LeetCode()- Min Stack

題目：

Design a stack that supports push, pop, top, and retrieving the minimum element in constant time.

push(x) -- Push element x onto stack.
pop() -- Removes the element on top of the stack.
top() -- Get the top element.
getMin() -- Retrieve the minimum element in the stack.

思路：

題意：給出四個函數API,構造一個stack，而且能夠返回最小值 用雙棧的策略，一個用來正常的存儲，一個用來存貯最小值 注意比較的時候。peek（）函數要用equals函數比較，因為彈出的是對象

代碼：

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

    public void pop() {
        if(stack == null){
            return;
        }
        if(min.peek().equals(stack.peek())){
            min.pop();
        }
        stack.pop();
    }

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

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