为什么在此C++代码中触发此 if 语句?

Question

这段代码应该反转链表。下面的代码返回一个空链表，即使提供了非空列表。

class Solution {
public:
ListNode* reverseList(ListNode* head) {
ListNode* curr, *prev, *next;
if (head == NULL)
{
return head;   
}   
curr = head;
prev = NULL;
while (curr != NULL)
{
next = curr -> next;
curr -> next = prev;
prev = curr;
curr = next;
}
head = prev;
return head;
}
};

虽然这段代码奇怪地工作，但我添加了一个 cout 语句只是为了检查是否触发了 else 语句。

class Solution {
public:
ListNode* reverseList(ListNode* head) {
ListNode* curr, *prev, *next;
if (head == NULL)
{
cout << "Triggered";
return head;   
}   
curr = head;
prev = NULL;
while (curr != NULL)
{
next = curr -> next;
curr -> next = prev;
prev = curr;
curr = next;
}
head = prev;
return head;
}
};

有人可以解释为什么会发生这种情况吗？

Answer 1

非常简单，您必须初始化pointers，否则会导致意外行为，包括根本不显示它或在触发初始化cout时仅显示它 - 但它不必做任何事情，这取决于您的编译器实现。

//cpp17
listNode* curr{}, *prev{}, *next{};
//before
listNode* curr = nullptr, *prev = nullptr, *next = nullptr;

它仍然没有按照您打算的相反顺序排列。

class Solution {
public:
ListNode* reverseList(ListNode* head) {
listNode* curr{}, *prev{}, *next{};
//ListNode* curr, *prev, *next;
if (head == NULL)
{
return head;   
}   
curr = head;
prev = NULL;
while (next != NULL)
{
next = curr -> next;
curr -> next = prev;
prev = curr;
curr = next;
}
head = prev;
return head;
}
};

欢呼:)

Answer 2

就像之前提到的，我抽出时间为另一种解决问题的方法编写解决方案，以通过类反转链表。为了更好地理解初学者，我跳过了三/五的规则，并在 main 函数中初始化了列表，而不是通过类中的构造函数：

#include <iostream>
class listElement
{
std::string data;
listElement* next;
listElement* last;

public:
void setData(std::string);
void append(std::string);
void displayElements();
void reverseDisplayElements(listElement*);
void freeMemory();
listElement* reverseList(listElement*);
};
void listElement::setData(std::string newData)
{
last = this;
data = newData;
next = nullptr;
}
void listElement::append(std::string newData)
{
// Double linked list
//    last->next = new listElement();
//    last->next->data = newData;
//    last->next->next = nullptr;
//    last = last->next;
// Singly linked list
//has next the value nullptr?
//If yes, next pointer
if (next == nullptr)
{
next = new listElement();
next->data = newData;
next->next = nullptr;
}
//else the method again
else
next->append(newData);
}
listElement* listElement::reverseList(listElement* head)
{
//return if no element in list
if(head == nullptr)
return nullptr;
//initialize temp
listElement* temp{};
while(head != nullptr){
listElement* next = head->next;
head->next = temp;
temp = head;
head = next;
}
return temp;
}
void listElement::displayElements()
{
//cout the first entry
std::cout << data << std::endl;
//if the end is not reached, call method next again
if (next != nullptr)
next->displayElements();
}
void listElement::reverseDisplayElements(listElement*head)
{
//recursiv from the last to the list beginning - stop
listElement *temp = head;
if(temp != nullptr)
{
if(temp->next != nullptr)
{
reverseDisplayElements(temp->next);
}
std::cout << temp->data << std::endl;
}
}
void listElement::freeMemory()
{
//If the end is not reached, call the method again
if (next != nullptr)
{
next->freeMemory();
delete(next);
}
}
int main ()
{
//Pointer to the Beginning of the list
listElement* linkedList;
//Creating the first element
linkedList = new listElement();
//Write data in the first element
linkedList->setData("Element 1");
//add more elements
linkedList->append("Element 2");
linkedList->append("Element 3");
linkedList->append("Element 4");
//display list
linkedList->displayElements();
//space divider
std::cout << "nPrint in reverse order:" << std::endl;
//display list in reverse order
//pass list beginning as stop point
linkedList->reverseDisplayElements(linkedList);
std::cout << std::endl;
linkedList->displayElements();
std::cout << "nReverse elements:" << std::endl;
linkedList = linkedList->reverseList(linkedList);
linkedList->displayElements();
std::cout << std::endl;
//destruct the list and free memory
linkedList->freeMemory();
delete(linkedList);
return 0;
}

顺便说一句，该任务有许多不同的解决方案。