[Lintcode]12. Min Stack /[Leetcode]155. Min Stack-优快云博客

本文详细介绍了如何设计和实现一个支持获取最小元素操作的栈，即MinStack。通过使用Python，我们展示了如何在O(1)时间内进行push、pop、top和getMin操作。文章还讨论了代码的思路和时间复杂度。

12. Min Stack / 155. Min Stack

本题难度: Medium/Easy
Topic: Data Structure

Description

import heapq
import collections
class MinStack:

def init(self):
"""
initialize your data structure here.
"""
self.stack = []

def push(self, x: 'int') -> 'None':
if len(self.stack) == 0:
self.stack.append((x,x))
return
if self.stack[-1][1]<x:
self.stack.append((x,self.stack[-1][1]))
else:
self.stack.append((x,x))
return

def pop(self) -> 'None':
return self.stack.pop()[0]

def top(self) -> 'int':
return self.stack[-1][0]

def getMin(self) -> 'int':
return self.stack[-1][1]

Your MinStack object will be instantiated and called as such:

obj = MinStack()

obj.push(x)

obj.pop()

param_3 = obj.top()

param_4 = obj.getMin()

我的代码

import heapq
import collections
class MinStack:

    def __init__(self):
        """
        initialize your data structure here.
        """
        self.stack = []

    def push(self, x: 'int') -> 'None':
        if len(self.stack) == 0:
            self.stack.append((x,x))
            return
        if self.stack[-1][1]<x:
            self.stack.append((x,self.stack[-1][1]))
        else:
            self.stack.append((x,x))
        return 

    def pop(self) -> 'None':
        return self.stack.pop()[0]
        
        

    def top(self) -> 'int':
        return self.stack[-1][0]
        

    def getMin(self) -> 'int':
        return self.stack[-1][1]
        


# Your MinStack object will be instantiated and called as such:
# obj = MinStack()
# obj.push(x)
# obj.pop()
# param_3 = obj.top()
# param_4 = obj.getMin()