实现push,pop,max为O(1)的栈

#原理
使用两个栈来实现要求: 栈的push,pop,max都为O(1)。

1. 一个栈用来保存入栈数据中的最大值
2. 另一个栈用来正常保存入栈和出栈的数据，它max方法用来调用第一个栈保存的当前最大值

#实现
##辅助栈
作用：用于保存当前最大值的栈

template<typename T>
class stackMax
{
public:
	stackMax():size(0), capacity(1), data(new T[capacity])
	{
	}
	~stackMax() 
	{
		if (data) {
			delete[] data;
		}
	}
	
	void push(T x) {
		if (size >= capacity) {
			resize();
		}
		data[size++] = x;
	}
	
	T top() {
		if (size <= 0) {
			return T();
		}
		return data[size-1];
	}
	void pop() {
		if (size <= 0) {
			return;
		} else if (size <= capacity/4) {
			resize();
		}
		size--;
	}
private:
	void resize(unsigned size) {
		if (size >= capacity) {
			capacity *= 2;
		} else if (size <= capacity/4) {
			capacity /= 2;
		}
		T *copy = new T[capacity];
		for (unsigned i = 0; i < size; i++) 
			copy[i] = data[i];
		delete[] data;
		data = copy;
	}
private:
	unsigned size;
	unsigned capacity;
	T *data;
};

##栈实现一

template<typename T>
class stack
{
public:
	stack():size(0), capacity(1), data(new T[capacity])
	{
	}
	~stack() 
	{
		if (data) {
			delete[] data;
		}
	}
	
	void push(T x) {
		if (size >= capacity) {
			resize();
		}
		
		if (size == 0 || x > maxStack.top()) {
			maxStack.push(x);
		} else {
			maxStack.push(maxStack.top());
		}
		data[size++] = x;
	}
	T top() {
		if (size <= 0) {
			return T();
		}
		return data[size-1];
	}
	void pop() {
		if (size <= 0) {
			return;
		} else if (size <= capacity/4) {
			resize();
		}
		size--;
		maxStack.pop();
	}
	int max() {
		return maxStack.top();
	}
private:
	void resize() {
		if (size >= capacity) {
			capacity *= 2;
		} else if (size <= capacity/4) {
			capacity /= 2;
		}
		T *copy = new T[capacity];
		for (unsigned i = 0; i < size; i++) {
			copy[i] = data[i];
		}
		delete[] data;
		data = copy;
	}
private:
	unsigned size;
	unsigned capacity;
	T *data;
	// 用于保存当前最大值得栈
	stackMax<T> maxStack;
};

#测试

int main()
{
	stack<int> si;
	si.push(50);
	si.push(30);
	si.push(60);
	si.push(80);
	si.push(40);
	cout << si.max() << endl; // 80
	si.pop();
	cout << si.max() << endl; // 80
	si.pop();
	cout << si.max() << endl; // 60
}
/*
	stack           maxStack
	|----|          |----|
	| 40 |          | 80 |
	| 80 |          | 80 |
	| 60 |          | 60 |
	| 30 |          | 50 |
	| 50 |          | 50 |
	|----|          |----|
*/

#缺点
这里的实现使用动态增长内存的方式，在push时有可能无法实现O(1)的时间复杂度。要想实现O(1)时间复杂度的push, pop，可以使用固定大小的数组来实现。

#参考
实现push,pop,max为O(1)的栈
一道IT技术面试题 带有max函数的栈

#补充
#栈实现二

template<typename T>
class stack
{
public:
	stack():size(0), capacity(1), data(new T[capacity])
	{
	}
	~stack() 
	{
		if (data) {
			delete[] data;
		}
	}
	
	void push(T x) {
		if (size >= capacity) {
			resize();
		}
		
		if (size == 0 || x < maxStack.top()) {
			maxStack.push(x);
		}  // x >= maxStack.top()时，不压栈
		data[size++] = x;
	}
	T top() {
		if (size <= 0) {
			return T();
		}
		return data[size-1];
	}
	void pop() {
		if (size <= 0) {
			return;
		} else if (size <= capacity/4) {
			resize();
		}
		
		if (data[size-1] == maxStack.top()) { 
			maxStack.pop();
		}
		size--;
	}
	int max() {
		return maxStack.top();
	}
private:
	void resize() {
		if (size >= capacity) {
			capacity *= 2;
		} else if (size <= capacity/4) {
			capacity /= 2;
		}
		T *copy = new T[capacity];
		for (unsigned i = 0; i < size; i++) {
			copy[i] = data[i];
		}
		delete[] data;
		data = copy;
	}
private:
	unsigned size;
	unsigned capacity;
	T *data;
	// 用于保存当前最大值得栈
	stackMax<T> maxStack;
};