「剣指Offer(第二版)」面接問題30.min関数を含むスタック

        ，                      min       ，   min、push   pop          O(1)。

  :
MinStack minStack = new MinStack();
minStack.push(-2);
minStack.push(0);
minStack.push(-3);
minStack.min();   -->    -3.
minStack.pop();
minStack.top();      -->    0.
minStack.min();   -->    -2.
 
  ：
             20000  

class MinStack:

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

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

    def pop(self) -> None:
        num = self.stack.pop()
        if num == self.min_stack[-1]:
            _ = self.min_stack.pop()

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

    def min(self) -> int:
        if len(self.min_stack) == 0:
            return []
        else:
            return self.min_stack[-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()