所以我试图做一个双链表，我可以搜索的名字，也搜索的人只低于一定的年龄，但我不知道如何做到这一点。到目前为止，我只得到了一个，但我遇到了麻烦。
#include <iostream>
#include <string>
using namespace std;
typedef string Elem; // list element type
class DNode { // doubly linked list node
private:
Elem elem; // node element value
int age;
DNode* prev; // previous node in list
DNode* next; // next node in list
friend class DLinkedList; // allow DLinkedList access
};
class DLinkedList { // doubly linked list
public:
DLinkedList(); // constructor
~DLinkedList(); // destructor
bool empty() const; // is list empty?
const Elem& front() const; // get front element
const Elem& back() const; // get back element
void addFront(const Elem& name); // add to front of list
void addBack(const Elem& name); // add to back of list
void removeFront(); // remove from front
void removeBack(); // remove from back
void displayViaAge(string age);
void displayViaName(const Elem& name);
void removeName(const Elem& name);
private: // local type definitions
DNode* header; // list sentinels
DNode* trailer;
protected: // local utilities
void add(DNode* v, const Elem& name); // insert new node before v
void remove(DNode* v); // remove node v
};
DLinkedList::DLinkedList() { // constructor
header = new DNode; // create sentinels
trailer = new DNode;
header->next = trailer; // have them point to each other
trailer->prev = header;
}
DLinkedList::~DLinkedList() { // destructor
while (!empty()) removeFront(); // remove all but sentinels
delete header; // remove the sentinels
delete trailer;
}
// insert new node before v
void DLinkedList::add(DNode* v, const Elem& name) {
DNode* u = new DNode;
u->elem = name; // create a new node and set name
u->next = v; // make v the successor of u
u->prev = v->prev; // set u's predecessor to v's current predecessor
u->prev->next = u; // make u the successor of v's predecessor
v->prev = u; // finally make u the predecessor of v
}
/*
void DLinkedList::removeName(const Elem& name) {
DNode* u = new DNode; u->elem = name; // create a new node and set name
u->next = v; // make v the successor of u
u->prev = v->prev; // set u's predecessor to v's current predecessor
u->prev->next = u; // make u the successor of v's predecessor
v->prev = u; // finally make u the predecessor of v
}
*/
void DLinkedList::addFront(const Elem& name) // add to front of list
{ add(header->next, name); }
void DLinkedList::addBack(const Elem& name) // add to back of list
{ add(trailer, name); }
void DLinkedList::remove(DNode* v) { // remove node v
DNode* u = v->prev; // predecessor
DNode* w = v->next; // successor
u->next = w; // unlink v from list
w->prev = u;
delete v;
}
void DLinkedList::removeFront() // remove from font
{ remove(header->next); }
void DLinkedList::removeBack() // remove from back
{ remove(trailer->prev); }
bool DLinkedList::empty() const // is list empty?
{ return (header->next == trailer); }
const Elem& DLinkedList::front() const // get front element
{ return header->next->elem; }
const Elem& DLinkedList::back() const // get back element
{ return trailer->prev->elem; }
void DLinkedList::displayViaAge(string age) { //Displays person via age
//int check = 0;
DNode*temp = header;
while(temp!=trailer)
{
// age = str.find(temp->elem);
// if(check == 1){
cout << temp->elem <<endl;
temp = temp -> next;
// }
}
cout << temp->elem<<endl;
}
void DLinkedList::displayViaName(const Elem& name) { //Displays person via age
int check = 0;
DNode*temp = header;
while(temp!=trailer)
{
if(temp->elem == name){
cout << "Yes that Person is in out system" << endl;
check = 1;
}
temp = temp -> next;
}
if(temp->elem == name){
cout << "Yes that Person is in out system" << endl;
check = 1;
}
if(check == 0){ cout << "Sorry that person is not in our system" << endl;}
check = 0;
}
class Person {
public:
void print();
string getName();
int getAge();
private:
string name;
int age;
};
int main(){
char input = 'z';
string entry;
int age;
DLinkedList person;
person.addFront("Takkun Bradly 19");
person.addFront("Devindra Ardnived 18");
person.addFront("SeboY Wang 20");
person.addFront("DoubleX Slash 31");
person.addFront("Uncle Jelly 17");
person.addFront("test 12");
cout << "What would you like to do?" << endl;
cout << "Enter 'A' to: Add a new person" << endl;
cout << "Enter 'B' to: Remove a person" << endl;
cout << "Enter 'C' to: Search for people via age" << endl;
cout << "Enter 'D' to: Search for people via name" << endl;
cout << "Enter 'E' to: Average all the total ages" << endl;
cout << "Enter 'F' to: Quit" << endl;
while(input != 'f') {
cin >> input;
cout << endl;
while ((input != 'a')&&(input != 'b')&&(input != 'c')&&(input != 'd')&&(input != 'e')&&(input != 'f')) {
cout << "Please enter a valid selection" << endl;
cin >> input;
}
if ((input == 'a')){
cout << "Please enter their name and age: ";
cin.ignore();
getline(cin, entry);
person.addFront(entry);
}
if ((input == 'b')){
cout << "Who would you like to remove: ";
cin.ignore();
getline(cin, entry);
person.removeFront();
}
if ((input == 'c')){
cout << "What is the age of the person you are looking for?: ";
cin.ignore();
getline(cin, entry);
person.displayViaAge(entry);
}
if ((input == 'd')){
cout << "What is the name of the person you are looking for?: ";
cin.ignore();
getline(cin, entry);
person.displayViaName(entry);
}
if ((input == 'e')){
cout << "The total average of ages are: " << endl;
}
cout << endl;
}
}
你把几件你不知道怎么做的事情，拼凑成一个你不知道如何解决的问题。
你似乎认为一个名字有两个部分。好:
if ((input == 'a')){
cout << "Please enter their name and age: ";
cin.ignore();
cin >> firstName >> secondName >> age;
...
}
如果您真的想将名称存储为单个字符串，
name = firstName + " " + secondName;
现在创建list的新成员:
person.addFront(name, age);
:
void DLinkedList::add(DNode* v, const Elem& name, int age)
这里的潜在问题是，您知道应该将name和age作为单独的变量来处理，但是您不知道如何处理，因此，您没有找出如何编写一个拼凑程序，而是知道如何做一些事情。很快你就被困住了。