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…

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.…

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 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…

题目 带最小值操作的栈 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. 解题 可以定义一个数组或者其他的存储最小值,第i个元素,表示栈中前i个元素的最小值. 定义两个ArrayList来存储栈,一个ArrayList存储当前栈中的元素,一个ArrayList存储最小栈,并且其第i个元素表示栈中前i个元素的最小值,这样两个栈的长度是始终一样的 入栈:最小栈需要加入的元素是 当前要入的元…

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…

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…

设计一个支持 push,pop,top 操作,并能在O(1)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. 示例: MinStack minStack = new MinStack(); minStack.push(-2); minStack.push(0); minStack.push(-3); minStack.getMin(); --> 返回 -3.…

设计一个支持 push,pop,top 操作,并能在常量时间内检索最小元素的栈.    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.示例:MinStack minStack = new MinStack();minStack.push(-2);minStack.push(0);minStack.push(-3);minStack.getMin();   --> 返回…

题意:实现栈的四个基本功能.要求:在get最小元素值时,复杂度O(1). 思路:链表直接实现.最快竟然还要61ms,醉了. class MinStack { public: MinStack(){ head.next=; head.t=; m=0x7FFFFFFF; } void push(int x) { node *p=(node *)new(node); p->t=x; p->next=head.next; head.next=p; head.t++; if(x < m) //要更…