https://leetcode.com/problems/min-stack/#/solutions

Design a stack that supports push, pop, top, and retrieving the minimum element in constant time.

  • push(x) -- Push element x onto stack.
  • pop() -- Removes the element on top of the stack.
  • top() -- Get the top element.
  • getMin() -- Retrieve the minimum element in the stack.

Example:

MinStack minStack = new MinStack();
minStack.push(-2);
minStack.push(0);
minStack.push(-3);
minStack.getMin(); --> Returns -3.
minStack.pop();
minStack.top(); --> Returns 0.
minStack.getMin(); --> Returns -2.
 
Sol:
 
 AC
 
class MinStack(object):

    def __init__(self):
"""
initialize your data structure here.
"""
self.stack = []
# set two stacks
self.minStack = [] def push(self, x):
self.stack.append(x)
if len(self.minStack) and x == self.minStack[-1][0]:
self.minStack[-1] = (x, self.minStack[-1][1] + 1)
elif len(self.minStack) == 0 or x < self.minStack[-1][0]:
self.minStack.append((x, 1)) def pop(self):
#如果 栈顶值 == 最小值栈顶值
if self.top() == self.getMin():
#如果 最小值栈顶元素次数 > 1
if self.minStack[-1][1] > 1:
#最小值栈顶元素次数 - 1
self.minStack[-1] = (self.minStack[-1][0], self.minStack[-1][1] - 1)
else:
#最小值栈顶元素弹出
self.minStack.pop()
return self.stack.pop() def top(self):
return self.stack[-1] def getMin(self):
return self.minStack[-1][0] # 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()
 

155. Min Stack - Unsolved的更多相关文章

  1. leetcode 155. Min Stack 、232. Implement Queue using Stacks 、225. Implement Stack using Queues

    155. Min Stack class MinStack { public: /** initialize your data structure here. */ MinStack() { } v ...

  2. leetcode 155. Min Stack --------- java

    Design a stack that supports push, pop, top, and retrieving the minimum element in constant time. pu ...

  3. Java [Leetcode 155]Min Stack

    题目描述: Design a stack that supports push, pop, top, and retrieving the minimum element in constant ti ...

  4. 155. Min Stack

    题目: Design a stack that supports push, pop, top, and retrieving the minimum element in constant time ...

  5. Java for LeetCode 155 Min Stack

    Design a stack that supports push, pop, top, and retrieving the minimum element in constant time. pu ...

  6. 【leetcode】155 - Min Stack

    Design a stack that supports push, pop, top, and retrieving the minimum element in constant time. pu ...

  7. LeetCode题解 #155 Min Stack

    写一个栈,支持push pop top getMin 难就难在在要在常量时间内返回最小的元素. 一开始乱想了很多东西,想到了HashMap,treeMap,堆什么的,都被自己一一否决了. 后来想到其实 ...

  8. [LeetCode] 155. Min Stack 最小栈

    Design a stack that supports push, pop, top, and retrieving the minimum element in constant time. pu ...

  9. LC 155 Min Stack

    问题描述 Design a stack that supports push, pop, top, and retrieving the minimum element in constant tim ...

随机推荐

  1. git图片

  2. 二叉树的深度优先遍历与广度优先遍历 [ C++ 实现 ]

    深度优先搜索算法(Depth First Search),是搜索算法的一种.是沿着树的深度遍历树的节点,尽可能深的搜索树的分支. 当节点v的所有边都己被探寻过,搜索将回溯到发现节点v的那条边的起始节点 ...

  3. HDU-1004.Let the ballon Rise(STL-map)

    2019-02-28-08:56:03 初次做本题是用字符串硬钢,最近校队训练时又遇到才知道用map是真的舒服.需要注意的是map的用法. clear : 清除map中的所有元素,map.clear( ...

  4. 快速了解和使用Photon Server

    https://blog.csdn.net/qq_36565626/article/details/78710787

  5. 不要62(数位DP)

    不要62 http://acm.hdu.edu.cn/showproblem.php?pid=2089 Time Limit: 1000/1000 MS (Java/Others)    Memory ...

  6. Python常用库大全,看看有没有你需要的

    作者:史豹链接:https://www.zhihu.com/question/20501628/answer/223340838来源:知乎著作权归作者所有.商业转载请联系作者获得授权,非商业转载请注明 ...

  7. 问题1:jquery实现全选功能,第二次失效(已解决)

    问题:使用了attr("checked",true”)设置子复选框的被选状态,第一次执行功能正常,但第二次失效. 解决方案:将attr("checked",tr ...

  8. [Java学习]面向对象-多态

    多态 多态发生条件 发生在有继承关系的类型中. 向上转型(自动类型转换)与向下转型(强制类型转换) //向上转型 //编译阶段a1被编译器看作是Animal类型,所以a1引用绑定的是Animal类中的 ...

  9. 【网络编程一】主机字节序与网络字节序以及ip地址转换函数

    在计算机设计之初,对内存中数据的处理也有不同的方式,(低位数据存储在低位地址处或者高位数据存储在低位地址处),然而,在通信的过程中(ISO/OSI模型和TCP/IP四层模型中),数据被一步步封装(然后 ...

  10. Django xadmin 根据登录用户过滤数据

    在adminx.py文件对应的的class中添加如下代码: def queryset(self): qs = super(taskAdmin, self).queryset() if self.req ...