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…

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…

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…

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…

Description: 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…

class MinStack { public: MinStack() { coll.resize(2); } void push(int x) { if(index == coll.size()-1) coll.resize(2*coll.size()); coll[++index]=x; if(x<min) { min=x; flag=index; } } void pop() { index--; } int top() { return coll[index]; } int getMin…

原题地址:https://oj.leetcode.com/problems/min-stack/ 解题思路:开辟两个栈,一个栈是普通的栈,一个栈用来维护最小值的队列. 代码: class MinStack: # @param x, an integer def __init__(self): self.stack1 = [] self.stack2 = [] # @return an integer def push(self, x): self.stack1.append(x) if len(…

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…

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…