我一直在学习Vector并且无法完成某些部分。我的代码和下面的描述。 输出应该是这样的：
b1 size = 0
b1 beginning capacity = 10
b1 empty? 1
b2 size = 0
b2 beginning capacity = 10
b2 empty? 1
b1 : []
b2 : [ 2 4 6 8 10 12 14 16 18 20 ]
b1 + b2 : [ 2 4 6 8 10 12 14 16 18 20 ] // not working
b2 duplicate : [ 1 2 2 4 4 6 6 8 8 10 12 14 16 18 20 ] // not working
b2 remove duplicate : [ 1 2 4 6 8 10 12 14 16 18 20 ] // not working
b1 : [ 1 2 3 4 5 6 7 8 9 ]
b1 size : 9
b1 capacity : 10
b2 : [ 1 2 4 6 8 10 12 14 16 18 20 ]
b2 size = 11
b2 capacity = 20
b1+b2 : [ 1 2 3 4 5 6 7 8 9 1 2 4 6 8 10 12 14 16 18 20 ] // nw
b1-b2 : [ 3 5 7 9] // nw
b1 union b2 : [ 1 2 3 4 5 6 7 8 9 10 12 14 16 18 20 ] // nw
b1 intersect b2 : [ 1 2 4 6 8 ] // nw
b3+b4 : [ 3 5 7 9 3 5 7 9 ] // nw
b3-b4 : [] // nw
b3 union b4 : [ 3 5 7 9 ] // nw
b3 intersect b4 : [ 3 5 7 9 ] // nw
骨架部分：
#include RunMyNumber.cpp
int main() {
MyNumber<int> b1;
cout << "b1 size = " << b1.size() << endl;
cout << "b1 beginning capacity = " << b1.capacity() << endl;
cout << "b1 empty? " << b1.empty() << endl << endl;
MyNumber<int> b2;
cout << "b2 size = " << b2.size() << endl;
cout << "b2 beginning capacity = " << b2.capacity() << endl;
cout << "b2 empty? " << b2.empty() << endl << endl;
for(int i=1;i<=10;i++)
{
b2.push_back(i*2);
}
cout << "b1 : " << b1.toString() << endl;
cout << "b2 : " << b2.toString() << endl;
cout << "b1 + b2 : " << (b1 + b2).toString() << endl << endl;
for(int i=1;i<10;i++)
{
b1.push_back(i);
}
for(int i=0;i<4;i++){
b2.duplicate(i+i,1);
}
b2.insert(0,1);
cout << "b2 duplicate : " << b2.toString() << endl;
b2.removeDuplicate();
cout << "b2 remove duplicate : " << b2.toString() << endl << endl;
cout << "b1 : " << b1.toString() << endl;
cout << "b1 size = " << b1.size() << endl;
cout << "b1 capacity = " << b1.capacity() << endl;
cout << "b2 : " << b2.toString() << endl;
cout << "b2 size = " << b2.size() << endl;
cout << "b2 capacity = " << b2.capacity() << endl << endl;
cout << "b1 + b2 : " << (b1 + b2).toString() << endl;
cout << "b1 - b2 : " << (b1 - b2).toString() << endl;
cout << "b1 union b2 : " << b1.myUnion(b2).toString() << endl;
cout << "b1 intersect b2 : " << b1.myIntersect(b2).toString() << endl << endl;
MyNumber<int> b3;
b3 = b1 - b2;
MyNumber<int> b4(b3);
cout << "b3 : " << b3.toString() << endl;
cout << "b4 : " << b4.toString() << endl << endl;
MyNumber<int> b5;
b5 = b3 + b4;
cout << "b3 + b4 : " << b5.toString() << endl;
b5 = b3 - b4;
cout << "b3 - b4 : " << b5.toString() << endl;
cout << "b3 union b4 : " << b3.myUnion(b4).toString() << endl;
cout << "b3 intersect b4 : " << b3.myIntersect(b4).toString() << endl << endl;
return 0;
}
以下是目前的代码：
#include "MyNumber.cpp"
#include <iostream>
#include <string>
#include <sstream>
#include <stdexcept>
#include <algorithm>
using namespace std;
template <typename B>
class MyNumber
{
private :
static const size_t BEGINNING_CAPACITY =10;
size_t _capacity;
size_t _size;
B* _data; // array' element
public :
// Constructor
MyNumber<B>() : _capacity(BEGINNING_CAPACITY),
_size(0),
_data(new B[BEGINNING_CAPACITY])
{}
//Destructor
~MyNumber<B>()
{
delete[] _data;
}
//Copy Constructor
MyNumber<B>(const MyNumber<B>& OtherNumber) :
_capacity(OtherNumber._capacity),
_size(OtherNumber._size),
_data(new B[_capacity])
{
for(size_t i = 0; i < _size; i++)
_data[i] = OtherNumber._data[i];
}
// template function swap STL algorithm
void swap(MyNumber<B>& OtherNumber)
{
swap(_size, OtherNumber._size);
swap(_capacity, OtherNumber._capacity);
swap(_data, OtherNumber._data);
}
MyNumber<B>& operator= (const MyNumber<B>& OtherNumber)
{
MyNumber<B> copy(OtherNumber);
exchange(copy);
return *this;
}
// Operator indexing []
B& operator[] (size_t index)
{
if(index < 0 || index >= _size)
{
throw out_of_range("Index operator [] out of range");
}
return _data[index];
}
//Function for adding new element
void push_back(const B& elemen)
{
if(_size == _capacity)
{
expand(2 *_capacity);
}
_data[_size] = elemen;
_size++;
}
//Function for inserting
void insert(size_t index, const B& elemen)
{
if(index < 0 || index > _size)
{
throw out_of_range("index insert out of range");
}
if(_size == _capacity)
{
expand(2 * _capacity);
}
for(size_t i = _size; i > index; i--)
{
_data[i] = _data[i-1];
}
_data[index] = elemen;
_size++;
}
//Function for expanding the size of capacity
void expand(size_t newCapacity)
{
_capacity = newCapacity;
B* newData = new B[newCapacity];
for(size_t i = 0; i < _size; i++)
newData[i] = _data[i];
delete[] _data;
_data = newData;
}
//Funtion for returning capacity
size_t capacity()
{
return _capacity;
}
//Function for returning size
size_t size()
{
return _size;
}
//Function for checking the vector
bool empty()
{
return _size == 0;
}
//Function for representing the vector
string toString()
{
ostringstream oss;
oss << "[ ";
for(int i = 0; i < _size; ++i)
oss << _data[i] << " ";
oss << "]";
return oss.str();
}
//Function for searching particular element
int find(const B& elemen){
for(int i=0;i<_size;i++){
if(_data[i] == elemen)
return i;
}
string exception = "Data not found!";
throw exception;}
坚持到这里： 在运算符+（合并两个向量），operator-（显示第一个向量上第二个不存在的数字），union，intersect。另外，我不知道如何制作新的矢量对象（第3和第4个）作为b1和b2之前操作的结果。
MyNumber<B>& operator+ (MyNumber<B>& OtherNumber)
{auto cmp = [] ( const B& b1, const B& b2 ) { return b1.value() < b2.value() ; } ;
sort( MyNumber.begin(), MyNumber.end(), cmp ) ; // sort first on ascending v
sort( OtherNumber.begin(), OtherNumber.end(), cmp ) ; // sort second on ascending v
merge( MyNumber.begin(), MyNumber.end(), OtherNumber.begin(), OtherNumber.end(),
back_inserter(b3??), cmp ) ; // merge first and second into third
// append contents of second to first
MyNumber.insert( MyNumber.end(), OtherNumber.begin(), OtherNumber.end() ) ;
// sort first on descending length of str
sort( MyNumber.begin(), MyNumber.end(),
[] ( const B& b1, const B& b2 ) { return b1.str().size() > b2.str().size() ; } ) ;
}
MyNumber<B>& operator- (MyNumber<B>& OtherNumber)
{
}
void duplicate(size_t index, size_t n)
{
}
void removeDuplicate()
{
}
MyNumber<B>& myUnion(MyNumber<B>& OtherNumber)
{
vec_union(MyNumber<B> &, OtherNumber)
{
if ( OtherNumber.empty() )
return MyNumber;
MyNumber<B>::iterator found = find(MyNumber.begin(), MyNumber.end(), OtherNumber.back());
if ( found == MyNumber.end() )
{
// all good, element not already in v1, so insert it.
B value = OtherNumber.back();
MyNumber.push_back(value);
OtherNumber.pop_back(); // remove back element
return vec_union(MyNumber, OtherNumber);
}
else
{
// element was already in v1, skip it and call vec_union again
OtherNumber.pop_back(); // remove back element
return vec_union(MyNumber, OtherNumber);
}
}
}
MyNumber<B>& myIntersect(MyNumber<B>& OtherNumber){
}
};
答案 0 :(得分：2)
如果向量总是排序，那么可以使用以下算法在O（N）时间内合并它们。
初始化I1 = 0，I2 = 0.
while ( I1 < V1.size() || I2 < V2.size() )
if ( V1[I1] == V2[I2] ) {
newVec.push_back( V1[I1] ); // if duplicates are to be entered, then push back twice.
I1++;
I2++;
} else if ( V1[I1] < V2[I2] ) { // or I2 == V2.size()
newVec.push_back( V1[I1] );
I1++;
} else { // or I1 == V1.size()
newVec.push_back( V2[I2] );
I2++;
}
对于减去V1-V2，也可以应用类似的算法 警告：我没有检查过这些算法，他们可能需要一些调试或小调整
初始化I1 = 0，I2 = 0.
while ( I1 < V1.size() ) //|| I2 < V2.size() ) (No need to iterate till end of V2)
if ( V1[I1] == V2[I2] ) {
//newVec.push_back( V1[I1] ); // Don't push back if element is common.
I1++;
I2++;
} else if ( V1[I1] < V2[I2] ) { // or I2 == V2.size()
newVec.push_back( V1[I1] );
I1++;
} else { // or I1 == V1.size()
//newVec.push_back( V2[I2] );
I2++;
}
答案 1 :(得分：0)
如上所述，您希望实现Set操作概念。确定是否仍需要保留元素和重复的原始顺序（如示例输出所示）。如果是这样，请保持矢量不变，并在操作期间构造临时集（排序的MyNumber） - 这将是低效的，但是有效。为了处理效率，您可能需要更复杂的数据结构。否则只需保持MyNumber排序并在find（）中使用二进制搜索。
接下来，从运算符+开始 - 根据operator + =（类似于operator =）
实现它MyNumber<B> operator+ (const MyNumber<B>& OtherNumber)
{
MyNumber<B> result(*this);
result += OtherNumber;
return result;
}
然后使用push_back实现+ =
MyNumber<B>& operator+= (MyNumber<B>& OtherNumber)
{
// todo expand() once
for (size_t i = 0; i < OtherNumber.size(); ++i)
push_back(OtherNumber[i]);
return *this;
}
请注意，operator + =返回MyNumber＆amp;但是operator +返回MyNumber
最好不要从find（）中抛出，因为如果值不在容器中，则不是错误。最好返回无效索引，如C中的负数或stl的大小（）。