Implement a stack with min() function, which will return the smallest number in the stack.

It should support push, pop and min operation all in O(1) cost.

Solution: 用两个stack。push时，minstack push Math.min(minStack.peek(), number)

    private Stack<Integer> stack;
    private Stack<Integer> minStack;
    public MinStack() {
        stack = new Stack<Integer>();
        minStack = new Stack<Integer>();
    }

    public void push(int number) {
        stack.push(number);
        if(minStack.isEmpty()){
            minStack.push(number);
        } else{
            minStack.push(Math.min(minStack.peek(), number));
        }
    }

    public int pop() {
        minStack.pop();
        return stack.pop();
    }

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