APT校园面面经

                                                              ／**  第一次发帖+第一次one on one interview  **/


first half 问简历：
为什么学cs， 喜欢什么方向， 实习干的啥， 最喜欢的项目是啥， 有没有什么项目别人分派给你结果出于兴趣或其他你又做了多余的深入挖掘。。。
从来没有被问过这么多没有用的问题

second half minStack：<

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完

楼主这个有0（1） space O（1） time的解的。
http://www.geeksforgeeks.org/design-a-stack-that-supports-getmin-in-o1-time-and-o1-extra-space/

Push(x) : Inserts x at the top of stack.
        If stack is empty, insert x into the stack and make minEle equal to x.
        If stack is not empty, compare x with minEle. Two cases arise:
        If x is greater than or equal to minEle, simply insert x.
        If x is less than minEle, insert (2*x – minEle) into the stack and make minEle equal to x. For example, let previous minEle was 3. Now we want to insert 2. We update minEle as 2 and insert 2*2 – 3 = 1 into the stack.
Pop() : Removes an element from top of stack.
        Remove element from top. Let the removed element be y. Two cases arise:
        If y is greater than or equal to minEle, the minimum element in the stack is still minEle.
        If y is less than minEle, the minimum element now becomes (2*minEle – y), so update (minEle = 2*minEle – y). This is where we retrieve previous minimum from current minimum and its value in stack. For example, let the element to be removed be 1 and minEle be 2. We remove 1 and update minEle as 2*2 – 1 = 3.

期待后续

zoeymiao 发表于 2017-10-26 08:12
期待后续

sorry。。没后续了