堆栈实现

时间:2013-02-10 18:22:59

标签: c++ arrays stack

我无法理解push和pop如何使用堆栈。我理解他们是如何工作的,因为他们将一个数字推到堆栈上,推出的最后一个数字会被弹出。我也理解指针背后的逻辑以及它们是如何工作的。我不明白的是如何编写代码。

我的程序应该让用户创建一个堆栈(确定它有多大)然后选择在堆栈上推送内存(数字)或将其弹出。

这是我到目前为止所得到的,我被困住了。我已经通过cplusplus.com进行了研究并阅读了关于这些内容的所有内容,但仍然无法弄清楚程序应该如何布局以及如何运行。

#include<iostream>
#include<cstring>
#include<cmath>
#include<cstdlib>

using namespace std;

int choice;
int * arrPtr = NULL;
int * stack;
int * top;
int val, capacity, size; 
//stack would point to the location of the first element,
// top would point to the location where the next element will be stored,
// capacity is the number of elements that can be stored in the stack,
// size would be the number of elements actually stored in the stack

int main()
{
//This is the menu
cout << "1. Create" << endl;
cout << "2. Push" << endl;
cout << "3. Pop" << endl;
cout << "4. Count" << endl;
cout << "5. Display" << endl;
cout << "6. Exit\n" << endl;
cin >> choice;

//I figured i would use choices for the menu and went ahead and wrote it out
switch(choice)
{
    case 1:
        create();
        break;
    case 2:
        push();
        break;
    case 3:
        pop();
        break;
    case 4:
        count(0);
        break;
    case 5:
        display();
        break;
    case 6:
        exit();
    default:
        cout << "Please enter correct choice (1-4)!";
        break;
    }

    return 0;   
}   //end main

void create()
{
    cout << "Enter the size of the stack you wish to create: ";
    int capacity = 0;
    cin >> capacity;

    arrPtr = new int[capacity];
} //end create function

//From here it went wrong, I cannot figure it out.           

void push(){
    for (int i = 0; i < capacity; i++)
    {
        cout << "Enter the number you wish to put on the stack: ";
        cin >> val;
        push(val)
    }//end for
}//end push

请帮助我理解这一点。

3 个答案:

答案 0 :(得分:1)

创建std::array。创建迭代器。将迭代器设置为数组的末尾。 (结束意思是array.end()

推送:递减迭代器,插入值。
Pop:返回值,增量迭代器 计数:不是堆栈上的标准,但你可以通过从末尾减去当前的迭代器来得到它 Peek:返回值

显然你想确保你没有从阵列前面推开,或者从后面弹出,所以你应该添加一些支票。

堆栈非常简单。希望这会有所帮助。

编辑:实现,未经测试的代码,有点思考这个

template <typename T, std::size_t N>
class Stack {
public:
  Stack();
  void push(T value);
  T pop();
  T peek() const;
  std::size_t count() const;

private:
  std::array<T, N> m_baseArray;
  std::array<T>::iterator m_it;
};

template <typename T, std::size_t N>
Stack::Stack() : m_baseArray(), m_it(m_baseArray.end()) { }

template <typename T, std::size_t N>
void Stack::push(T value) {
  if (m_it == m_baseArray.begin())
    throw std::exception();
  --m_it;
  *m_it = value;
}

template <typename T, std::size_t N>
T Stack::pop() {
  if (m_it == m_baseArray.end())
    throw std::exception();
  T res = *m_it;
  ++m_it;
  return res;
}

template <typename T, std::size_t N>
T Stack::peek() const {
  if (m_it == m_baseArray.end())
    throw std::exception();
  return *m_it;
}

template <typename T, std::size_t N>
std::size_t Stack::count() const {
  return m_baseArray.end() - m_it;
}

答案 1 :(得分:0)

使用std::vector

的最简单堆栈
#include <iostream>
#include <vector>
using namespace std;

template<typename T>
class Stack{
private: 
  vector<T> theArray;
public:
  Stack(){ theArray.clear(); }
  void push(T data) { theArray.push_back(data); }
  T pop() { T retData = theArray.back(); theArray.pop_back(); return retData; }
  void display() {
    for(size_t i = theArray.size() - 1; i != (size_t)-1; i--){
      cout << theArray[i] << '\n';
    }
  }
};

int main(int argc, char *argv[])
{
  Stack<int> s;
  s.push(10);
  s.push(20);
  s.push(30);
  s.display();

  int ret = s.pop();
  cout << "After popping : " << ret << '\n';
  s.display();
  return 0;
}

答案 2 :(得分:0)

我已经在Mac上编译了这段代码,但它应该可以在安装了bash的PC上运行。

#include<iostream>
#include<cstring>
#include<cmath>
#include<cstdlib>

using namespace std;

int choice;
int capacity; /*size of stack*/ 

int stack[1000];/*THIS IS WHERE ALL THE STACK ENTRIES WILL BE STORED*/
int top=0; /* THIS WILL KEEP CHECK OF THE INDEX OF TOP MOST ELEMENT */

//stack would point to the location of the first element,
// top would point to the location where the next element will be stored,
// capacity is the number of elements that can be stored in the stack,
// size would be the number of elements actually stored in the stack

void create() //Set size of stack
{
    cout << "Enter the size of the stack you wish to create: ";
    cin >> capacity;

} //end create function


void push(int n) //Enter data into stack

{

/* Ensures that there isn't a stack overflow */

    if(top<capacity)
    {
        stack[top++]=n; 
    }


    else
    {
        cout<<"Stack FULL!!\n";
    }

    //end for
}//end push

void pop() //Remove data from stack
{
    /* Ensures that there isn't a stack underflow */
    if(top>0)
    {
        cout<<"Popped entry is:"<<stack[top-1]<<"\n";
        top--;   
    }

    else
    {
        cout<<"Stack is EMPTY!!\n";
    }


}

void count() //Count number of elements
{
    cout<<"Stack size is:"<<top<<"\n"; 
}

void display() //Print elements from lowest stack entry to highest 
{
    int i=0;
    cout<<"Your stack is:\n";
    while(i<top)
    {
        cout<<i+1<<") "<<stack[i]<<"\n";
        i++;
    }
}



int main()
{
    system("clear");  //Start with blank screen



    int exitCheck=1,entry;

    //I figured i would use choices for the menu and went ahead and wrote it out -- This is good approach!!
    cout<<"Welcome user \n";

    create(); /*Size of stack can be set only once to prevent inconsistency */

    while(exitCheck == 1)  /* This is the menu */
    {

        cout << "1. Push" << endl;
        cout << "2. Pop" << endl;
        cout << "3. Count" << endl;
        cout << "4. Display" << endl;
        cout << "5. Exit\n" << endl;
        cin >> choice; //Choice should be placed here as we need userinput during each turn 

        switch(choice)
        {
            case 1:
                cout<< "Enter your data: ";
                cin>>entry;
                push(entry);
                break;
            case 2:
                pop();
                break;
            case 3:
                count();
                break;
            case 4:
                display();
                break;
            case 5:
                {
                    exitCheck=1;
                    break;
                } /*exit in this case wont achieve a proper result in a concole based app thats why i replaced it with loop exit condition */

            default:
                cout << "Please enter correct choice (1-5)!\n";
                break;
            }

        cout<< "Enter 1 to continue else anything else to quit:\n";
        cin>> exitCheck;


    }

    cout<<"Thanks for using this code!!\n";    
return 0;   
}
  //end main