python 的 list 效率底下？

题目

Design a stack that supports push, pop, top, and retrievingthe 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.

答案：

class MinStack:
    def __init__(self):
        self.L=[]
    def pop(self):
        return self.L.pop()
    def push(self,x):
        self.L.append(x)
    def top(self):
        return self.L[-1]
    def getMin(self):
        tmpL = self.L[:]
        tmpL.sort()
        return tmpL[0]

每次提示答案都超时…表吐槽，表示刚学python