题解 | #包含min函数的栈#

# -*- coding:utf-8 -*-
class Solution:
    def __init__(self):
        self.stack1 = []
        self.stack2 = []
    def push(self, node):
        # write code here
        if self.stack2 == []:
            self.stack2.append(node)
        else:
            
            if node < self.stack2[-1]:
                self.stack2.append(node)
            else:
                self.stack2.append(self.stack2[-1]) 
            # stack2的长度要和stack1保持一致，因为有可能stack1栈顶就是最小值
            # 所以每当stack1进行出栈操作，stack2也要同步进行出栈
        self.stack1.append(node)
    def pop(self):
        # write code here
        self.stack2.pop()
        return self.stack1.pop()
    def top(self):
        # write code here
        return self.stack1[-1]
    def min(self):
        # write code here
        return self.stack2[-1]