数据结构C++版-栈

一、概念

 

二、应用实例

1.进制转换

#include <stdlib.h>
#include <iostream>
#include <string>

#include "MyStack.h"
#include "Coordinate.h"
using namespace std;

#define  BINARY 2
#define  OCTONARY 8
#define  HEXADECIMAL 16

void main ( )
{ 
    //MyStack<int> s(30);
    MyStack<char> s(30);

    char num[]="0123456789ABCDEF";

    int N=1348;
    int mod=0;
    while (N!=0)
    {
        mod=N%16;
        s.push(num[mod]);
        N=N/16;
    }

    s.stackTraverse(false);

    


    /*
        int elem=0;

        for (int i=s.stackLength()-1;i>=0;i--)
    {
        s.pop(elem);
        cout<<num[elem];
    }*/


    /*while(!s.stackEmpty())
    {
        s.pop(elem);
        cout<<num[elem];
    }*/


    system("pause");
    
} 

2.括号匹配

MyStack<char> s(30);   //存放未匹配的括号
    MyStack<char> s1(30);  //存放栈顶括号的另一半

    char str[] = "([])[";    //存放待匹配文本目标，要求无空格 e.g.   [()()]  ([]([]))
    char current=0;  //当前括号需要匹配的另一半

    for (int i=0;i<strlen(str);i++)
    {
        if (current!=str[i])
        {
            s.push(str[i]);
            switch(str[i])
            {
            case  '[':    //case 后面数据类型是int，单个字符会转换成其ASC码

                if (current!=0)
                {
                    s1.push(current);
                }
                current=']';
                break;

            case '(':

                if (current!=0)
                {
                    s1.push(current);
                }
                current=')';
                break;

            default:
                cout<<"括号不匹配."<<endl;  //default后面语句可以注释掉，因为current不等于str[i]时str[i]就会入栈，第一个栈不为零匹配就会失败

            }
            
        } 
        else
        {
            char elem;
            s.pop(elem);

            if(!s1.pop(current))
            {
                current=0;
            }
        }
    }



if (s.stackEmpty())
    {
        cout<<"括号匹配"<<endl;
    } 
    else
    {
        cout<<"括号不匹配"<<endl;
    }

    system("pause");