【发布时间】：2020-05-23 13:08:56
【问题描述】：
我正在用 C++ 编写双向链表的代码。我是 C++ 的初学者，所以请原谅我问这个问题。我正在用数字（1、2、3、...、n）填充双向链表。我首先使用搜索功能并找到包含我要删除的值的节点。然后我调用 remove 函数来删除节点。我的删除功能似乎不仅要删除一个节点，而且我一直在绞尽脑汁想弄清楚。我认为这可能是因为我的 DList 类没有析构函数，但我不确定如何实现它，因为我要删除的结构是 ListNode。任何见解将不胜感激。谢谢。
例如。测试用例 n = 8
删除前的列表：1 2 3 4 5 6 7 8
删除后的列表：7
DLIST.H
#ifndef DLIST_H
#define DLIST_H
struct ListNode
{
/* define your list node type */
int val;
ListNode* next;
ListNode* prev;
};
class DList
{
public:
DList();
/* implement copy constructor, assignment, destructor if needed */
void add_to_front(int value);
void add_to_back(int value);
int first();
int last();
void remove(ListNode* node);
void display();
ListNode* previous(ListNode* node);
ListNode* next(ListNode* node);
ListNode* search_value(int value);
private:
/* declare your data */
ListNode* head;
ListNode* tail;
};
#endif
DLIST.CC
#include "dlist.h"
#include <cstddef>
#include <cstdlib>
#include <cstdio>
#include <iostream>
//ListNode* head;
//ListNode* tail;
DList::DList(){
head = NULL;
tail = NULL;
}
void DList::add_to_front(int value){
if (head != NULL){
//std::cout<<"addfront1\n";
ListNode *newhead = new ListNode();
newhead->val = value;
//std::cout<<"addfront1 val" << newhead->val << "\n";
newhead->next = head;
newhead->prev = NULL;
head->prev = newhead;
head = newhead;
}else{
//std::cout<<"addfront2\n";
ListNode *newhead = new ListNode();
newhead->val = value;
//std::cout<<"addfront2 val" << newhead->val << "\n";
newhead->prev = NULL;
newhead->next = NULL;
head = newhead;
tail = newhead;
}
}
void DList::add_to_back(int value){
if (tail != NULL){
//std::cout<<"addback1\n";
ListNode *newtail = new ListNode();
newtail->val = value;
//std::cout<<"addback1 val" << newtail->val << "\n";
newtail->next = NULL;
newtail->prev = tail;
tail->next = newtail;
tail = newtail;
}else{
//std::cout<<"addback2\n";
ListNode *newtail = new ListNode();
newtail->val = value;
//std::cout<<"addback2 val" << newtail->val << "\n";
newtail->prev = NULL;
newtail->next = NULL;
tail = newtail;
head = newtail;
}
}
int DList::first(){
return head->val;
}
int DList::last(){
return tail->val;
}
void DList::remove(ListNode* node){
std::cout << "want to del value is " << node->val << "\n";
ListNode *toDelete = new ListNode();
if (head == NULL || tail == NULL){
return;
}
if(head->val == node->val){
toDelete = head;
head = node->next;
if (head != NULL){
head->prev = NULL;
}
std::cout << "del1 value is " << toDelete->val << "\n";
delete(toDelete);
}else if(tail->val == node->val){
toDelete = tail;
tail = node->prev;
std::cout << "del2 newtail value is " << node->prev->val << "\n";
if (tail != NULL){
tail->next = NULL;
}
std::cout << "del2 value is " << toDelete->val << "\n";
delete(toDelete);
}else if (node != NULL){
std::cout << "node1 value is " << node->val << "\n";
node->prev->next = node->next;
node->next->prev = node->prev;
std::cout << "node2 value is " << node->val << "\n";
delete(node);
}
}
ListNode* DList::previous(ListNode* node){
if(node->prev != NULL){
return node->prev;
}else{
return node;
}
}
ListNode* DList::next(ListNode* node){
if(node->next != NULL){
return node->next;
}else{
return node;
}
}
ListNode* DList::search_value(int value){
//std::cout << "head.0\n";
//std::cout << "head value is " << head->val << "\n";
while(head->next != NULL){
//std::cout << "head.1\n";
if(head->next->val == value){
//std::cout << "head.2\n";
return head->next;
}else{
//std::cout << "head.3\n";
head = head->next;
}
}
if (head->next == NULL){
if (head->val == value){
return head;
}else{
//std::cout << "error\n";
}
}
}
void DList::display(){
std::cout<<"Element In The Linked List Are : ";
ListNode *disp= new ListNode;
disp = head;
while(disp!=NULL)
{
std::cout<<" "<<disp->val;
if(disp->next==NULL)
{
std::cout<<"\n";
break;
}
disp=disp->next;
}
}
DLIST_TEST.CC
#include <cassert>
#include <cstdlib>
#include <cstdio>
#include <ctime>
#include <iostream>
#include <iomanip>
#include "dlist.h"
using namespace std;
int main (int argc, char* argv[])
{
int N = -1;
if (argc == 2) {
N = atoi (argv[1]);
assert (N > 0);
}
//cout << "N is " << N << "\n";
DList testList = DList();
int i = 0;
while(i<N){
//cout << "i value is "<< i << "\n";
testList.add_to_back(i+1);
//cout << "second\n";
i++;
}
int randn = rand() % N + 1;// randn in the range 1 to N
cout << "randn is "<< randn <<"\n";
testList.display();
clock_t t1,t2;
t1=clock();
ListNode* loc = testList.search_value(randn);
cout << "loc value is "<< loc->val <<"\n";
testList.remove(loc);
t2=clock();
float diff ((float)t2-(float)t1);
float seconds = diff / CLOCKS_PER_SEC;
cout<<"\nTime taken by function: "<<seconds<<" seconds\n" << endl;
testList.display();
testList.add_to_front(N);
testList.display();
return 0;
}
【问题讨论】：
-
第一个错误：
ListNode *toDelete = new ListNode();与“删除”功能无关，尤其是因为它是最终成为内存泄漏的开始。 -
同样的
ListNode *disp= new ListNode;错误在display()。 -
当要删除的
node在列表中间但恰好与头或尾节点具有相同的值时，您的remove()将从列表中删除错误的节点.这是您必须修复的另一个错误。要找出错误的原因，您需要使用调试器。您是否使用过调试器来调试您的程序，如果没有，为什么不呢？ -
@SamVarshavchik，插入到列表中的值是递增的 (1,2,3,....,n)，所以我觉得不需要检查重复项，因为没有两个节点会有相同的值。我不太确定调试器，因为我真的没有使用它们的经验，但我认为这将是我学习的下一步。
-
@WhozCraig，我的理解是在函数中创建的 toDelete 只会持续函数的生命周期，因此在函数调用结束时不会有问题。不是这样吗？