150-155. 最小栈

设计一个支持 push ，pop ，top 操作，并能在常数时间内检索到最小元素的栈。(第一个我写的,但是在获取最小值上面时间耗时太久了)
class MinStack1(object):

    def __init__(self):
        """
        initialize your data structure here.
        """
        self.stack = []
        self.stack_copy = []

    def push(self, x):
        """
        :type x: int
        :rtype: None
        """
        if x is not None:
            self.stack.append(x)
            self.stack_copy.append(x)

    def pop(self):
        """
        :rtype: None
        """
        if self.stack:
            self.stack_copy.remove(self.stack.pop())

    def top(self):
        """
        :rtype: int
        """
        if self.stack:
            return self.stack[-1]

    def getMin(self):
        """
        :rtype: int
        """
        if self.stack:
            return sorted(self.stack_copy)[0]


class MinStack(object):

    def __init__(self):
        """
        initialize your data structure here.
        """
        self.stack = []
        self.min_stack = []

    def push(self, x):
        """
        :type x: int
        :rtype: None
        """
        self.stack.append(x)
        if len(self.min_stack) == 0:
            self.min_stack.append(x)
        else:
            if self.min_stack[-1] >= x:
                self.min_stack.append(x)

    def pop(self):
        """
        :rtype: None
        """
        x = self.stack.pop()
        if x == self.min_stack[-1]:
            self.min_stack.pop()
        return x

    def top(self):
        """
        :rtype: int
        """
        return self.stack[-1]

    def getMin(self):
        """
        :rtype: int
        """
        return self.min_stack[-1]


if __name__ == '__main__':
    s1 = MinStack()
    s1.push(-2)
    s1.push(0)
    s1.push(-10)
    s1.push(-3)
    print(s1.getMin())
    s1.pop()
    print(s1.top())
    print(s1.getMin())