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

Implement the MinStack class:

• MinStack() initializes the stack object.
• void push(int val) pushes the element val onto the stack.
• void pop() removes the element on the top of the stack.
• int top() gets the top element of the stack.
• int getMin() retrieves the minimum element in the stack.

Example 1:

Input
["MinStack","push","push","push","getMin","pop","top","getMin"]
[[],[-2],[0],[-3],[],[],[],[]]

Output
[null,null,null,null,-3,null,0,-2]

Explanation
MinStack minStack = new MinStack();
minStack.push(-2);
minStack.push(0);
minStack.push(-3);
minStack.getMin(); // return -3
minStack.pop();
minStack.top();    // return 0
minStack.getMin(); // return -2 

Constraints:

• -231 <= val <= 231 - 1
• Methods pop, top and getMinoperations will always be called on non-empty stacks.
• At most 3 * 104 calls will be made to push, pop, top, and getMin.

解题思路

Stack进行push和pop操作都非常简单，困难的是如何操作getMin。

getMin即拿到stack中所有未出栈数据中的最小值，因此我们在push和pop时可以给每个数据附加上一个属性，即当前的最小值。

each element is (x,x), the first value if the real value and the second value if the current min value

有几个注意点：

1. 考虑边界条件，比如说stack中的最后一个数据被pop
2. 每次pop之后需要把当前最小值变为top数据的最小值。

class MinStack(object):

    def __init__(self):
        """
        initialize your data structure here.
        """
        self.data=[] 
        self.current_min=None
        # each element is (x,x), the first value if the real value and the second value if the current min value


    def push(self, x):
        """
        :type x: int
        :rtype: None
        """
        if self.current_min==None:
            self.current_min=x
            self.data.append((x,self.current_min))
        elif x>=self.current_min:
            self.data.append((x,self.current_min))
        elif x<self.current_min:
            self.current_min=x
            self.data.append((x,self.current_min))
        return None

    def pop(self):
        """
        :rtype: None
        """
        self.data.pop(-1)
        if len(self.data)==0:
            self.current_min=None
        else:
            self.current_min=self.data[-1][1]
        return None


    def top(self):
        """
        :rtype: int
        """
        return self.data[-1][0]


    def min(self):
        """
        :rtype: int
        """
        return self.data[-1][1]




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

性能：

执行用时：108 ms, 在所有 Python 提交中击败了85.47%的用户

内存消耗：17 MB, 在所有 Python 提交中击败了5.14%的用户

其它可能解法：

辅助栈

取巧的方法可以直接使用min函数