举例让抽象问题具体化：包含min函数的栈

定义栈的数据结构，请在该类型中实现一个能够得到栈最小元素的min函数。在该栈中，调用min、push及pop的时间复杂度都是O(1)。

import java.util.Stack;

public class Solution {

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

    public void push(int node) {
        Integer temp = new Integer(node);
        dataStack.push(temp);
        if (minStack.empty()) {
            minStack.push(temp);
        } else {
            int min = minStack.peek().intValue();
            if (node > min) {
                minStack.push(minStack.peek());
            } else {
                minStack.push(temp);
            }
        }
    }

    public void pop() {
        dataStack.pop();
        minStack.pop();
    }

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

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