题解 | #包含min函数的栈#

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

    def push(self, node):
        # write code here
        if not self.stack:
            self.stack.append((node, node))
        else:
            # 将当前最小值 保存了起来
            self.stack.append((node, min(node, self.stack[-1][1])))
    def pop(self):
        # write code here
        self.stack.pop()
    def top(self):
        # write code here
        return self.stack[-1][0]
    def min(self):
        # write code here
        return self.stack[-1][1]