压缩还原，大佬思路的Python实现

这题有两种解法
一、借助辅助栈，保存一系列的最小值
当有比当前最小值还小的元素进来，压进辅助栈，当执行pop()操作则要对比栈与辅助栈的元素大小，判断辅助栈是否需要同样执行pop()操作
二、压缩还原
栈保存该元素与最小值的差值，更新最小值

class Solution:
    def __init__(self):
        self.stack = []
        self.min_num = 0
    def push(self, node):
        if not self.stack:
            self.stack.append(0)
            self.min_num = node
        else:
            self.stack.append(node - self.min_num)
            if self.min_num > node:
                self.min_num = node
    def pop(self):
        # write code here
        if self.stack:
            temp = self.stack.pop()
            if temp >= 0:
                return self.min_num + temp
            else:
                self.min_num = self.min_num - temp
                return self.min_num + temp
    def top(self):
        if self.stack:
            temp = self.stack[-1]
            if temp >= 0:
                return self.min_num + temp
            else:
                return self.min_num
    def min(self):
        return self.min_num