题解如代码,需要使用一个辅助栈,用来存当前最小值:

import java.util.Stack;

public class Solution {

    Stack<Integer> stack  = new Stack<>();
    //需要一个辅助栈来存放最小元素,该辅助栈栈顶则为当前最小元素
    Stack<Integer> minStack = new Stack<>();

    int min = Integer.MAX_VALUE;

    public void push(int node) {
        if(node < min) {
            minStack.push(node);
            min = node;
        }
        stack.push(node);
    }

    public void pop() {
        int t = stack.pop();
        //判断stack pop的元素是否是当前最小值,是的话需要都pop
        if(t == min && minStack.size() != 0){
            minStack.pop();
        }
    }

    public int top() {
        return stack.peek();
    }

    public int min() {
        return minStack.peek();
    }
}