuff*/
    cin >> T;

    while (T--) {
        generated_addr = NULL;
        struct Node *head = NULL, *tail = NULL;

        cin >> n >> k;
        for (i = 1; i <= n; i++) {
            cin >> l;
            append(&head, &tail, l);
        }

        for (int i = 0; i < k; i++) {
            int a, b;
            cin >> a >> b;

            Node *tempA = head;
            int count = -1;

            while (tempA != NULL) {
                count++;
                if (count == a - 1) break;
                tempA = tempA->next;
            }
            Node *tempB = head;
            count = -1;

            while (tempB != NULL) {
                count++;
                if (count == b - 1) break;
                tempB = tempB->next;
            }

            // when both a is greater than N
            if (a <= n) tempA->arb = tempB;
        }
        /*read finished*/

        generated_addr = head;
        Node *res = copyList(head);

        Node *cloned_addr = res;

        cout << validation(head, res, cloned_addr, generated_addr) << endl;
    }

    return 0;
}
// } Driver Code Ends


/*  the node structure is as follows

struct Node {
    int data;
    Node *next;
    Node *arb;

    Node(int x) {
        data = x;
        next = NULL;
        arb = NULL;
    }
};
*/

// Should return the head of the copied linked list the
// output will be 1 if successfully copied
Node *copyList(Node *head) {
    if(!head)
    return nullptr;
    
    Node*q=head;
    Node*clone=new Node(q->data);
    clone->next=0;
    clone->arb=q->arb;
    Node*p=clone;
    Node*r=q;
    q=q->next;
    while(q)
    {
        r->next=p;
        p->next=new Node(q->data);
        p=p->next;
        p->next=0;
        p->arb=q->arb;
        r=q;
        q=q->next;
    }
    r->next=p;
    p=clone;
    while(p)
    {
        if(p->arb)
        p->arb=p->arb->next;
        p=p->next;
    }
    
    return clone;

    
}

Answer 1

在构造克隆列表本身之前，无法分配列表中的指针，因为在此之前，指向的节点将不存在。
因此，您需要进行两次迭代:第一次迭代以克隆列表并创建将原始节点与克隆关联的字典，第二次迭代以更新指针。代码如下所示:

Node *copyList(Node *head) {
    if(!head)
    return nullptr;
    
    Node* it1 = head;
    Node* clone = new Node;
    Node* it2 = clone;
    
    std::map<Node*, Node*> nodeDict;
    nodeDict[nullptr] = nullptr;
    
    // first iteration: create the list and the values
    while(it1){
        it2->data = it1->data;
        nodeDict[it1] = it2;
        
        it1 = it1->next;
        it2->next = it1 ? new Node: nullptr;
        it2 = it2->next;
    }
    // second iteration: connect the nodes
    it1 = head;
    it2 = clone;
    while(it1){
        it2->arb = nodeDict[it1->arb];
        it1 = it1->next;
        it2 = it2->next;
    }
    
    return clone;
}