实现一个包含min函数的栈（O(1)的时间复杂度）

微信图片_20191221152107.jpg

class Solution:
    def __init__(self):
        self.stack = []
        self.min_stack = []
    def push(self, node):
        # write code here
        self.stack.append(node)
        if self.min_stack == []:
            self.min_stack.append(node)
        else:
            if node <= self.min_stack[-1]:
                self.min_stack.append(node)
            else:
                self.min_stack.append(self.min_stack[-1])
    def pop(self):
        # write code here
        if self.stack:
            self.min_stack.pop()
            return self.stack.pop()
    def top(self):
        # write code here
        if self.stack:
            return self.stack[-1]
    def min(self):
        # write code here
        if self.min_stack:
            return self.min_stack[-1]