Q3.2 Stack with Min

最新推荐文章于 2024-08-11 01:59:25 发布

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本文介绍了一种特殊的数据结构——栈，除了支持基本的push和pop操作外，还实现了min函数，用于返回栈中当前的最小元素，所有操作均在O(1)时间内完成。通过使用两个栈s1和s2，s1用于常规的元素存取，而s2则保持非递增顺序，确保min操作快速响应。

Q; How would you design a stack which, in addition to push and pop, also has a function min which returns the minimum element? Push, pop and min should all operate in O(1) time.

A: 用两个栈s1, s2, s1正常push, pop,. s2保存非递增序列，即当元素值小于等于s2.top()的时候，才会push进去

#include <iostream>
#include <stack> 
using namespace std;

const int MAX_INT = ~(1<<31);

class stackWithMin{
	private:
		stack<int> s1;
		stack<int> s2;
		
	public:
		void push(int val) {
			s1.push(val);
			if (s2.empty() || val <= s2.top()) {
				s2.push(val);
			}
		}
		
		void pop() {
			if (s1.top() == s2.top()) {
				s2.pop();
			}
			s1.pop();
		}
		
		int top() {
			return s1.top();
		}
		
		bool empty() {
			return s1.empty();
		}
		int min() {
			if (s2.empty()) {
				return MAX_INT;
			}
			return s2.top();
		}
};

int main(){
    stackWithMin mystack;
    for(int i=0; i<20; ++i)
        mystack.push(i);
    cout<<mystack.min()<<" "<<mystack.top()<<endl;
    mystack.push(-100);
    mystack.push(-100);
    cout<<mystack.min()<<" "<<mystack.top()<<endl;
    mystack.pop();
    cout<<mystack.min()<<" "<<mystack.top()<<endl;
    return 0;
}