❤leetcode,python2❤最小栈

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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。

push(x) – 将元素 x 推入栈中。
pop() – 删除栈顶的元素。
top() – 获取栈顶元素。
getMin() – 检索栈中的最小元素。
示例:

MinStack minStack = new MinStack();
minStack.push(-2);
minStack.push(0);
minStack.push(-3);
minStack.getMin(); --> 返回 -3.
minStack.pop();
minStack.top(); --> 返回 0.
minStack.getMin(); --> 返回 -2.

class MinStack(object):

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

    def push(self, x):
        """
        :type x: int
        :rtype: None
        """
        self.all.append(x)

    def pop(self):
        """
        :rtype: None
        """
        del self.all[-1]

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

    def getMin(self):
        """
        :rtype: int
        """
        return min(self.all)
        


# Your MinStack object will be instantiated and called as such:
# obj = MinStack()
# obj.push(x)
# obj.pop()
# param_3 = obj.top()
# param_4 = obj.getMin()