Min Stack

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.

Example:

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

My solution:

class MinStack:

    def __init__(self):
        """
        initialize your data structure here.
        """
        self.stack = list()
        self.min = None

    def push(self, x):
        """
        :type x: int
        :rtype: void
        """
        self.stack.append(x)
        if self.min is None:
            self.min = x
        else:
            self.min = min(x, self.min)

    def pop(self):
        """
        :rtype: void
        """
        num = self.stack.pop()
        if self.min == num:
            self.min = None
            for i in range(len(self.stack)):
                if self.min is None:
                    self.min = self.stack[i]
                else:
                    self.min = min(self.stack[i], self.min)

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

    def getMin(self):
        """
        :rtype: int
        """
        if self.min is not None:
            return self.min

# 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()