数据结构探险—栈篇
什么是栈?
古代栈就是牲口棚的意思。
栈是一种机制:后进先出 LIFO(last in first out)
电梯
空栈。栈底,栈顶。没有元素的时候,栈顶和栈底指向同一个元素,如果加入新元素,栈顶不断升高。取出数据时栈顶不断地降低。栈顶和栈底都称之为栈要素。
- 通过demo说明栈的基本原理
- 热身运动-进制转换:十进制转换到二进制,八进制,十六进制
N = (N div d) * d + N mod d
- 步步为营- 括号匹配检测:检测一个字符串中的各种括号是否匹配
[()] [()()] [()[()]]
实例介绍
mystack.h:
#ifndef MYSTACK_H
#define MYSTACK_H
class MyStack
{
public:MyStack(int size); //分配内存初始化栈空间,设定栈容量,栈顶~MyStack(); //回收栈空间内存bool stackEmpty(); //判断栈是否为空bool stackFull(); //判断栈是否为满void clearStack(); //清空栈int stackLength(); //栈中元素的个数bool push(char elem); //将元素压入栈中,栈顶上升bool pop(char &elem); //将元素推出栈,栈顶下降void stackTraverse(bool isFromButtom); //遍历栈中元素并输出
private:int m_iTop; //栈顶,栈中元素个数int m_iSize; //栈容量char *m_pBuffer; //栈空间指针
};#endif
mystack.cpp:
#include "Mystack.h"
#include <iostream>
using namespace std;MyStack::MyStack(int size)
{m_iSize = size;m_pBuffer = new char[size];m_iTop = 0;
}
MyStack::~MyStack()
{delete[]m_pBuffer;m_pBuffer = NULL;}
bool MyStack::stackEmpty()
{if (m_iTop == 0)//if(0 == m_iTop){return true;}else{return false;}
}
bool MyStack::stackFull()
{if ( m_iTop == m_iSize)//>={return true;}else{return false;}
}void MyStack::clearStack()
{m_iTop = 0;//原栈中所有值无效
}int MyStack::stackLength()
{return m_iTop;
}bool MyStack::push(char elem)//放入栈顶
{if (stackFull()){return false;}m_pBuffer[m_iTop] = elem;m_iTop++;return true;
}
bool MyStack::pop(char &elem)
{if (stackEmpty()){return false;}m_iTop--;//因为入栈时做了++,使栈顶指向下一个空位置elem = m_pBuffer[m_iTop];return true;
}//char MyStack::pop()
//{
// if (stackEmpty())
// {
// throw 1;
// }
// else
// {
// m_iTop--;
// return m_pBuffer[m_iTop];
// }
//}void MyStack::stackTraverse(bool isFromButtom)
{if (isFromButtom){for (int i = 0; i < m_iTop; i++){cout << m_pBuffer[i] << ",";}}else{for (int i = m_iTop - 1; i >= 0; i--){cout << m_pBuffer[i] << ",";}}}
main.cpp:
#include "Mystack.h"
#include <iostream>
#include <stdlib.h>
using namespace std;
int main(void)
{MyStack *pStack = new MyStack(5);pStack->push('h');//底pStack->push('e');pStack->push('l');pStack->push('l');pStack->push('o');//顶pStack->stackTraverse(true);char elem = 0;pStack->pop(elem);cout << endl;cout << elem << endl;//pStack->clearStack();pStack->stackTraverse(false);cout << pStack->stackLength() << endl;if (pStack->stackEmpty()){cout << "栈为空" << endl;}if (pStack->stackFull()){cout << "栈为满" << endl;}delete pStack;pStack = NULL;system("pause");return 0;
}
运行结果:
案例改造。
要求:
#ifndef MYSTACK_H
#define MYSTACK_H
#include "Coordinate.h"
class MyStack
{
public:MyStack(int size); //分配内存初始化栈空间,设定栈容量,栈顶~MyStack(); //回收栈空间内存bool stackEmpty(); //判断栈是否为空bool stackFull(); //判断栈是否为满void clearStack(); //清空栈int stackLength(); //栈中元素的个数bool push(Coordinate elem); //将元素压入栈中,栈顶上升bool pop(Coordinate &elem); //将元素推出栈,栈顶下降void stackTraverse(bool isFromButtom); //遍历栈中元素并输出
private:int m_iTop; //栈顶,栈中元素个数int m_iSize; //栈容量Coordinate *m_pBuffer; //栈空间指针
};
#endif#include "Mystack.h"
#include <iostream>
using namespace std;MyStack::MyStack(int size)
{m_iSize = size;m_pBuffer = new Coordinate[size];m_iTop = 0;
}
MyStack::~MyStack()
{delete[]m_pBuffer;m_pBuffer = NULL;}
bool MyStack::stackEmpty()
{if (m_iTop == 0)//if(0 == m_iTop){return true;}else{return false;}
}
bool MyStack::stackFull()
{if ( m_iTop == m_iSize)//>={return true;}else{return false;}
}void MyStack::clearStack()
{m_iTop = 0;//原栈中所有值无效
}int MyStack::stackLength()
{return m_iTop;
}bool MyStack::push(Coordinate elem)//放入栈顶
{if (stackFull()){return false;}m_pBuffer[m_iTop] = elem;//因为这里的coordinate是一个简单的复制。所以使用默认拷贝函数就可以了m_iTop++;return true;
}
bool MyStack::pop(Coordinate &elem)
{if (stackEmpty()){return false;}m_iTop--;//因为入栈时做了++,使栈顶指向下一个空位置elem = m_pBuffer[m_iTop];return true;
}//char MyStack::pop()
//{
// if (stackEmpty())
// {
// throw 1;
// }
// else
// {
// m_iTop--;
// return m_pBuffer[m_iTop];
// }
//}void MyStack::stackTraverse(bool isFromButtom)
{if (isFromButtom){for (int i = 0; i < m_iTop; i++){//cout << m_pBuffer[i] << ",";m_pBuffer[i].printCoordinate();}}else{for (int i = m_iTop - 1; i >= 0; i--){//cout << m_pBuffer[i] << ",";m_pBuffer[i].printCoordinate();}}}
#ifndef COORDINATE_H
#define COORDINATE_H
class Coordinate
{
public:Coordinate(int x=0,int y=0);void printCoordinate();
private:int m_iX;int m_iY;
};
#endif#include "Coordinate.h"
#include <iostream>
using namespace std;Coordinate::Coordinate(int x, int y)
{m_iX = x;m_iY = y;
}
void Coordinate::printCoordinate()
{cout << "(" << m_iX << "," << m_iY << ")" << endl;
}
main.cpp:
#include "Mystack.h"
#include <iostream>
#include <stdlib.h>
using namespace std;
int main(void)
{MyStack *pStack = new MyStack(5);pStack->push(Coordinate(1,2));//底pStack->push(Coordinate(3, 4));pStack->stackTraverse(true);pStack->stackTraverse(false);cout << pStack->stackLength() << endl;delete pStack;pStack = NULL;system("pause");return 0;
}
运行结果:
经过改造我们使栈满足了coordinate对象的入栈出栈。
将普通栈改为类模板栈。使其可以适用于任何数据类型
上面我们实现过两遍对于栈的实现。一次是实现char数组的栈。一次是实现coordinate对象的。两次除过数据类型。差别不是很大。所以本次我们使用类模板实现适用任何数据类型的栈
mystack.h:(因为编译器不支持类模板分开编译。所以cpp为空)
#ifndef MYSTACK_H
#define MYSTACK_H
#include <iostream>
using namespace std;
template <typename T>
class MyStack
{
public:MyStack(int size); //分配内存初始化栈空间,设定栈容量,栈顶~MyStack(); //回收栈空间内存bool stackEmpty(); //判断栈是否为空bool stackFull(); //判断栈是否为满void clearStack(); //清空栈int stackLength(); //栈中元素的个数bool push(T elem); //将元素压入栈中,栈顶上升bool pop(T &elem); //将元素推出栈,栈顶下降void stackTraverse(bool isFromButtom); //遍历栈中元素并输出
private:int m_iTop; //栈顶,栈中元素个数int m_iSize; //栈容量T *m_pBuffer; //栈空间指针
};template <typename T>
MyStack<T>::MyStack(int size)
{m_iSize = size;m_pBuffer = new T[size];m_iTop = 0;
}
template <typename T>
MyStack<T>::~MyStack()
{delete[]m_pBuffer;m_pBuffer = NULL;}
template <typename T>
bool MyStack<T>::stackEmpty()
{if (m_iTop == 0)//if(0 == m_iTop){return true;}else{return false;}
}
template <typename T>
bool MyStack<T>::stackFull()
{if (m_iTop == m_iSize)//>={return true;}else{return false;}
}
template <typename T>
void MyStack<T>::clearStack()
{m_iTop = 0;//原栈中所有值无效
}
template <typename T>
int MyStack<T>::stackLength()
{return m_iTop;
}
template <typename T>
bool MyStack<T>::push(T elem)//放入栈顶
{if (stackFull()){return false;}m_pBuffer[m_iTop] = elem;//因为这里的coordinate是一个简单的复制。所以使用默认拷贝函数就可以了m_iTop++;return true;
}
template <typename T>
bool MyStack<T>::pop(T &elem)
{if (stackEmpty()){return false;}m_iTop--;//因为入栈时做了++,使栈顶指向下一个空位置elem = m_pBuffer[m_iTop];return true;
}//char MyStack::pop()
//{
// if (stackEmpty())
// {
// throw 1;
// }
// else
// {
// m_iTop--;
// return m_pBuffer[m_iTop];
// }
//}
template <typename T>
void MyStack<T>::stackTraverse(bool isFromButtom)
{if (isFromButtom){for (int i = 0; i < m_iTop; i++){cout << m_pBuffer[i];//m_pBuffer[i].printCoordinate();}}else{for (int i = m_iTop - 1; i >= 0; i--){cout << m_pBuffer[i];//m_pBuffer[i].printCoordinate();}}}
#endif
#ifndef COORDINATE_H
#define COORDINATE_H
#include <ostream>
using namespace std;
class Coordinate
{friend ostream &operator<<(ostream &out, Coordinate &coor);
public:Coordinate(int x=0,int y=0);void printCoordinate();
private:int m_iX;int m_iY;
};
#endif#include "Coordinate.h"
#include <iostream>
using namespace std;Coordinate::Coordinate(int x, int y)
{m_iX = x;m_iY = y;
}
void Coordinate::printCoordinate()
{cout << "(" << m_iX << "," << m_iY << ")" << endl;
}ostream &operator<<(ostream &out, Coordinate &coor)
{out << "(" << coor.m_iX << "," << coor.m_iY << ")" << endl;return out;
}
main.cpp:
#include "Mystack.h"
#include <iostream>
#include <stdlib.h>
#include "Coordinate.h"
using namespace std;
int main(void)
{MyStack<Coordinate> *pStack = new MyStack<Coordinate>(5);pStack->push(Coordinate(1,2));//底pStack->push(Coordinate(3, 4));pStack->stackTraverse(true);pStack->stackTraverse(false);cout << pStack->stackLength() << endl;MyStack<char> *pStack2 = new MyStack<char>(5);pStack2->push('h');//底pStack2->push('e');pStack2->push('l');pStack2->push('l');pStack2->push('o');//顶pStack2->stackTraverse(true);delete pStack;pStack = NULL;system("pause");return 0;
}
可以看到我们的类模板已经将栈改造成了通用数据类型的栈。
栈应用-进制转换
短除法。不停除以进制数。保留余数。然后商继续除以进制保留余数。直到商为0
栈的应用:将每次的余数4 0 5 2 入栈。然后从栈顶开始打印。
#ifndef MYSTACK_H
#define MYSTACK_H
#include <iostream>
using namespace std;
template <typename T>
class MyStack
{
public:MyStack(int size); //分配内存初始化栈空间,设定栈容量,栈顶~MyStack(); //回收栈空间内存bool stackEmpty(); //判断栈是否为空bool stackFull(); //判断栈是否为满void clearStack(); //清空栈int stackLength(); //栈中元素的个数bool push(T elem); //将元素压入栈中,栈顶上升bool pop(T &elem); //将元素推出栈,栈顶下降void stackTraverse(bool isFromButtom); //遍历栈中元素并输出
private:int m_iTop; //栈顶,栈中元素个数int m_iSize; //栈容量T *m_pBuffer; //栈空间指针
};template <typename T>
MyStack<T>::MyStack(int size)
{m_iSize = size;m_pBuffer = new T[size];m_iTop = 0;
}
template <typename T>
MyStack<T>::~MyStack()
{delete[]m_pBuffer;m_pBuffer = NULL;}
template <typename T>
bool MyStack<T>::stackEmpty()
{if (m_iTop == 0)//if(0 == m_iTop){return true;}else{return false;}
}
template <typename T>
bool MyStack<T>::stackFull()
{if (m_iTop == m_iSize)//>={return true;}else{return false;}
}
template <typename T>
void MyStack<T>::clearStack()
{m_iTop = 0;//原栈中所有值无效
}
template <typename T>
int MyStack<T>::stackLength()
{return m_iTop;
}
template <typename T>
bool MyStack<T>::push(T elem)//放入栈顶
{if (stackFull()){return false;}m_pBuffer[m_iTop] = elem;//因为这里的coordinate是一个简单的复制。所以使用默认拷贝函数就可以了m_iTop++;return true;
}
template <typename T>
bool MyStack<T>::pop(T &elem)
{if (stackEmpty()){return false;}m_iTop--;//因为入栈时做了++,使栈顶指向下一个空位置elem = m_pBuffer[m_iTop];return true;
}//char MyStack::pop()
//{
// if (stackEmpty())
// {
// throw 1;
// }
// else
// {
// m_iTop--;
// return m_pBuffer[m_iTop];
// }
//}
template <typename T>
void MyStack<T>::stackTraverse(bool isFromButtom)
{if (isFromButtom){for (int i = 0; i < m_iTop; i++){cout << m_pBuffer[i];//m_pBuffer[i].printCoordinate();}}else{for (int i = m_iTop - 1; i >= 0; i--){cout << m_pBuffer[i];//m_pBuffer[i].printCoordinate();}}}
#endif#include "Mystack.h"
#include <iostream>
#include <stdlib.h>
using namespace std;#define BINARY 2
#define OCTONARY 8
#define HEXADECIMAL 16int main(void)
{MyStack<int> *pStack = new MyStack<int>(30);int N = 1348;int mod = 0;while (N !=0){mod = N % BINARY;pStack->push(mod);N = N / BINARY;}pStack->stackTraverse(false);delete pStack;pStack = NULL;system("pause");return 0;
}
二进制和8进制都没有问题了,16进制还需要进一步改造。
运行结果:
!运行结果](http://upload-images.jianshu.io/upload_images/1779926-c6c62f86ff27da42.png?imageMogr2/auto-orient/strip%7CimageView2/2/w/1240)
16进制改造
mystack.h与原来一致。
#include "Mystack.h"
#include <iostream>
#include <stdlib.h>
using namespace std;#define BINARY 2
#define OCTONARY 8
#define HEXADECIMAL 16int main(void)
{char num[] = "0123456789ABCDEF";MyStack<char> *pStack = new MyStack<char>(30);int N = 2016;int mod = 0;while (N !=0){mod = N % HEXADECIMAL;pStack->push(num[mod]);N = N / HEXADECIMAL;}pStack->stackTraverse(false);/*for (int i=pStack->stackLength()-1;i>=0;i--){num[pStack[i]]}*//*int elem = 0;while (!pStack->stackEmpty()){pStack->pop(elem);cout << num[elem];}*/delete pStack;pStack = NULL;system("pause");return 0;
}
如果仍使栈为int型。则可以使用注释部分打印出内容。修改为char之后。可使用pStack->push(num[mod]);
栈应用括号匹配
从前往后扫描。左方括号入栈,左圆括号入栈,当遇到右括号则左圆括号出栈。当遇到右方括号,左方括号出栈。字符串扫描完毕时栈为空则全部匹配。栈中还有东西则不是全部匹配
#include "Mystack.h"
#include <iostream>
#include <stdlib.h>
using namespace std;int main(void)
{MyStack<char> *pStack = new MyStack<char>(30);//已存入的字符MyStack<char> *pNeedStack = new MyStack<char>(30);//需要的字符。char str[] = "[()]]";char currentNeed = 0;for (int i=0;i<strlen(str);i++){if (str[i] != currentNeed)//如果此时扫描到的字符不是我们所需要的。{pStack->push(str[i]);//那么将这个字符存入“已存入字符”switch (str[i])//对于这个字符,生成它的currentneed{case '[':if (currentNeed !=0)//如果currentneed已经有值,不为初值。{pNeedStack->push(currentNeed);//将当前的需要字符入栈。}currentNeed = ']';//生成当前需要。break;case '(':if (currentNeed != 0){pNeedStack->push(currentNeed);}currentNeed = ')';break;default:cout << "字符串不匹配" << endl;system("pause");return 0;}}else{char elem;pStack->pop(elem);if (pNeedStack->pop(currentNeed)){currentNeed = 0;}}}if (pStack->stackEmpty()){cout << "字符串括号匹配" << endl;}delete pStack;pStack = NULL;delete pNeedStack;pNeedStack = NULL;system("pause");return 0;
}
运行过程:
最开始:currentneed为0.
- str[0]为"[",此时需要的currentneed为0,不相等。
- 进入if内部。将"[" 存入栈1。进入switch的case内部。匹配到case:"["
- 此时判断到当前的currentneed = 0.不满足if。则生成currentneed "]"。并break
出循环。 - str[1]为"(",此时需要的currentneed是"]",不相等。
- 进入if内部将"("存入栈1.进入switch的case内部。匹配到case:"("
- 此时判断到当前的currentneed ="]"不等于0.将该字符存入需要栈,因为下面就要对他进行覆盖了、
- 生成新的的currentneed")",并break出循环
- str[2]为")",正好与我们当前的currentneed一致。
- 那么我们将栈一的"("弹出。并将needstack里的上一个急需的赋值给currentneed。
- 进入下一次循环。
也就是currentneed变量里面存放的是当前下一次循环刚开始急需匹配的。
need栈里存放的是历史需要的。
当当前需要的和正在扫描的一致。则将栈1中出栈。