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…

［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…

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.…

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…

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…

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…

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 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…

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…