class MinStack { public: void push(int x) { if(values.empty()) { values.push_back(x); min_indices.push_back(); } else { if( x < values[min_indices.back()] ) min_indices.push_back(values.size()); values.push_back(x); } } void pop() { values.pop_back()…

错误原因: 1.在递归的时候,递归函数中忘记加返回return. 1.1做题时遇到一个奇葩错误,把它记到这里,看代码: 代码1:错误,用c++提交wrong answer,但是用g++提交却accepted. int set_find(int d) { if(set[d]<0) return d; set_find(set[d]);//这里递归时没有返回值,只有在结束条件时返回(即最后一层递归有返回值,前面的没有返回值) } 上述代码在进行递归的时候,只有在递归结束时的最后一层返回一个值d,但是…

1- 问题描述 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 m…

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 e…

原标题:https://oj.leetcode.com/problems/min-stack/ 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…

这是悦乐书的第177次更新,第179篇原创 01 看题和准备 今天介绍的是LeetCode算法题中Easy级别的第36题(顺位题号是155).设计一个支持push,pop,top和在恒定时间内检索最小元素的堆栈. push(x) - 将元素x推入堆栈. pop() - 删除堆栈顶部的元素. top() - 获取顶部元素. getMin() - 检索堆栈中的最小元素. 例如: MinStack minStack = new MinStack(); minStack.push(-2); minSta…

LeetCode 155:最小栈 Min Stack 设计一个支持 push,pop,top 操作,并能在常数时间内检索到最小元素的栈. push(x) -- 将元素 x 推入栈中. pop() -- 删除栈顶的元素. top() -- 获取栈顶元素. getMin() -- 检索栈中的最小元素. Design a stack that supports push, pop, top, and retrieving the minimum element in constant time. pu…

题目 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 minimu…

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 e…

题目描述: 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 min…

155. Min Stack class MinStack { public: /** initialize your data structure here. */ MinStack() { } void push(int x) { if(s1.empty() && s2.empty()){ s1.push(x); s2.push(x); } else{ if(x < s2.top()){ s1.push(x); s2.push(x); } else{ s1.push(x); s2…

Min Stack My Submissions Question Solution 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…

Implement a stack with min() function, which will return the smallest number in the stack. It should support push, pop and min operation all in O(1) cost. Notice min operation will never be called if there is no number in the stack. Have you met this…

3.2 How would you design a stack which, in addition to push and pop, also has a function min which returns the minimum element? Push, pop and min should all operate in O(1) time. LeetCode上的原题,请参见我之前的博客Min Stack 最小栈.…

题目: 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 minim…

Min Stack 本题收获: 1.可以利用两个栈操作. 2.栈的基本操作. 题目: 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…

Min Stack 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…

［抄题］: 实现一个带有取最小值min方法的栈,min方法将返回当前栈中的最小值. 你实现的栈将支持push,pop 和 min 操作,所有操作要求都在O(1)时间内完成. ［思维问题］: ［一句话思路］: 用一个minstack来辅助实现 ［输入量］:空: 正常情况:特大:特小:程序里处理到的特殊情况:异常情况(不合法不合理的输入): ［画图］: ［一刷］: 主函数中,数据结构为类+名 类是Stack<Integer> minStack.empty() == true 或者isempty都可…

遇到个好玩的问题,就是用一个stack实现min stack,什么意思呢,就是我实现stack,但是能以O(1)的时间复杂度和空间复杂度去找到我stack里面的最小值. 常规的方法是:用一个变量存放当前最小值,但是会出现这种情况,就是当我的stack pop掉的值刚好是最小值时候,后面就没法知道当前的最小值了. 怎么办呢?可以考虑在push阶段做个改变,就是在我每次往stack里面push数据的时候,跟当前的最小值比较,如果比当前最小值还小的话,那么将当前最小值入栈,再把最小值修改为这个值.在p…

/Go/src/container/list/list.go:10…

Crazy Bobo Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others) Total Submission(s): 218    Accepted Submission(s): 60 Problem Description Bobo has a tree,whose vertices are conveniently labeled by 1,2,...,n.Each node…

原因分析 CDH 集群环境没有对 Container分配足够的运行环境(内存) 解决办法 需要修改的配置文件,将具体的配置项修改匹配集群环境资源.如下: 配置文件 配置设置 解释 计算值(参考) yarn-site.xml yarn.nodemanager.resource.memory-mb 分配给容器的物理内存数量 = 52 * 2 =104 G yarn-site.xml yarn.scheduler.minimum-allocation-mb 容器可以请求的最小物理内存量(以 MiB 为…

http://wiki.postgresql.org/wiki/Tuning_Your_PostgreSQL_Server IpcMemoryCreate: shmget(key=5432001, size=415776768, 03600) failed: Invalid argument This error usually means that PostgreSQL's request for a shared memory segment exceeded your kernel's S…

docker info 指令报若下错误:WARNING: No memory limit support 或WARNING: No swap limit support 解决方法: 1.打开/etc/default/grub文件,添加如下内容: GRUB_CMDLINE_LINUX="cgroup_enable=memory swapaccount=1" 或执行sed  -i  's#GRUB_CMDLINE_LINUX=""#GRUB_CMDLINE_LINUX=…

原语句: db.carMongoDTO.aggregate({}}}, {}}}) 报错: Exceeded memory limit for $group, but didn't allow external sort. Pass allowDiskUse:true to opt in 原因是聚合的结果必须要限制在16M以内操作,(mongodb支持的最大影响信息的大小),否则必须放在磁盘中做缓存(allowDiskUse=True) 修改为: db.carMongoDTO.aggregate…

直接上代码: public class Test001 { public static void main(String[] args) { //java.lang.StackOverflowError 栈溢出错误, 这个是error 不是异常,因为StackOverflowError 是Error的子类 // 栈溢出, 递归方法,调方法 m1(); } public static void m1(){ m1(); } @Test public void test02(){ // java.la…

[题目] 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 mini…

作者: 负雪明烛 id: fuxuemingzhu 个人博客:http://fuxuemingzhu.cn/ 目录 题目描述 解题方法 栈同时保存当前值和最小值 辅助栈 同步栈 不同步栈 日期 题目地址:https://leetcode.com/problems/min-stack/description/ 题目描述 Design a stack that supports push, pop, top, and retrieving the minimum element in constan…

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 e…

题目描述: 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…