了解栈的实现问题的声明 [英] Understanding Stack implementation problem statement

问题描述

我需要计算偷看中间元件也问题陈述用于实现该方法如下： -

I need to compute the peek mid element also the problem statement for implementing this method is as follows:-

*回报对象，它有中的所有对象的中间值，而不从栈中删除。 *返回其具有下列顺序（尺寸（）/ 2）1的值的对象 *例如。 *当堆栈具有以下值（1，2，5,4，2,6）*此方法返回4和不删除该对象。

*returns object which has the middle value among the all objects without removing it from the stack. *returns the object which has the value of following order (size()/2)+1 *e.g. *When the stack has the following values (1, 2, 5, 4, 2, 6) *this method returns 4 and doesn't remove the object.

所以我的查询是： -

so my query is:-

我应该考虑的排序堆的中间元素获得元素中旬以后=栈[尺寸（）/ 2 + 1] 或者我应该考虑它的价值计算，即中期= MAX +分/ 2

should i consider the middle element in terms of position i.e. after sorting the elements of the stack the mid element is obtained as mid = stack[size()/2+1] or should i consider it in terms of value i.e. mid= max+min/2

在上述问题都情况是正确的（以我的观点来看），即

as in above problem both the situations are correct( in my point of view) i.e.

栈[尺寸（）/ 2 + 1] =堆栈[6/2 + 1] = 4

和 MAX +分/ 2 = 6 + 1/2 = 3.5 和四舍五入将等于4

and max+min/2=6+1/2=3.5 and rounding off will be equal to 4

好心帮我理解问题陈述

推荐答案

一个堆栈是一个数据结构，因此在最一般的情况下应当能够存储任何类型的数据。你正在处理int类型时，这只是一个简化您的任务。数据结构明智的，它使您考虑中间元素，而不是执行的元素值的任何计算（即太具体的数据结构）的感觉。

A stack is a data structure, and as such in the most generic case should be able to store any data type. The fact that you are dealing with ints is just a simplification for your assignment. Data structure wise, it makes sense that you consider the middle element, and not perform any computation on element values (that is too specific for a data structure).

好像你想要的是在（（N / 2）+ 1）个元件，所以在本实施例折射率（n / 2）元素

It seems like what you want is the ((n/2) + 1) th element, therefore the element at index (n/2) in this example.

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