创建两个链表的并集
[英]Creating a Union of two linked lists
问题描述
我正在尝试创建一个将两个链表联合起来的函数。 我创建了一个函数，它完全符合我的要求，但作为一组集合而不是我将在下面提供的节点。
我的限制如下：我可以使用 contains() 和 Node 类访问器和修改器，但除此之外，我不能在我的联合函数的定义中使用任何函数调用 必须在不调用任何其他函数的情况下对操作进行编码帮助。
任何指导或帮助走上正确的轨道将不胜感激。
这是我尝试编码为链表的函数的数组集版本：
template <class ItemType>
ArraySet<ItemType> ArraySet <ItemType> ::setUnion(const ArraySet<ItemType> set2)
{
ArraySet<ItemType> unionSet;
for (int i = 0; i < getCurrentSize(); i++)
{
unionSet.add(items[i]);
}
for (int i = 0; i < set2.getCurrentSize(); i++)
{
unionSet.add(set2.items[i]);
if (items[i] == set2.items[i])
unionSet.remove(set2.items[i]);
}
return unionSet;
}
这是我的 LinkedSet 类：
#ifndef LINKED_SET_
#define LINKED_SET_
#include "SetInterface.h"
#include "Node.h"
namespace cs_set {
template<class ItemType>
class LinkedSet : public SetInterface<ItemType>
{
private:
Node<ItemType>* headPtr;
int itemCount;
// Returns either a pointer to the node containing a given entry
// or nullptr if the entry is not in the bag.
Node<ItemType>* getPointerTo(const ItemType& target) const;
void clone(const LinkedSet<ItemType>& copyMe); //Copy function
public:
class ItemNotFoundError {};
class DuplicateItemError {};
LinkedSet(); //Constructor #Big3
LinkedSet operator=(const LinkedSet<ItemType>& right); //Assignment operator
LinkedSet(const LinkedSet<ItemType>& aSet); //Copy Constructor #Big3
virtual ~LinkedSet(); //Destructor #Big3
int getCurrentSize() const;
bool isEmpty() const;
void add(const ItemType& newEntry);
void remove(const ItemType& anEntry);
void clear();
bool contains(const ItemType& anEntry) const;
LinkedSet<ItemType> setUnion(const LinkedSet<ItemType>* set2);
LinkedSet<ItemType> setIntersection(const LinkedSet<ItemType>* set2);
LinkedSet<ItemType> setDifference(const LinkedSet<ItemType>* set2);
//int getFrequencyOf(const ItemType& anEntry) const;
std::vector<ItemType> toVector() const;
};
}
#include "LinkedSet.cpp"
#endif
链接集函数：
#include "Node.h"
#include "LinkedSet.h"
#include <cstddef>
namespace cs_set {
template<class ItemType>
LinkedSet<ItemType>::LinkedSet() {
headPtr = nullptr;
itemCount = 0;
}
template<class ItemType>
void LinkedSet<ItemType>::clone(const LinkedSet<ItemType>& copyMe) {
itemCount = copyMe.itemCount;
Node<ItemType>* origChainPtr = copyMe.headPtr;
if (origChainPtr == nullptr) {
headPtr = nullptr;
}
else {
// Copy first node
headPtr = new Node<ItemType>();
headPtr->setItem(origChainPtr->getItem());
// Copy remaining nodes
Node<ItemType>* newChainPtr = headPtr;
origChainPtr = origChainPtr->getNext();
while (origChainPtr != nullptr) {
// Get next item from original chain
ItemType nextItem = origChainPtr->getItem();
// Create a new node containing the next item
Node<ItemType>* newNodePtr = new Node<ItemType>(nextItem);
// Link new node to end of new chain
newChainPtr->setNext(newNodePtr);
// Advance pointer to new last node
newChainPtr = newChainPtr->getNext();
// Advance original-chain pointer
origChainPtr = origChainPtr->getNext();
}
newChainPtr->setNext(nullptr);
}
}
/* BACKUP JUST INCASE I MESS THINGS UP
template<class ItemType>
void LinkedSet<ItemType>::clone(const LinkedSet<ItemType>& aSet) {
itemCount = aSet.itemCount;
Node<ItemType>* origChainPtr = aSet.headPtr;
if (origChainPtr == nullptr) {
headPtr = nullptr;
} else {
// Copy first node
headPtr = new Node<ItemType>();
headPtr->setItem(origChainPtr->getItem());
// Copy remaining nodes
Node<ItemType>* newChainPtr = headPtr;
origChainPtr = origChainPtr->getNext();
while (origChainPtr != nullptr) {
// Get next item from original chain
ItemType nextItem = origChainPtr->getItem();
// Create a new node containing the next item
Node<ItemType>* newNodePtr = new Node<ItemType>(nextItem);
// Link new node to end of new chain
newChainPtr->setNext(newNodePtr);
// Advance pointer to new last node
newChainPtr = newChainPtr->getNext();
// Advance original-chain pointer
origChainPtr = origChainPtr->getNext();
}
newChainPtr->setNext(nullptr);
}
}
*/
template<class ItemType>
LinkedSet<ItemType>::~LinkedSet() {
clear();
}
template<class ItemType>
bool LinkedSet<ItemType>::isEmpty() const {
return itemCount == 0;
}
template<class ItemType>
int LinkedSet<ItemType>::getCurrentSize() const {
return itemCount;
}
template<class ItemType>
void LinkedSet<ItemType>::add(const ItemType& newEntry) {
for (Node<ItemType>* nodePtr = headPtr; nodePtr; nodePtr = nodePtr->getNext())
{
if (nodePtr->getItem() == newEntry)
throw DuplicateItemError();
}
Node<ItemType>* nextNodePtr = new Node<ItemType>();
nextNodePtr->setItem(newEntry);
nextNodePtr->setNext(headPtr);
headPtr = nextNodePtr; // New node is now first node
itemCount++;
}
template<class ItemType>
std::vector<ItemType> LinkedSet<ItemType>::toVector() const {
std::vector<ItemType> setContents;
Node<ItemType>* curPtr = headPtr;
while ((curPtr != nullptr)) {
setContents.push_back(curPtr->getItem());
curPtr = curPtr->getNext();
}
return setContents;
}
template<class ItemType>
void LinkedSet<ItemType>::remove(const ItemType& anEntry) {
Node<ItemType>* entryNodePtr = getPointerTo(anEntry);
if (entryNodePtr == nullptr) {
throw ItemNotFoundError();
} else {
// replace data of node to delete with data from first node
entryNodePtr->setItem(headPtr->getItem());
// Delete first node
Node<ItemType>* nodeToDeletePtr = headPtr;
headPtr = headPtr->getNext();
delete nodeToDeletePtr;
itemCount--;
}
}
template<class ItemType>
void LinkedSet<ItemType>::clear() {
Node<ItemType>* nodeToDeletePtr = headPtr;
while (headPtr != nullptr) {
headPtr = headPtr->getNext();
delete nodeToDeletePtr;
nodeToDeletePtr = headPtr;
}
headPtr = nullptr;
itemCount = 0;
}
/*
template<class ItemType>
int LinkedSet<ItemType>::getFrequencyOf(const ItemType& anEntry) const {
int frequency = 0;
int counter = 0;
Node<ItemType>* curPtr = headPtr;
while ((curPtr != nullptr) && (counter < itemCount)) {
if (anEntry == curPtr->getItem()) {
frequency++;
}
counter++;
curPtr = curPtr->getNext();
}
return frequency;
}
*/
template<class ItemType>
bool LinkedSet<ItemType>::contains(const ItemType& anEntry) const {
return (getPointerTo(anEntry) != nullptr);
}
// private
// Returns either a pointer to the node containing a given entry
// or nullptr if the entry is not in the bag.
template<class ItemType>
Node<ItemType>* LinkedSet<ItemType>::getPointerTo(const ItemType& anEntry) const {
bool found = false;
Node<ItemType>* curPtr = headPtr;
while (!found && (curPtr != nullptr)) {
if (anEntry == curPtr->getItem()) {
found = true;
} else {
curPtr = curPtr->getNext();
}
}
return curPtr;
}
template <class ItemType>
LinkedSet<ItemType> LinkedSet<ItemType>::operator=(const LinkedSet<ItemType>& right) {
if (this != &right) {
clear();
clone(right);
}
return *this;
}
template<class ItemType>
LinkedSet<ItemType> ::LinkedSet(const LinkedSet<ItemType>& aSet)
{
clone(aSet);
}
/* This function is responsible for merging two sets together
to create one big set by using for loops to add the contents
of the array sets and also detecting / removing duplicates
@param set2 used to define the set being used
@return it returns the merged set
*/
template <class ItemType>
LinkedSet<ItemType> LinkedSet <ItemType> ::setUnion(const LinkedSet<ItemType>* set2)
{
}
/*
This function is responsible for detecting if matching
values of two sets are detected. For example, if one set
has {1,2,3) and the second has the same then there are
three elements that intersect.
@param set2 used to define the set
@retunr returns the intersected values
*/
template <class ItemType>
LinkedSet<ItemType> LinkedSet<ItemType>::setIntersection(const LinkedSet<ItemType> set2)
{
}
/*
This function is responsible for detecting values that are present
in the first set and not the second for example, if set one has values
(1,2,3,4,5,6,7) and set2 has values (1,2,3) then the difference is
the elements 1,2,3
@param set2 used to define the set
@return returns the differences aka numbers that do not appear in the second set
*/
template<class ItemType>
LinkedSet<ItemType> LinkedSet<ItemType>::setDifference(const LinkedSet<ItemType> set2)
{
}
}
节点类：
#ifndef NODE_
#define NODE_
namespace cs_set {
template<class ItemType>
class Node {
private:
ItemType item;
Node<ItemType>* next;
public:
Node(const ItemType& anItem = ItemType(), Node<ItemType>* nextNodePtr = nullptr);
void setItem(const ItemType& anItem);
void setNext(Node<ItemType>* nextNodePtr);
ItemType getItem() const ;
Node<ItemType>* getNext() const ;
};
}
#include "Node.cpp"
#endif
节点类函数：
#include "Node.h"
namespace cs_set {
template<class ItemType>
Node<ItemType>::Node(const ItemType& anItem, Node<ItemType>* nextNodePtr) {
item = anItem;
next = nextNodePtr;
}
template<class ItemType>
void Node<ItemType>::setItem(const ItemType& anItem) {
item = anItem;
}
template<class ItemType>
void Node<ItemType>::setNext(Node<ItemType>* nextNodePtr) {
next = nextNodePtr;
}
template<class ItemType>
ItemType Node<ItemType>::getItem() const {
return item;
}
template<class ItemType>
Node<ItemType>* Node<ItemType>::getNext() const {
return next;
}
}
设置接口类：
#ifndef SET_INTERFACE
#define SET_INTERFACE
#include <vector>
#include <algorithm>
#include <iterator>
namespace cs_set {
template<class ItemType>
class SetInterface
{
public:
/** Gets the current number of entries in this bag.
@return The integer number of entries currently in the set. */
virtual int getCurrentSize() const = 0;
/** Sees whether this set is empty.
@return True if the set is empty, or false if not. */
virtual bool isEmpty() const = 0;
/** Adds a new entry to this set.
@post If successful, newEntry is stored in the set and
the count of items in the set has increased by 1.
@param newEntry The object to be added as a new entry.
@return True if addition was successful, or false if not. */
virtual void add(const ItemType& newEntry) = 0;
/** Removes one occurrence of a given entry from this set,
if possible.
@post If successful, anEntry has been removed from the set
and the count of items in the bag has decreased by 1.
@param anEntry The entry to be removed.
@return True if removal was successful, or false if not. */
virtual void remove(const ItemType& anEntry) = 0;
/** Removes all entries from this set.
@post set contains no items, and the count of items is 0. */
virtual void clear() = 0;
/** Counts the number of times a given entry appears in this set.
@param anEntry The entry to be counted.
@return The number of times anEntry appears in the set. */
// virtual int getFrequencyOf(const ItemType& anEntry) const = 0;
/** Tests whether this set contains a given entry.
@param anEntry The entry to locate.
@return True if bag contains anEntry, or false otherwise. */
virtual bool contains(const ItemType& anEntry) const = 0;
/** Empties and then fills a given vector with all entries that
are in this set.
@return A vector containing all the entries in the bag. */
virtual std::vector<ItemType> toVector() const = 0;
/** Destroys this set and frees its assigned memory. (See C++ Interlude 2.) */
virtual ~SetInterface() { }
};
}
#endif
我试图实现的输出是：
套装 1 包含：一二三
套装 2 包含：四五六
两个集合的并集是：
一二三四五六
1 个解决方案
解决方案1
1 已采纳 2020-09-16 10:11:24
如何统一两个列表有多种选择。
- 最容易编码的解决方案是连接两个列表，然后检查每个元素是否在列表中还有另一个相等的元素，如果它不是第一个删除它。
- 一种中间解决方案是连接列表，对组合列表进行排序，然后遍历列表并用元素的一个实例替换相等元素的连续运行。
- 一般来说，计算上最简单的解决方案是使用散列将两个列表转换为一个集合。 然后你就完成了，可以返回集合的元素。
我想你会选择 3。使用 std 你会这样做：
template<typename T>
std::list<T> unify(const std::list<T>& a, const std::list<T>& b) {
std::unordered_set<T> combined_set;
combined_set.insert(a.begin(),a.end());
combined_set.insert(b.begin(),b.end());
std::list<T> result;
for (auto& element : combined_set) result.push_back(element);
return result;
}
在 C++ 中使用多重函数的两个链表的联合
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