题解 | #包含min函数的栈#

import java.util.*;
import java.util.Stack;

public class Solution {

    Stack<Integer> stack = new Stack<Integer>();
    Stack<Integer> montonicityStack = new Stack<Integer>();

    public void push(int node) {
        if(montonicityStack.isEmpty()||montonicityStack.peek()>=node){
            stack.push(node);
            montonicityStack.push(node);
        }
        stack.push(node);
    }
    
    public void pop() {
        int min = montonicityStack.peek();
        int temp = stack.peek();
        if(min==temp){
            stack.pop();
            montonicityStack.pop();
        }
        stack.pop();
    }
    
    public int top() {
        return stack.peek();
    }
    
    public int min() {
        return montonicityStack.peek();
    }
}

通过对传入的node值比较，使得单调栈的栈顶的值一直是当前栈的最小值