题解 | 包含min函数的栈

# -*- coding:utf-8 -*-
class Solution:
    def __init__(self):
        self.stack = []
        self.min_stack = []

    def push(self, node):
        # write code here
        self.stack.append(node)
        if not self.min_stack or node <= self.min_stack[-1]:
            self.min_stack.append(node)
    def pop(self):
        # write code here
        if self.stack:
            poped = self.stack.pop()
            if poped == self.min_stack[-1]:
                self.min_stack.pop()
            return poped
    def top(self):
        # write code here
        return self.stack[-1] if self.stack else None
    def min(self):
        # write code here
        return self.min_stack[-1] if self.stack else None