/**
1. 因为在维护栈先进后出的性质下要, 实时维护栈中的最小的最小值, 所以要用另一个栈存最小值
2. 操作
2.1 当 push(x)   时, 如果 stack2 是空的 或 x <= stack2.top(), stack2.push(x) , 一定要<= , 如果< 的话,重复的最小元素可能被提前弹出
2.2 当 y = pop() 时, 如果 stack2 不为空 且 x == stack2.top(), stack2.pop()
*/

class MinStack {

    /** initialize your data structure here. */

    Stack<Integer> stack, min;
    public MinStack() {
        stack = new Stack<>();
        min = new Stack<>();
    }

    public void push(int x) {
        if (min.isEmpty() || x <= min.peek()) min.push(x);
        stack.push(x);
    }

    public void pop() {
        int y  = stack.pop();
        if (!min.isEmpty() && y == min.peek()) min.pop();
    }

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

    public int getMin() {
        return min.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();
 */