【题目描述】

AcWing 41. 包含min函数的栈

【思路】
使用单调栈（minstack）维系原基础栈(data)的最小值。原栈栈顶数值只有大于单调栈的栈顶元素才push进单调栈。

class MinStack {

    /** initialize your data structure here. */
    /*
    两个栈 data基础栈存放原数据 minstack最小栈存放最小值
    */
    Stack<Integer> data;
    Stack<Integer> minstack;

    public MinStack() {
        data = new Stack<Integer>();
        minstack = new Stack<Integer>();
    }

    public void push(int x) {
        if( data.empty() ){
            data.push(x);
            minstack.push(x);
            //要return
            return;
        }

        int min = minstack.peek();
        if(x > min) minstack.push(min);
        else minstack.push(x);
        data.push(x);

    }

    public void pop() {
        minstack.pop();
        data.pop();
    }

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

    public int getMin() {
        return minstack.peek();
    }
}

/**
 * Your MinStack object will be instantiated and called as such:
 * MinStack obj = new MinStack();
 * obj.push(x);
 * obj.pop();
 * int param_3 = obj.top();
 * int param_4 = obj.getMin();
 */