我试图通过遍历其节点并通过list.h添加它们来创建已经存在的有序双链表，但是当我打印有序列表时，很少有节点丢失。
#ifnded LIST_H
#define LIST_H
struct Node {
// data member
std::string value;
// constructors
Node (std::string v) : value(v) { }
Node (const Node& src) : value(src.value) { }
// overloaded operators
friend std::ostream& operator<< (std::ostream& os, const Node& g);
friend bool operator< (const Node& lhs, const Node& rhs);
friend bool operator> (const Node& lhs, const Node& rhs);
};
//====================================================================================
// doubly linkes list accepting any object type as its data member (value)
template<class T>
class Link {
public:
// public data member
T value;
// constructor
Link<T> (T v, Link<T>* p = 0, Link<T>* s = 0) : value(v), prev(p), succ(s) { }
// copy constructor
Link<T> (const Link<T>& l) : value(l.value), prev(l.next()), succ(l.previous()) { }
// non-modifying members
Link<T>* next () const { return succ; }
Link<T>* previous () const { return prev; }
// modifying members
Link<T>* insert (Link<T>* n);
Link<T>* add (Link<T>* n);
Link<T>* add_ordered (Link<T>* n);
private:
// private data members
Link<T>* prev;
Link<T>* succ;
};
#include "list.cpp"
#endif
：
list.cpp std::ostream& operator<<(std::ostream& os, const Node& g) {
os << g.value ;
return os;
}
bool operator< (const Node& lhs, const Node& rhs) {
return lhs.value < rhs.value;
}
bool operator> (const Node& lhs, const Node& rhs) {
return lhs.value > rhs.value;
}
//=======================================================================================
// It inserts the node passed as a parameter before the node currently pointed to by this.
template<class T>
Link<T>* Link<T>::insert (Link<T>* n) {
if (!n) return this;
if (!this) return n;
n->succ = this;
n->prev = prev;
if (prev) prev->succ = n;
prev = n;
return n;
}
// It inserts the node passed as a parameter after the node currently pointer to by this.
template<class T>
Link<T>* Link<T>::add (Link<T>* n) {
if (!n) return this;
if (!this) return n;
// n->succ = nullptr;
n->succ = succ;
n->prev = this;
succ = n;
return n;
}
/*
It inserts the new node such that
current node and new node in
lexicographical order; returns
pointer to last node.
First node in ordered list contains the
lexicographically smaller value.
It assumes argument is the tail.
*/
template<class T>
Link<T>* Link<T>::add_ordered (Link<T>* n) {
if (!n) return this;
if (!this) { std::cout <<"new node first\n"; return n; }
Link<T>* tail = this;
if (n->value > tail->value) {
std::cout <<"new node largest \n";
add(n); // add after current (last) node
return n;
}
Link<T>* current_node = this;
while (current_node) {
if (n->value > current_node->value) {
std::cout <<"new node spliced \n";
add(n);
return tail;
}
std::cout <<"advance to head\n";
current_node = current_node->previous();
}
insert(n); // insert before current (first) node
std::cout << "new node smallest\n";
return tail;
}
：
main.cpp #include <iostream>
#include <string>
#include <list.h>
template<class T>
void create_ordered_list (Link<T>* src, Link<T>*& dest) {
while (src){
Link<T> l(src->value, nullptr, nullptr);
dest = dest->add_ordered(new Link<T>(l));
src = src->next();
}
}
template<class T>
void print(Link<T>* n) {
Link<T>* tail = n;
if (tail->next()) {
std::cout <<"[";
while (tail) {
std::cout << tail->value;
if (tail = tail->next()) std::cout <<"\n";
}
std::cout <<"]";
} else if (tail->previous()) {
std::cout <<"[";
while (tail) {
std::cout << tail->value;
if (tail = tail->previous()) std::cout <<"\n";
}
std::cout <<"]";
}
getchar();
}
//===============================================================
int main () {
Link<Node>* head = new Link<Node>(Node("Zeus"));
head = head->insert(new Link<Node>(Node("Hera")));
head = head->insert(new Link<Node>(Node("Athena")));
head = head->insert(new Link<Node>(Node("Ares")));
head = head->insert(new Link<Node>(Node("Poseidon")));
print<Node>(head);
std::cout <<"\nOrdered lists\n";
Link<Node>* ordered_list = nullptr;
create_ordered_list(head, ordered_list);
print(ordered_list);
}
：
class UserSignupType extends AbstractType
{
public function buildForm(FormBuilderInterface $builder, array $options)
{
$builder
->add('userType', ChoiceType::class, array(
'choices' => array(
"Subscriber" => "Subscriber",
"Friend" => "Friend"
),
'expanded' => true,
'mapped' => false
));
$builder->addEventListener(
FormEvents::PRE_SET_DATA,
function (FormEvent $event) {
$form = $event->getForm();
$usertype = $form->get('userType')->getData(); //updated per JBaffords answer
if($userType == "Subscriber")
{
$builder->add('agency', EntityType::class, array(
"class" => "\AppBundle\Entity\Agency",
"label" => "name"));
}
elseif($userType == "Friend")
{
$builder->add('phoneNumber', PhoneNumberType::class, array(
'default_region' => 'US',
'format' => PhoneNumberFormat::NATIONAL));
}
}
);
}
// ...
}
订购后的输出：
[宙斯
赫拉
雅典娜]
预期产出：
[宙斯
海神
赫拉
雅典娜
战神]
答案 0 :(得分：1)
首先，您必须更正add()功能：
template<class T>
Link<T>* Link<T>::add (Link<T>* n) {
...
n->succ = succ;
n->prev = this;
succ = n;
if (n->succ) n->succ->prev = n; ///!!!
return n;
}
然后你必须在add()函数中插入一些insert()和add_ordered()并调整一些回报：
template<class T>
Link<T>* Link<T>::add_ordered (Link<T>* n) {
if (!n) return this;
if (!this) { std::cout <<"new node first\n"; return n; }
Link<T>* tail = this;
if (n->value > tail->value) {
std::cout <<"new node largest \n";
return insert(n); // No: you need to insert before current node !!!
}
Link<T>* current_node = this;
while (current_node) {
if (n->value > current_node->value) {
std::cout <<"new node spliced \n";
add(n);
return this;
}
std::cout <<"advance to head\n";
current_node = current_node->previous();
}
add(n); // no: here we have to add at the end !
std::cout << "new node smallest\n";
return this;
}
然后它应该工作：live demo。