题解 | #包含min函数的栈#

import java.util.Stack;

public class Solution {

    Stack<Integer> dataStack = new Stack<>();
    Stack<Integer> minStack = new Stack<>();


    public void push(int node) {

        dataStack.push(node);

        if (minStack.isEmpty() || minStack.peek() > node) {
            minStack.push(node);
        } else {
            //放入自己 因为要保证对应dataStack中的最小元素要在minStack中存在, 
            //一旦dataStack中a元素移除 即使a是最小的,
            //也要保证minStack中下一个弹出的元素应该是,dataStack中除去a下一个最小的元素
            minStack.push(minStack.peek());
        }
    }

    public void pop() {

        if (!dataStack.isEmpty()) {
            dataStack.pop();
        }

        if (!minStack.isEmpty()) {
            minStack.pop();
        }
    }

    public int top() {
        if (!dataStack.isEmpty()) {
           return dataStack.peek(); 
        }
        return 0;
    }

    public int min() {
        if (!minStack.isEmpty()) {
            return minStack.peek();
        }
        return 0;
    }
}