c++ - 无法创建或返回反向链接列表

Question

这里使用函数returnReverseLinkedList返回给定链表的反向链表。但是这种方法的问题是我丢失了原始链表。createReversedLinkedList因此，我创建了另一个函数来调用copy of the original linked list并reverse the copy保持对两者的拥有。不幸createReversedLinkedList的是给予Runtime error。显然我的最终目标是检查if the given linked list is palindrome or not。这个问题只是一个垫脚石。有人能告诉我为什么吗？

//Check if a linked list is a palindrome
#include <iostream>
using namespace std;
class node
{
public:
    int data;
    node *next;
    node(int data)
    {
        this->data = data;
        this->next = NULL;
    }
};

node *returnReverseLinkedList(node *head)
{
    // Will Lose original Linked List
    if (head == NULL)
        return NULL;
    else if (head != NULL && head->next == NULL)
        return head;

    node *prev = NULL;
    node *curr = head;
    node *tempNext = head->next;

    while (tempNext != NULL)
    {
        curr->next = prev;
        prev = curr;
        curr = tempNext;
        tempNext = tempNext->next;
    }

    curr->next = prev;
    return curr;
}
node *createReversedLinkedList(node *head)
{
    if (head == NULL)
        return NULL;
    else if (head != NULL && head->next == NULL)
        return NULL;
    else
    {
        node *temp = head;

        node *newHead = NULL;
        node *newTail = NULL;

        while (temp != NULL)
        {
            node *newNode = new node(temp->data);
            if (newHead == NULL)
            {
                newHead = newNode;
                newTail = newNode;
            }
            else
            {
                newTail->next = newNode;
                newTail = newNode;
            }
        }
        return returnReverseLinkedList(newHead);
    }
}

bool check_palindrome(node *head)
{
    node *original = head;
    node *reverse = returnReverseLinkedList(head);

    while (original->next != NULL || reverse->next != NULL)
    {
        if (original->data != reverse->data)
            return false;
        cout << "debug 2" << endl;
        original = original->next;
        reverse = reverse->next;
    }

    return true;
}
// #include "solution.h"
node *takeinput()
{
    int data;
    cin >> data;
    node *head = NULL, *tail = NULL;
    while (data != -1)
    {
        node *newnode = new node(data);
        if (head == NULL)
        {
            head = newnode;
            tail = newnode;
        }
        else
        {
            tail->next = newnode;
            tail = newnode;
        }
        cin >> data;
    }
    return head;
}
void print(node *head)
{
    node *temp = head;
    while (temp != NULL)
    {
        cout << temp->data << " ";
        temp = temp->next;
    }
    cout << endl;
}
int main()
{
    node *head = takeinput();
    node *revese2 = createReversedLinkedList(head);
    print(revese2);
    // bool ans = check_palindrome(head);
    // if (ans)
    //     cout << "true";
    // else
    //     cout << "false";
    // return 0;
}

标签： c++recursiondata-structureslinked-list

Answer 1

正如 OP 所要求的那样，构建反向链接只需像构建堆栈（例如 LIFO）一样简单地完成，而不是复制相同的原始正向链。例如：

node *createReversedLinkedList(const node *head)
{
    node *newHead = NULL;
    for (; head; head = head->next)
    {
        node *p = new node(head->data)
        p->next = newHead;
        newHead = p;
    }
    return newHead;
}

请注意，我们不会将复制的节点挂在新列表的尾部；他们挂在新名单的首位，并随着每次增加而成为新的负责人。就是这样。无需制作相同的列表，然后将其反转；您可以在开始构建副本时将其反转。

---

关于代码其余部分的注释。你有一个可怕的内存泄漏，即使你修复了我上面显示的反转生成。在您的check_palindrome函数中，您永远不会释放动态反向副本（实际上，您不能因为您在第一次遍历后丢弃了指向其头部的原始指针：

bool check_palindrome(node *head)
{
    node *original = head;
    node *reverse = returnReverseLinkedList(head); // only reference to reversed copy

    while (original->next != NULL || reverse->next != NULL)
    {
        if (original->data != reverse->data)
            return false; // completely leaked entire reversed copy
        original = original->next;
        reverse = reverse->next; // lost original list head
    }

    return true;
}

对抗这种可怕泄漏的最明显方法是记住原始列表并使用不同的指针进行迭代，并且在释放副本之前不要离开函数。

bool check_palindrome(const node *head)
{
    bool result = true;
    node *reverse = returnReverseLinkedList(head);

    for (node *p = reverse; p; p = p->next, head = head->next)
    {
        if (p->data != head->data)
        {
            result = false;
            break;
        }
    }

    while (reverse)
    {
        node *tmp = reverse;
        reverse = reverse->next;
        delete tmp;
    }

    return result;
}