題目描述
定義棧的數據結構,請在該類型中實現一個能夠得到棧中所含最小元素的min函數(時間複雜度應爲O(1))。
注意:保證測試中不會當棧爲空的時候,對棧調用pop()或者min()或者top()方法。
# -*- coding:utf-8 -*-
class Solution:
def __init__(self):
self.s_stack = []
self.min_stack = []
def push(self, node):
self.s_stack.append(node)
if self.min_stack:
if self.min_stack[-1] > node:
self.min_stack.append(node)
else:
self.min_stack.append(self.min_stack[-1])
else:
self.min_stack.append(node)
# write code here
def pop(self):
# write code here
if self.s_stack:
self.s_stack.pop()
self.min_stack.pop()
def top(self):
# write code here
if self.s_stack:
return self.s_stack[-1]
if self.min_stack:
return self.min_stack[-1]
def min(self):
# write code here
if self.min_stack:
return self.min_stack[-1]
return None