我在编写添加和删除功能方面遇到了一些困难。 这里有一些用于显示列表，搜索和删除列表的代码。
我知道如何迭代地实现它，但是通过递归，我遇到了一些问题。
#include <iostream>
#include <string>
using namespace std;
struct AddressBook
{
string name;
string surname;
long long phone;
AddressBook* next;
};
void delPerson(AddressBook*& head)
{
if (head == NULL)
{
cout << "There are no persons in my book\n";
}
else
{
string pName;
cout << "Enter name of person: ";
getline(cin, pName);
AddressBook* temp = head;
AddressBook* curr = NULL;
while (temp != NULL)
{
if (temp->name == pName)
{
break;
}
else
{
curr = temp;
temp = temp->next;
}
}
if (temp == NULL)
{
cout << "There are no person with this name\n";
}
else
{
if (temp == head)
{
head = head->next;
delete temp;
}
else
{
curr->next = temp->next;
delete temp;
}
cout << pName << " was deleted from the book\n";
}
}
}
void display(AddressBook* head)
{
if (head == NULL)
{
cout << "My book is empty\n";
}
else
{
AddressBook* temp = head;
while (temp != NULL)
{
cout << endl;
cout << "Name: " << temp->name << endl;
cout << "Surname: " << temp->surname << endl;
cout << "Phone: " << temp->phone << endl;
temp = temp->next;
}
}
}
void search(AddressBook* head)
{
if (head == NULL)
{
cout << "My book is empty\n";
}
else
{
cin.get();
string pName;
cout << "Enter name of person: ";
getline(cin, pName);
AddressBook* temp = head;
while (temp != NULL)
{
if (temp->name == pName)
{
cout << temp->name << " is found\n\n";
cout << "Name: " << temp->name << endl;
cout << "Surname: " << temp->surname << endl;
cout << "Phone: " << temp->phone << endl;
break;
}
else
{
temp = temp->next;
}
}
if (temp == NULL)
{
cout << pName << " isn't found in my book\n";
}
}
}
void delMemory(AddressBook* head)
{
while (head != NULL)
{
AddressBook* temp = head;
head = head->next;
delete temp;
}
}
void addPerson(AddressBook*& head)
{
string pName;
string sName;
long long pPhone = 0;
while (true)
{
cin.ignore();
cout << endl;
cout << "Enter the name(press '0' to end): ";
getline(cin, pName);
if (pName == "0")break;
cout << "Enter surname: ";
getline(cin, sName);
cout << "Enter phone: ";
cin >> pPhone;
AddressBook* bleah = new AddressBook;
bleah->name = pName;
bleah->surname = sName;
bleah->phone = pPhone;
bleah->next = NULL;
if (head == NULL)
{
head = bleah;
}
else
{
AddressBook* temp = head;
while (temp->next != NULL)
{
temp = temp->next;
}
temp->next = bleah;
}
}
}
int main()
{
cout << "=============== My Address Book ===============\n\n";
cout << "1. To add person\n";
cout << "2. To display all\n";
cout << "3. To delete person\n";
cout << "4. To search person\n";
cout << "5. Exit\n";
AddressBook* head = NULL;
int choice = 0;
while (true)
{
cin >> choice;
switch (choice)
{
case 1: addPerson(head);
break;
case 2: display(head);
break;
case 3: cin.get();
delPerson(head);
break;
case 4: search(head);
break;
case 5: return 0;
default: return 0;
}
cout << "=============== My Address Book ===============\n\n";
cout << "1. To add person\n";
cout << "2. To display all\n";
cout << "3. To delete person\n";
cout << "4. To search person\n";
cout << "5. Exit\n";
cout << endl;
}
delMemory(head);
return 0;
}
这是递归解决方案，但我在实现添加和删除功能方面遇到了麻烦
#include <iostream>
#include <string>
using namespace std;
struct AddressBook
{
string name;
string surname;
long long phone;
AddressBook* next;
};
AddressBook* delPerson(AddressBook*& head)
{
}
AddressBook* display(AddressBook* head)
{
if (head!= NULL)
{
cout << endl;
cout << "Name: " << head->name << endl;
cout << "Surname: " << head->surname << endl;
cout << "Phone: " << head->phone << endl;
return display(head->next);
}
return head;
}
AddressBook* search(AddressBook* head, string pName)
{
if (head == NULL)
{
cout << pName << " isn't found in my book\n";
return head;
}
if (head->name == pName)
{
cout << head->name << " is found\n\n";
cout << "Name: " << head->name << endl;
cout << "Surname: " << head->surname << endl;
cout << "Phone: " << head->phone << endl;
return head;
}
else
{
return search(head->next, pName);
}
}
void delMemory(AddressBook* head)
{
if (head != NULL)
{
delMemory(head->next);
}
delete head;
}
AddressBook* allMem(AddressBook*& head)
{
}
int main()
{
cout << "=============== My Address Book ===============\n\n";
cout << "1. To add person\n";
cout << "2. To display all\n";
cout << "3. To delete person\n";
cout << "4. To search person\n";
cout << "5. Exit\n";
AddressBook* head = NULL;
string pName;
int choice = 0;
while (true)
{
cin >> choice;
switch (choice)
{
case 1: allMem(head);
break;
case 2: display(head);
break;
case 3: cin.get();
delPerson(head);
break;
case 4: cin.get();
cout << "Enter name of person: ";
getline(cin, pName);
search(head, pName);
break;
case 5: return 0;
default: return 0;
}
cout << "=============== My Address Book ===============\n\n";
cout << "1. To add person\n";
cout << "2. To display all\n";
cout << "3. To delete person\n";
cout << "4. To search person\n";
cout << "5. Exit\n";
cout << endl;
}
delMemory(head);
return 0;
}
答案 0 :(得分：0)
这是一个非常简单的示例，说明如何使用递归来添加链接列表。您可以将此技术应用于上面的示例。
struct node{
int i;
node* next;
};
void add(node* cur, int i){
if(cur->next == nullptr){
node* n = new node;
n->i = i;
cur->next = n;
}
else
add(cur->next, i);
}
您只需检查下一个节点是否为空。如果是，您已到达列表的末尾，您可以在那里添加新节点。如果不是，则递归调用该函数，提供当前的下一个节点。
答案 1 :(得分：0)
试试这个：
typedef struct Nodetype {
int key;
struct Node * next;
} Node;
Node * newNode(int data){
Node *temp = (Node*)malloc(sizeof(Node));
temp->key = data;
temp->next = NULL;
return temp;
}
Node* add(Node * head, int val){
if(!head)
return newNode(val);
head->next = add(head->next, val);
return head;
}
Node * delete(Node *head, int val){
if(!head)
return NULL;
Node * temp;
if(head->key == val){
temp = head;
head = head->next;
free(temp);
}
else{
head->next = delete(head->next, val);
}
return head;
}
您可以根据需要更改代码。我假设一切都是整数。
答案 2 :(得分：0)
已发布的内容错过了大部分问题：在递归函数中询问用户输入只会导致痛苦。要么在每次迭代中要求用户输入，要么需要传递额外的控制信息以告诉函数不要求用户输入。要么是糟糕的计划。
而是插入另一个或两个函数。从软件工程的角度来看，这是有益的：功能应该绝对最小化以完成一项工作。只做一件事的函数很容易理解，很容易调试，而且往往很短。
因此我们需要一个函数来从用户那里获取AddressBook：
AddressBook*createAddressBook()
{
AddressBook* bleah = new AddressBook; // terrible name, by the way
// since we got the bleah first, we can directly assign to it. No need for
// extra temporary variables.
cout << "Enter the name: ";
getline(cin, bleah->name);
cout << "Enter surname: ";
getline(cin, bleah->surname);
cout << "Enter phone: ";
cin >> bleah->phone; // bad idea to store phone number as a number.
// For example a number can't record preceding zeros
bleah->next = NULL;
return bleah;
}
这是一个简单的愚蠢的功能，只做一个AddressBook。说实话，这应该是一个AddressBook构造函数，但我们可以保存它以用于将来的问题。另请注意，0废话上没有出口。此函数生成AddressBook。期。如果您不想使用AddressBook，则不要调用此函数。让菜单句柄退出。
接下来，我们需要一个函数来查找AddressBoook的添加位置。看起来这总是列表的末尾，所以
AddressBook*& findEnd(AddressBook *& head)
{
if (head != NULL)
{
return findend(head->next); // keep looking for the end
}
return head; // return the end.
}
同样，一个简单的，愚蠢的函数除了找到列表中的最后一个之外什么都不做。请注意，您将获得对AddressBook指针的引用。这允许您愉快地返回指向NULL的下一个并使用新的AddressBook
这将我们带回addPerson
AddressBook* addPerson(AddressBook*& head)
{
findEnd(head) = createAddressBook();
return head;
}
简单，愚蠢的函数，调用另外两个简单的愚蠢函数，并将一个人添加到列表中。
将相似的流程应用于delPerson。
注意：我没有运行任何此类操作。可能是一两个错字。