【发布时间】:2023-03-13 11:03:01
【问题描述】:
所以,我尝试将我的 Java 版本的链表移植到 C++ 中,但我在使用指针指向正确的对象时遇到了很大的困难。 remove遍历和reverse traverse的目的就是简单的把链表的值依次打印出来,然后倒序打印出来。这是代码。我发布大部分而不是某些部分的原因是因为我相信您需要它的上下文。
控制台输出: 他们是平等的 删除 5 遍历 -858993460
它打印出它们相等,因为它在 remove(data) 中 while 循环的第一次迭代时立即中断。
#include <iostream>
#include <string>
#include <string.h>
using namespace std;
class Node{
public:
int data;
Node *pNext;
Node();
Node(int x);
};
Node::Node(){
data = NULL;
pNext = NULL;
}
Node::Node(int x){
data = x;
pNext = NULL;
}
class List{
private:
Node *head = NULL;
public:
void insert(int data){
Node temp(data);
temp.pNext = head;
head = &temp;
}
Node removeHead(){
if (head != NULL){
Node *temp = head;
head = head->pNext;
return *temp;
}
else{
cout << "Empty List" << endl;
}
return NULL;
}
Node remove(int data){
Node *previousLink = head;
Node *currentLink = head;
while (currentLink->data != data){
if (currentLink == currentLink->pNext){
cout << "They are equal" << endl;
break;
}
previousLink = currentLink;
if (currentLink->data == NULL)
currentLink = currentLink->pNext;
cout << "Current " << currentLink->data << " Previous "
<< previousLink->data <<endl;
}
if (head->data == data){
head = head->pNext;
}
else{
previousLink->pNext = currentLink->pNext;
}
return *currentLink;
}
void traverse(){
traverse(head);
}
void reverseTraverse(){
reverseTraverse(head);
}
private:
void traverse(Node *link){
cout << link->data << endl;
if (link != link->pNext)
traverse(link->pNext);
}
void reverseTraverse(Node *link){
if (link != link->pNext)
traverse(link->pNext);
cout << link->data << endl;
}
};
int main(){
cout << " Hello World\n";
List list;
for (int i = 0; i <= 9; ++i)
list.insert(i);
list.remove(5);
cout << "removed 5\n";
cout << "traverse" << endl;
list.traverse();
cout << "reverse" << endl;
list.reverseTraverse();
system("PAUSE");
return 0;
}
【问题讨论】:
-
我不太清楚为什么不使用内置的 STL 容器。也许使用Standard Containers 会更容易,比如
List<T>。
标签: c++ pointers reference linked-list traversal