数据结构实战：栈与队列-优快云博客

本文深入探讨了数据结构中的栈与队列概念，通过C++代码实现了数组和链表栈，模拟了浏览器进退功能，并解决了LeetCode括号匹配问题。文章还提供了斐波那契数列和数据集合全排列的算法实现。

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1.用数组实现一个栈,用了一个多小时····不过还算顺利，全部代码如下

#include<iostream>
using namespace std;

template<typename T >
class arrayStack{
public:
	arrayStack(){
		capacity = 5;
		size = 0;
		list = new T[capacity];
	}
	~arrayStack(){
		delete list;
		list = nullptr;
	}
	void push(const T& elem);
	void pop();
	T& top() const;
	int getSize() const;
	bool empty() const;
	void checkCapacity();
private:
	T* list;
	int capacity;
	int size;
};
template<typename T> void arrayStack<T>::push(const T& elem){
	checkCapacity();
	list[size] = elem;
	size++;
}
template<typename T> void arrayStack<T>::checkCapacity(){
	if (size == capacity)
	{
		T* newlist = new T[capacity + 5];
		for (int i = 0; i < size; i++)
		{
			newlist[i] = list[i];
		}
		list = newlist;
	}
}
template<typename T> void arrayStack<T>::pop(){
	if (size == 0)
		return;
	size--;
}
template<typename T> T& arrayStack<T>::top() const{
	if (size == 0)
		throw  "EmptyStackException";
	return(list[size - 1]);
}
template<typename T> int  arrayStack<T>::getSize() const{
	return size;
}
template<typename T> bool arrayStack<T>::empty() const{
	return(0 == size);
}
void main(){
	arrayStack<char*> stack;
	stack.push("a");
	cout << stack.top() << endl;
	stack.push("g");
	stack.push("e");
	stack.push("a");
	stack.push("g");
	stack.push("e");
	cout << stack.getSize();
	cout << stack.top();
	stack.push("y");
	cout << stack.top();
	system("pause");
	return;
}

不过有小伙伴知道arrayStack<char*> stack;这句话为什么是char* 而不是 char呢
下一步用链表实现一个栈
2.存储任意类型的栈，不过这个程序在指针上出现了问题，编译能通过，由于在这道任务上花费了很长时间，这个有待后续解BUG。有兴趣的小伙伴可以帮忙解决一下这个问题。

#include<iostream>
using namespace std;

template<typename T>
class stackNode{
public:
	T data;
	stackNode *next;
	stackNode(){};
	stackNode(T e){
		data = T;
	}
};

template<typename T>
class linkStack
{
public:
	linkStack();
	~linkStack();
	void pop();
	void push(const T& elem);
	bool stackEmpty()const;
	int getSize()const;
	T& getTop() const;
private:
	int size;
	stackNode<T>* buttom;
	stackNode<T>* top;
};
template<typename T> linkStack<T>::linkStack(){
	size = 0;
	stackNode<T>* buttom = top = NULL;
}
template<typename T> linkStack<T>::~linkStack(){
	stackNode<T> *ptr1 = buttom;
	stackNode<T> *ptr2 = buttom;
	while (ptr1 != NULL)
	{
		ptr2 = ptr1;
		ptr1 = ptr1->next;
		delete ptr2;
	}
	delete ptr1;
	delete buttom;
	delete top;
	ptr1 = ptr2 = buttom = top = NULL;
	size = 0;
}
template<typename T>void linkStack<T>::pop(){
	if (stackEmpty())
		return;
	if (getSize() == 1)
	{
		buttom = top = NULL;
		size = 0;
	}
	stackNode<T> *ptr = buttom;
	while (ptr->next != top)
	{
		ptr = ptr->next;
	}
	delete top;
	top = ptr;
	size--;
	delete ptr;
	ptr = NULL;
}
template<typename T> void linkStack<T>::push(const T& elem){
	stackNode<T> *ptr = new stackNode<T>;
	if (ptr == NULL)
		return;
	ptr->data = elem;
	ptr->next = NULL;
	if (top == NULL)
	{
		buttom = top = ptr;
		return;
	}
	top->next = ptr;
	top = ptr;
	delete ptr;
	ptr = NULL;
	size++;
}
template<typename T>bool linkStack<T>::stackEmpty()const{
	return (0 == getSize());
}
template<typename T>int linkStack<T>::getSize()const{
	return size;
}
template<typename T>T& linkStack<T>::getTop() const{
	return top->data;
}
void main(){
	linkStack<char*> stack;
	stack.push("4");
	/*stack.push(5);
	stack.push(5);
	cout << stack.getSize()<<endl;
	cout << stack.getTop()<<endl;
	stack.push(8);
	stack.push(3);
	stack.push(9);
	stack.push(3);
	stack.pop();
	stack.pop();
	cout << stack.getTop()<<endl;
	cout << stack.getSize()<<endl;*/
	cout << "hello" << endl;
	system("pause");
	return;
}

3.模拟浏览器的进退
这个也出现了上面的问题，在访问内存上有差错，有待解决

#include<iostream>
#include<stack>
#include<string>

using namespace std;
class browser{
public:
	browser(){
		stack<string>* currentPage = new stack<string>;
		currentPage->push("This is page D");
		currentPage->push("This is page C");
		currentPage->push("This is page B");
		currentPage->push("This is page A");
		stack<string>* currentPage2 = new stack<string>;
		cout << currentPage->top();
	}
	~browser()
	{
		delete currentPage;
		delete currentPage2;
		currentPage = nullptr;
		currentPage2 = nullptr;
	}
	void forward(){
		/*if ((currentPage->size()) == 1)
		{
			cout << "this is the end" << endl;
			return;
		}*/
		string m = currentPage->top();
		currentPage2->push(m);
		currentPage->pop();
		cout << currentPage->top();
		return;
	}
	void back(){	
		if (currentPage2->size() == 1)
		{
			cout << "this is the begin" << endl;
			return;
		}
		currentPage->push(currentPage2->top());
		currentPage2->pop();
		cout << currentPage->top();
	}
private:
	stack<string>* currentPage;
	stack<string>* currentPage2;
};
class A{
public:
	A()
	{
		i = 1;
	}
	int i;
};
void main(){
/*	A* a = new A;
	if (a == NULL)
		cout << "aaaa" << endl;
	cout << a->i;*/

	browser* Browser = new browser;
	if (Browser == NULL)
		cout<<"sssss";
	Browser->forward();
	/*Browser->forward();
	Browser->forward();
	Browser->forward();
	Browser->back();
	Browser->back();
	Browser->back();
	Browser->back();
	delete Browser;
	Browser = nullptr;*/
	cout << "hello" << endl;
	system("pause");
}

关于队列的编程真的没有时间再手打了，队列的知识与栈还有链表相似。
4.斐波那契数列
这个不难

#include<iostream>
using namespace std;
int solution(int num){
	if (num < 1)
		return 0;
	if (num == 1 || num == 2)
		return 1;
	int count = 2;
	int pre = 1;
	int result = 1;
	while (count < num)
	{
		int mid = result;
		result += pre;
		pre = mid;
		count++;
	}
	return result;
}
void main()
{
	int k = solution(6);
	int i = solution(5);
	int j = solution(4);
	cout << k << i << j << endl;
	system("pause");
}

求n的阶乘就不打了，挺简单的
5.求一个数据集合的全排列
我选择了字符型集合数组作为参数
采用递归的思想，当cur = 0时，把字符串中所有元素都换到第一个元素中去，再组合上后面的所有可能，往后推导类似，代码如下

#include<iostream>
using namespace std;
void swap(char *ptr1, char *ptr2)
{
	char tmp = *ptr1;
	*ptr1 = *ptr2;
	*ptr2 = tmp;
}
void solution(char *data, int cur, int length)
{
	if (cur == length){
		for (int i = 0; i < length; i++)
		{
			cout << data[i];
		}
		cout << endl;
	}
	for (int j = cur; j < length; j++){
		swap(&data[cur], &data[j]);
		solution(data, cur + 1, length);
		swap(&data[cur], &data[j]);
	}
}
void main(){
	char Data[] = { "fenhy" };
	solution(Data, 0, 5);
	system("pause");
}

6.LeetCode习题
给定一个只包括 ‘(’，’)’，’{’，’}’，’[’，’]’ 的字符串，判断字符串是否有效。
有效字符串需满足：
左括号必须用相同类型的右括号闭合。
左括号必须以正确的顺序闭合。
注意空字符串可被认为是有效字符串。
很明显这道题是运用栈的经典题目，使用匹配的规则，代码如下

#define _SCL_SECURE_NO_WARNINGS
#include<iostream>
#include<string>
#include<stack>
#include<algorithm>

using namespace std;

class Solution {
public:
	bool isValid(string s) {
		stack<char> Stack;
		int length = s.length();
		char* arr = new char[length+1];
		s.copy(arr, length);
		arr[length] = '\0';
		for (int i = 0; i < length; i++)
		{
			if (Stack.size() == 0)
				Stack.push(arr[i]);
			else if (isRight(Stack.top(), arr[i])){
				Stack.pop();
			}
			else
			{
				Stack.push(arr[i]);
			}
		}
		return Stack.size() == 0;
	}
	bool isRight(char a, char b){
		return (a == '(' && b == ')') || (a == '[' && b == ']') || (a == '{' && b == '}');
	}
};
void main(){
	string str1("()[]{}");
	string str2("{}(()");
	Solution sol;
	int a = sol.isValid(str1);
	int b = sol.isValid(str2);
	cout << a <<b<< endl;
	system("pause");
}

任务2实现