Ch3-2: implement min() for a stack with O(1)

Here will show the thinking flow from bad to good.

idea 1:

const int MAX_INT = ~(1<<31);//2147483647
struct node{
    int val, min;
};

class StackWithMin{
public:
    StackWithMin(int size){
        buf = new node[size];
        buf[0].min = MAX_INT;
        cur = 0;
    }
    ~StackWithMin(){
        delete[] buf;
    }
    void push(int val){
        buf[++cur].val = val;
        //buf[ptop].min = (val > buf[--ptop].min)?buf[--ptop].min:val;
        buf[cur].min = (val>(buf[cur-1].min))?buf[cur-1].min:val;
    }
    void pop(){
        --cur;
    }
    int top(){
        return buf[cur].val;
    }
    bool empty(){
        return cur==0;
    }
    int min(){
        return buf[cur].min;
    }
    void print(){
    	while(--cur){
			cout << buf[cur].val << ", ";
		}
	}
private:
    node *buf;
    int cur;
};

注：正如 http://www.learncpp.com/cpp-tutorial/69-dynamic-memory-allocation-with-new-and-delete/所述：

new[] 在这里是“ use the array form of new and delete (often called new[] and delete[]):”

而“buf是用来access这个node型的array的”， 所以“ Note that array access is done the same way with dynamically allocated arrays as with normal arrays. While this might look slightly funny, given that pnArray is explicitly declared as a pointer, remember that arrays are really just pointers in C++ anyway.”

output:

Executing the program....
$demo 
0 19
20 + 20 - 20
18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 0 manhattan
-100 -100
-100 -100

Whole solution with ans1 and ans2:

ans use 2 stack instead of 1 so save more space in average.

// solution for CC150: Ch3-2.cpp
// idea 0: iterate the stack one by one, and use a min variable to store the minimum number
//    	   this is problematic solution, for example, the poped top is the min value and then
//         the min variable is not the real min value.
// idea 0.1: how about use 2 min variable? NO...  Becuase if continuous pop, it won't work
// idea 1: use data structure: node(val, minTillnow), then each node has the current value
//		   and the min value to this node, so even pop, the rest nodes store the info about
//    	   min, but the space = 2X(idea 1).

#include 
  
   
using namespace std;
const int MAX_INT = ~(1<<31);//2147483647

struct node{
	int val;
	int min;  // min till this node
};

class StackWithMin1{
public:
	StackWithMin1(int size){
		buf = new node[size];
		buf[0].min = MAX_INT;
		//this->ptop = -1;
        ptop = 0;
	}

	~StackWithMin1(){
		delete[] buf;
	}

	void push(int val){
		buf[++ptop].val = val;
		buf[ptop].min = (val > buf[ptop-1].min)?buf[ptop-1].min:val;
	}

	void pop(){
		--ptop;
	}

	int top(){
		return buf[ptop].val;
	}

	bool empty(){
		return ptop==0;
	}

	int min(){
		return buf[ptop].min;
	}

	void print(){
        int ptr = this->ptop;
        cout << this->ptop<<" + " <
   
    <<" - "<< ptr << endl;
		while(ptr){
			cout << buf[--ptr].val << " ";
		}
        cout << "manhattan" << endl;
	}

private:
	node* buf;
	int ptop;
};


// idea 2: use 2 stack, 1st to store all the nodes, 2nd to store all the min value to this node
// 		   because push/pop is for connected nodes, so the min value will keep in order in stack1/2

class stack{
public:
	stack(int size=100){
		arr = new int[size];
		cur = -1;
	}
	~stack(){
		delete[] arr;
	}
	void push(int val){
		arr[++cur] = val;
	}
	void pop(){
		--cur;
	}
	int top(){
		return arr[cur];
	}
	bool empty(){
		return cur==-1;
	}
    void print(){
        
        while(cur>-1){
			cout << arr[cur--] << " ";
		}
        cout << "manhattan" << endl;
	}
private:
	int* arr;
	int cur;
};

class StackWithMin2{
public:
	StackWithMin2(){
		//
	}
	~StackWithMin2(){
		//
	}
	void push(int val){
		s1.push(val);
		if(val <= min())
			s2.push(val);
	}
	void pop(){
		if(s1.top()==min())
			s2.pop();
		s1.pop();
	}
	int top(){
		return s1.top();
	}

	bool empty(){
		return s1.empty();
	}
    
	int min(){
		if(s2.empty()) return MAX_INT;
		else return s2.top();
	}
    /*
    void print(){   // how to print s1,s2?
        while(s1.cur){
    		cout << arr[--s1.cur] << " ";
		}
        cout << "manhattan" << endl;
	}
    */
    void print(){
        s1.print();
        s2.print();
    }

private:
	stack s1,s2;
	//stack s2(100);
};


int main(){
    //cout<
    
     <
     Executing the program....
$demo 
20 90
90 20 30 40 59 58 57 56 55 54 53 52 51 50 manhattan  //s1
20 30 40 50 manhattan  // s2:here shows that s2 only stack the current min value
-100 -100
-100 -100