我正在使用这本教科书
第5章迭代器用法:
http://cpp.datastructures.net/source/ch05/CPP/IteratorPrint.cpp-DSACiterator.html
这是我如何实现它(ObjectIterator类)
#include <iostream>
using namespace std;
class PositionException {
protected:
string message;
public:
PositionException(const string &m) {
message = m;
}
void print() {
cout << message << endl;
}
};
template <typename T>
class NodeList {
protected:
struct Node {
Node *prev;
Node *next;
T item;
Node(T i = T(), Node *p = NULL, Node *n = NULL) : item(i), prev(p), next(n) {}
};
typedef Node * NodePtr;
public:
class Position {
protected:
NodePtr node;
public:
bool isNull() {
return node == NULL;
}
Position(NodePtr n = NULL) : node(n) {}
T& element() {
return node->item;
}
friend class NodeList;
};
class ObjectIterator {
protected:
NodePtr node;
public:
ObjectIterator(NodePtr n = NULL) : node(n) {}
bool hasNext() {
if (node->next == NULL) {
return false;
}
if (node->next->next == NULL) {
return false;
}
return true;
}
ObjectIterator next() {
return ObjectIterator(node->next);
}
T& element() {
return node->item;
}
friend class NodeList;
};
protected:
int sz;
NodePtr header;
NodePtr trailer;
public:
NodeList() {
header = new Node();
trailer = new Node();
header->next = trailer;
trailer->prev = header;
sz = 0;
}
bool isEmpty() const {
return size() == 0;
}
int size() const {
return sz;
}
bool isFirst(const Position& p) const {
return p.node->prev == header;
}
bool isLast(const Position& p) const {
return p.node->next = trailer;
}
Position first() const {
if (isEmpty()) {
throw PositionException("no first for empty list");
}
return Position(header->next);
}
Position last() const {
if (isEmpty()) {
throw PositionException("no last for emtpy list");
}
return Position(trailer->prev);
}
Position before(const Position& p) const{
if (p.node->prev == header) {
throw PositionException("already the first element, nothing before it");
}
return Position(p.node->prev);
}
Position after(const Position& p) const{
if (p.node->next == trailer) {
throw PositionException("already the last element, nothing after it");
}
return Position(p.node->next);
}
Position insertAfter(const Position& p, const T& o) {
NodePtr node = new Node(o, p.node, p.node->next);
p.node->next->prev = node;
p.node->next = node;
sz++;
return Position(node);
}
Position insertBefore(const Position& p, const T& o) {
NodePtr node = new Node(o, p.node->prev, p.node);
p.node->prev->next = node;
p.node->prev = node;
sz++;
return Position(node);
}
void remove(const Position& p) {
p.node->prev->next = p.node->next;
p.node->next->prev = p.node->prev;
sz--;
delete p.node;
}
void removeFirst() {
remove(first());
}
void removeLast() {
remove(last());
}
void replaceElement(const Position& p, const T& element) {
if (p.isNull()) {
throw PositionException("p is null");
}
p.node->item = element;
}
Position insertFirst(const T& o) {
NodePtr node = new Node(o, header, header->next);
header->next->prev = node;
header->next = node;
sz++;
return Position(node);
}
Position insertLast(const T& o) {
NodePtr node = new Node(o, trailer->prev, trailer);
trailer->prev->next = node;
trailer->prev = node;
sz++;
return Position(node);
}
void copyFrom(const NodeList<T>& nl) {
sz = nl.sz;
if (nl.sz > 0) {
Position p0 = nl.first();
Position p = insertFirst(p0.node->item);
while (!nl.isLast(p0)) {
p0 = nl.after(p0);
insertAfter(p, p0.node->item);
}
}
}
void emptyList() {
while (!isEmpty()) {
removeFirst();
}
}
~NodeList() {
emptyList();
delete header;
delete trailer;
}
NodeList<T>& operator=(const NodeList<T>& nl) {
emptyList();
copyFrom(nl);
}
NodeList(const NodeList<T>& nl) {
emptyList();
copyFrom(nl);
}
void print() {
cout << "size is: " << size() << endl;
if (size() > 0) {
ObjectIterator i = elements();
while (i.hasNext()) {
cout << i.element() << "\t";
i = i.next();
}
cout << endl;
}
}
ObjectIterator elements() {
if (isEmpty()) {
throw PositionException("iterator error: empty");
}
return ObjectIterator(header->next);
}
void swapItems(const Position& p1, const Position& p2) {
T temp = p1.node->item;
p1.node->item = p2.node->item;
p2.node->item = temp;
}
};
如果我不使用迭代器,下面的打印代码是正确的
void print() {
if (size() > 0) {
NodePtr n = header->next;
while (n != trailer) {
cout << n->item << "\t";
n = n->next;
}
cout << endl;
}
}
如果我使用迭代器作为打印功能
void print() {
cout << "size is: " << size() << endl;
if (size() > 0) {
ObjectIterator i = elements();
while (i.hasNext()) {
cout << i.element() << "\t";
i = i.next();
}
cout << endl;
}
}
其中一个节点丢失了。 本书没有提供实现迭代器的正确方法
答案 0 :(得分:2)
你的hasNext()函数对于双向链表是正确的,假设预告片的下一个是NULL。 错误是打印功能中的用法
void print() {
cout << "size is: " << size() << endl;
if (size() > 0) {
ObjectIterator i = elements();
while (true) {
cout << i.element() << "\t";
if (i.hasNext())
i = i.next();
else
break;
}
cout << endl;
}
}
编辑: 上面的代码只为你做调试。这不是你如何实现'迭代器'
首先,改变next()实现:
ObjectIterator next() {
return (*this = ObjectIterator(node->next));
}
现在,不是返回第一个元素,而是返回标题以进行迭代
ObjectIterator elements() {
return ObjectIterator(header);
}
现在你可以使用类似于书和STL的语法:
void print() {
cout << "size is: " << size() << endl;
ObjectIterator i = elements();
while (i.hasNext()) {
cout << i.next().element() << "\t";
}
cout << endl;
}
请注意,由于您返回标题,因此无需检查(size()&gt; 0)
编辑2: 在next()中,您可以推进迭代器的节点而不是迭代器容器本身
ObjectIterator next() {
node = node->next;
return *this;
}