题解 | #包含min函数的栈#

# -*- coding:utf-8 -*-
class Solution:
    stack1=[]
    stack2=[]    
    def push(self, node):
        # write code here
        self.stack1.append(node)
        if len(self.stack2)==0:
            self.stack2.append(node)
        else:
            if node>self.min():
                self.stack2.append(self.min())
            else:
                self.stack2.append(node)
    def pop(self):
        # write code here
        self.stack1.pop()
        self.stack2.pop()        
    def top(self):
        # write code here
        return self.stack1[len(self.stack1)-1]
    def min(self):
        # write code here
        return self.stack2[len(self.stack2)-1]